Determine whether each expression is a polynomial. If it is a polynomial, find the degree and determine whether it is a monomial, binomial, or trinomial.

1. 7ab + 6b2 – 2a

3

SOLUTION:

A polynomial is a monomial or the sum of monomials. 7ab + 6b2 – 2a

3 is the sum of 3 monomials, so it is a

polynomial. The degree of a polynomial is the greatest degree of any term in the polynomial.The degree of each term is 2, 2, and

3, so the degree of 7ab + 6b2 – 2a

3 is 3. The polynomial has three terms, so it is a trinomial.

2. 2y – 5 + 3y2

SOLUTION:

2y – 5 + 3y2

is the sum of monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 1, 0, and

2, so the degree of 2y – 5 + 3y2 is 2. The polynomial has three terms, so it is a trinomial.

3. 3x2

SOLUTION:

A polynomial is a monomial or the sum of monomials, so 3x2 is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of 3x2 is 2. The

polynomial has one term, so it is a monomial.

4.

SOLUTION: A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer exponents. It has

only one term. is a division of two monomials, so it is not a monomial.

5. 5m2p

3 + 6

SOLUTION:

A polynomial is a monomial or the sum of monomials. 5m2p

3 + 6

is the sum of 2 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 5 and 0,

so the degree of 5m2p

3 + 6 is 5. The polynomial has two terms, so it is a binomial.

6. 5q–4

+ 6q

SOLUTION: A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer exponents. It has

only one term. 5q-4 is equal to , which is a division of two monomials, so it is not a monomial.

Write each polynomial in standard form. Identify the leading coefficient.

7. –4d4 + 1 – d

2

SOLUTION: Find the degree of each term.

–4d4→ 4

1 → 0

–d2 → 2

The greatest degree is 4, from the term –4d4, so the leading coefficient of –4d

4 + 1 – d

2 is –4.

Rewrite the polynomial with each monomial in descending order according to degree.

–4d4 – d

2 + 1

8. 2x5 – 12 + 3x

SOLUTION: Find the degree of each term.

2x5 → 5

–12 → 0 3x → 1

The greatest degree is 5, from the term 2x5, so the leading coefficient of 2x

5 – 12 + 3x is 2.

Rewrite the polynomial with each monomial in descending order according to degree.

2x5 + 3x – 12

9. 4z – 2z2 – 5z

4

SOLUTION: Find the degree of each term. 4z → 1

– 2z2 → 2

– 5z4→ 4

The greatest degree is 4, from the term – 5z4, so the leading coefficient of 4z – 2z

2 – 5z

4 is –5.

Rewrite the polynomial with each monomial in descending order according to degree.

–5z4 – 2z

2 + 4z

10. 2a + 4a3 – 5a

2 – 1

SOLUTION: Find the degree of each term. 2a → 1

4a3→ 3

– 5a2 → 2

– 1 → 0

The greatest degree is 3, from the term 4a3, so the leading coefficient of 2a + 4a

3 – 5a

2 – 1 is 4.

Rewrite the polynomial with each monomial in descending order according to degree.

4a3– 5a

2 + 2a – 1

Find each sum or difference.

11. (6x3 − 4) + (−2x

3 + 9)

SOLUTION:

12. (g3 − 2g2 + 5g + 6) − (g2 + 2g)

SOLUTION:

13. (4 + 2a2 − 2a) − (3a

2 − 8a + 7)

SOLUTION:

14. (8y − 4y2) + (3y − 9y

2)

SOLUTION:

15. (−4z3 − 2z + 8) − (4z

3 + 3z2 − 5)

SOLUTION:

16. (−3d2 − 8 + 2d) + (4d − 12 + d2

)

SOLUTION:

17. (y + 5) + (2y + 4y2 – 2)

SOLUTION:

18. (3n3 − 5n + n2

) − (−8n2 + 3n

3)

SOLUTION:

19. CCSS SENSE-MAKING The total number of students T who traveled for spring break consists of two groups: students who flew to their destinations F and students who drove to their destination D. The number (in thousands) of students who flew and the total number of students who flew or drove can be modeled by the following equations,where n is the number of years since 1995. T = 14n + 21

F = 8n + 7 a. Write an equation that models the number of students who drove to their destination for this time period. b. Predict the number of students who will drive to their destination in 2012. c. How many students will drive or fly to their destination in 2015?

SOLUTION: a.

D = 6n + 14 b. n = 2012 – 1995 = 17

The number of students who will drive to their destination in 2012 is 116,000 students. c. n = 2015 – 1995 = 20

The number of students who will drive or fly to their destination in 2015 is 301,000 students.

Determine whether each expression is a polynomial. If it is a polynomial, find the degree and determine whether it is a monomial, binomial, or trinomial.

20.

SOLUTION: A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer exponents. It has

only one term. is a division of two monomials, so it is not a monomial.

21.

SOLUTION: A polynomial is a monomial or the sum of monomials. 21 is a monomial, so it is also a polynomial. The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of 21 is 0. The polynomial has only one term, so it is a monomial.

22. c4 – 2c

2 + 1

SOLUTION:

A polynomial is a monomial or the sum of monomials. c4 – 2c

2 + 1

is the sum of 3 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 4, 2, and

0, so the degree of c4 – 2c2 + 1 is 4. The polynomial has three terms, so it is a trinomial.

23. d + 3dc

SOLUTION: A polynomial is a monomial or the sum of monomials. A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer

exponents. 3dc

has a variable in the exponent, so it is not a monomial.

24. a – a2

SOLUTION:

A polynomial is a monomial or the sum of monomials. a – a2 is the sum of 2 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 1 and 2,

so the degree of a – a2 is 2. The polynomial has two terms, so it is a binomial.

25. 5n3 + nq

3

SOLUTION:

A polynomial is a monomial or the sum of monomials. 5n3 + nq3 is the sum of 2 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 3 and 4,

so the degree of 5n3 + nq3 is 4. The polynomial has two terms, so it is a binomial.

Write each polynomial in standard form. Identify the leading coefficient.

26. 5x2 – 2 + 3x

SOLUTION: Find the degree of each term.

5x2 → 2

– 2 → 0

3x → 1

The greatest degree is 2, from the term 5x2, so the leading coefficient of 5x

2 – 2 + 3x is 5.

Rewrite the polynomial with each monomial in descending order according to degree.

5x2 + 3x – 2

27. 8y + 7y3

SOLUTION: Find the degree of each term.

7y3 → 3

8y → 1

The greatest degree is 3, from the term 7y3, so the leading coefficient of 8y + 7y

3 is 7.

Rewrite the polynomial with each monomial in descending order according to degree.

7y3 + 8y

28. 4 – 3c – 5c2

SOLUTION: Find the degree of each term. 4 → 0 3c → 1

– 5c2 → 2

The greatest degree is 2, from the term – 5c2, so the leading coefficient of 4 – 3c – 5c

2 is –5.

Rewrite the polynomial with each monomial in descending order according to degree.

–5c2 – 3c + 4

29. –y3 + 3y – 3y

2 + 2

SOLUTION: Find the degree of each term.

–y3 → 3

3y → 1

3y2 → 2

2 → 0

The greatest degree is 3, from the term –y3, so the leading coefficient of –y

3 + 3y – 3y

2 + 2 is –1.

Rewrite the polynomial with each monomial in descending order according to degree. –y 3 – 3y 2 + 3y + 2

30. 11t + 2t2 – 3 + t

5

SOLUTION: Find the degree of each term. 11t → 1

2t2 → 2

–3 → 0

t5 → 5

The greatest degree is 5, from the term t5, so the leading coefficient of 11t + 2t2 – 3 + t

5 is 1.

Rewrite the polynomial with each monomial in descending order according to degree.

t5 + 2t

2 + 11t – 3

31. 2 + r – r3

SOLUTION: Find the degree of each term. 2 → 0 r → 1

– r3→ 3

The greatest degree is 3, from the term – r3, so the leading coefficient of 2 + r – r3

is –1. Rewrite the polynomial with each monomial in descending order according to degree.

–r3 + r + 2

32.

SOLUTION: Find the degree of each term.

→ 0

–3x4 → 4

7 → 0

The greatest degree is 4, from the term –3x4, so the leading coefficient of is –3.

Rewrite the polynomial with each monomial in descending order according to degree.

33. –9b2 + 10b – b

6

SOLUTION: Find the degree of each term.

–9b2 → 2

10b → 1

–b6→ 6

The greatest degree is 6, from the term –b6, so the leading coefficient of –9b2 + 10b – b6 is –1. Rewrite the polynomial with each monomial in descending order according to degree.

–b6 – 9b

2 + 10b

Find each sum or difference.

34. (2c2 + 6c + 4) + (5c – 7)

SOLUTION:

35. (2x + 3x2) − (7 − 8x

2)

SOLUTION:

36. (3c3 − c + 11) − (c2 + 2c + 8)

SOLUTION:

37. (z2 + z) + (z

2 − 11)

SOLUTION:

38. (2x − 2y + 1) − (3y + 4x)

SOLUTION:

39. (4a − 5b2 + 3) + (6 − 2a + 3b

2)

SOLUTION:

40. (x2y − 3x

2 + y) + (3y − 2x2y)

SOLUTION:

41. (−8xy + 3x2 − 5y) + (4x

2 − 2y + 6xy)

SOLUTION:

42. (5n − 2p2 + 2np) − (4p

2 + 4n)

SOLUTION:

43. (4rxt − 8r2x + x2

) − (6rx2 + 5rxt − 2x

2)

SOLUTION:

44. PETS From 1999 through 2009, the number of dogs D and the number of cats C (in hundreds) adopted from animal

shelters in the United States are modeled by the equations D = 2n + 3 and C = n + 4, where n is the number of yearssince 1999. a. Write an equation that models the total number T of dogs and cats adopted in hundreds for this time period. b. If this trend continues, how many dogs and cats will be adopted in 2013?

SOLUTION: a.

So, an equation that models the total number of dogs and cats adopted is T = 3n + 7. b. Evaluate the equation for the total number of dogs and cats for n = 2013 – 1999 = 14.

The number of cats and dogs adopted in 2013 will be 49 × 100 or 4900 cats and dogs.

Classify each polynomial according to its degree and number of terms.

45. 4x – 3x2 + 5

SOLUTION:

Find the degree of each term of 4x – 3x2 + 5.

4x → 1

– 3x2 → 2

5 → 0

The greatest degree is 2 and there are 3 terms, so 4x – 3x2 + 5 is a quadratic trinomial.

46. 11z3

SOLUTION: Find the degree of each term. 11z3

→ 3

The greatest degree is 3 and there is one term, so 11z3 is a cubic monomial.

47. 9 + y4

SOLUTION:

Find the degree of each term of 9 + y 4.

9 → 1

y4→ 4

The greatest degree is 4 and there are 2 terms, so 9 + y 4 is a quartic binomial.

48. 3x3 – 7

SOLUTION:

Find the degree of each term of 3x3 – 7.

3x3 → 3

–7 → 0

The greatest degree is 3 and there are 2 terms, so 3x3 – 7 is a cubic binomial.

49. –2x5 – x

2 + 5x – 8

SOLUTION:

Find the degree of each term of –2x5 – x2 + 5x – 8.

–2x5 → 5

–x2 → 2

5x → 1 – 8 → 0

The greatest degree is 5 and there are 4 terms, so –2x5 – x2 + 5x – 8 is a quintic polynomial.

50. 10t – 4t2 + 6t

3

SOLUTION:

Find the degree of each term of 10t – 4t2 + 6t

3.

10t → 1

4t2 → 2

6t3 → 3

The greatest degree is 3 and there are 3 terms, so 10t – 4t2 + 6t

3 is a cubic trinomial.

51. ENROLLMENT In a rapidly growing school system, the numbers (in hundreds) of total students N and

K-5 students P enrolled from 2000 to 2009 are modeled by the equations N = 1.25t2 – t + 7.5 and P =

0.7t2 – 0.95t + 3.8, where t is the number of years since 2000.

a. Write an equation modeling the number of 6-12 students S enrolled for this time period. b. How many 6-12 students were enrolled in the school system in 2007?

SOLUTION: a. To write an equation that represents the number of 6-12 students enrolled, subtract the equations that represent the total number of students and the number of K-5 students.

b. Replace t with 7 in the equation for S to determine the number of students enrolled in 6-12 in 2007.

In 2007 there were 30.3 hundreds or 3030 students enrolled in 6-12.

52. CCSS REASONING The perimeter of the figure shown is represented by the expression 3x2 − 7x + 2. Write a

polynomial that represents the measure of the third side.

SOLUTION:

53. GEOMETRY Consider the rectangle.

a. What does (4x2 + 2x – 1)(2x

2 – x + 3) represent?

b. What does 2(4x2 + 2x – 1) + 2(2x

2 – x + 3) represent?

SOLUTION:

a. (4x2 + 2x – 1)(2x

2 – x + 3) is a multiplication of the length and the width of the rectangle, which is the formula for

the area of a rectangle.

b. 2(4x2 + 2x – 1) + 2(2x

2 – x + 3) is the sum of twice the length and twice the width of a rectangle, which is the

formula for the perimeter of the rectangle

Find each sum or difference.

54. (4x + 2y − 6z) + (5y − 2z + 7x) + (−9z − 2x − 3y)

SOLUTION:

55. (5a2 − 4) + (a

2 − 2a + 12) + (4a2 − 6a + 8)

SOLUTION:

56. (3c2 − 7) + (4c + 7) − (c

2 + 5c − 8)

SOLUTION:

57. (3n3 + 3n − 10) − (4n

2 − 5n) + (4n3 − 3n

2 − 9n + 4)

SOLUTION:

58. FOOTBALL The National Football League is divided into two conferences, the American A and the National N. From 2002 through 2009, the total attendance T (in thousands) for both conferences and for the American Conference games are modeled by the following equations, where x is the number of years since 2002.

T = –0.69x3 + 55.83x

2 + 643.31x + 10,538

A = –3.78x3 + 58.96x

2 + 265.96x + 5257

Determine how many people attended National Conference football games in 2009.

SOLUTION:

Let x = 7 represent 2009, then find how many people attended a National Conference football game in 2009.

In 2009 the number of people who attended National Conference football games was about 8829 thousand. Multiply by 1000 to find the attendance in standard form. 8829 ×1000 = 8,829,000. So, about 8,829,000 people attended National Conference football games in 2009.

59. CAR RENTAL The cost to rent a car for a day is $15 plus $0.15 for each mile driven. a. Write a polynomial that represents the cost of renting a car for m miles. b. If a car is driven 145 miles, how much would it cost to rent? c. If a car is driven 105 miles each day for four days, how much would it cost to rent a car? d. If a car is driven 220 miles each day for seven days, how much would it cost to rent a car?

SOLUTION: a. The cost to rent a car is the daily rate and the mileage cost or 15 + 0.15m. b. Substitute 145 for m to find the daily cost to drive 145 miles.

The cost to rent the car would be $36.75. c. The expression represents the cost per day. For 4 days, multiply the entire expression by 4 to find the cost for a 4 day trip. Substitute 105 for m.

The cost to rent the car would be $123. d. The expression represents the cost per day. For 7 days, multiply the entire expression by 7 to find the cost for a 7 day trip. Substitute 220 for m.

The cost to rent the car would be $336.

60. MULTIPLE REPRESENTATIONS In this problem, you will explore perimeter and area. a. Geometric Draw three rectangles that each have a perimeter of 400 feet. b. Tabular Record the width and length of each rectangle in a table like the one shown below. Find the area of eachrectangle.

c. Graphical On a coordinate system, graph the area of rectangle 4 in terms of the length, x. Use the graph to determine the largest area possible. d. Analytical Determine the length and width that produce the largest area.

SOLUTION: a. For the perimeters to be 400, create lengths and widths that sum to 200. Note that some lengths are already provided in the table in part b.

b. The area is length multiplied by width. Make sure the units are squared for area. The sum of the length and width must be 200, so if the length is x, the width must be 200 – x.

c. The length will be the x-values on the horizontal and the area will be the y-values on the vertical. The length cannot pass 200 since the sum of the length and width is 200. Set the intervals for the x-axis to 25 feet. Extend the table of values to find more points to plot on the graph. It appears that 10,000 is the greatest area, so set the intervals

for the y-axis to 1000 ft2. After graphing, it appears that the highest point on the graph is at an area of 10,000 ft

2.

d. The associated x-value with the maximum area is x = 100, so the length must be 100 and the width must be 200 – 100, or 100. The length and width of the rectangle must be 100 feet each to have the largest area.

61. CCSS CRITIQUE Cheyenne and Sebastian are finding (2x2 − x) − (3x + 3x

2 − 2). Is either of them correct? Explain your reasoning.

SOLUTION:

Neither is correct. Cheyenne, did not distribute the negative to the 2nd and 3rd terms when she found the additive inverse. Sebastian did not distribute the negate to the 3rd terms when he found the additive inverse. To find the additive inverse, all terms should be multiplied by −1.

62. REASONING Determine whether each of the following statements is true or false . Explain your reasoning. a. A binomial can have a degree of zero. b. The order in which polynomials are subtracted does not matter.

SOLUTION: a. If a binomial has two terms that are each a degree of 0, then those terms can be combined and the binomial becomes a monomial. For example, 18 + 7 = 25. If one of the terms of the binomial does not have a degree of 0, then the binomial cannot have a degree of 0, since the degree of a polynomial is the greatest degree of any term in the polynomial. b. Subtraction is not commutative. While 2 + 5 = 5 + 2, 2 – 5 ≠ 5 – 2. This is also true for polynomials. Sample answer: (2x – 3) – (4x – 3) = –2x, but (4x – 3) – (2x – 3) = 2x

63. CHALLENGE Write a polynomial that represents the sum of an odd integer 2n + 1 and the next two consecutive odd integers.

SOLUTION:

64. WRITING IN MATH Why would you add or subtract equations that represent real-world situations? Explain.

SOLUTION:

65. WRITING IN MATH Describe how to add and subtract polynomials using both the vertical and horizontal formats.

SOLUTION: To add polynomials in a horizontal format, you combine like terms. For the vertical format, you write the polynomials in standard form, align like terms in columns, and combine like terms.

To subtract polynomials in a horizontal format you find the additive inverse of the polynomial you are subtracting, andthen combine like terms. For the vertical format, you write the polynomials in standard form, align like terms in columns, and subtract by adding the additive inverse.

66. Three consecutive integers can be represented by x, x + 1, and x + 2. What is the sum of these three integers? A x(x + 1)(x + 2)

B x3 + 3

C 3x + 3 D x + 3

SOLUTION:

The correct choice is C.

67. SHORT RESPONSE What is the perimeter of a square with sides that measure 2x + 3 units?

SOLUTION:

The perimeter of the square is 8x + 12 units.

68. Jim cuts a board in the shape of a regular hexagon and pounds in a nail at each vertex, as shown. How many rubber bands will he need to stretch a rubber band across every possible pair of nails?

F 15 G 14 H 12 J 9

SOLUTION: The first nail would connect to 5 others, the second to 4 others, the third to 3 others, etc.5 + 4 + 3 + 2 + 1 = 15 The correct choice is F.

69. Which ordered pair is in the solution set of the system of inequalities shown in the graph?

A (−3, 0) B (0, −3) C (5, 0) D (0, 5)

SOLUTION: Choice A is outside the shaded area for both inequalities. Choices B and D are inside the shaded area for only one inequality. Choice C is the only point in the solution for both inequalities. So, the correct choice is C.

70. COMPUTERS A computer technician charges by the hour to fix and repair computer equipment. The total cost of the technician for one hour is $75, for two hours is $125, for three hours is $175, for four hours is $225, and so on. Write a recursive formula for the sequence.

SOLUTION: Write out the terms. $75, $125, $175, $225, ... The first term is 75, and 50 is added to form each following term. Therefore, we have a1 = 75, an = an – 1 + 50, n ≥ 2.

Determine whether each sequence is arithmetic, geometric, or neither. Explain.71. 8, –32, 128, –512, ...

SOLUTION: Check for a common difference. –32 – 8 = –40 128 – (–32) = 160 There is no common difference. Check for a common ratio. –32 ÷ 8 = –4 128 ÷ (–32) = –4 Geometric; the common ratio is –4.

72. 25, 8, –9, –26, ...

SOLUTION: Check for a common difference. 8 – 25 = –17 –9 – 8 = –17 Arithmetic; the common difference is –17.

73.

SOLUTION: Check for a common difference.

There is no common difference. Check for a common ratio.

There is no common ratio, so the sequence is not arithmetic or geometric.

74. 43, 52, 61, 70, ...

SOLUTION: Check for a common difference. 52 – 43 = 9 61 – 52 = 9 Arithmetic; the common difference is 9.

75. –27, –16, –5, 6, ...

SOLUTION: Check for a common difference. –16 – (–27) = 11 –5 – (–16) = 11 Arithmetic; the common difference is 11.

76. 200, 100, 50, 25, …

SOLUTION: Check for a common difference. 100 – 200 = –100 50 – 100 = –50 There is no common difference. Check for a common ratio. 100 ÷ 200 = 0.5 50 ÷ 100 = 0.5

Geometric; the common ratio is 0.5 or .

77. JOBS Kimi received an offer for a new job. She wants to compare the offer with her current job. What is total amount of sales that Kimi must get each month to make the same income at either job?

SOLUTION: Let s be Kimi’s monthly sales.

Kimi must sell $80,000 each month to make the same income at either job.

Determine whether each sequence is an arithmetic sequence. If it is, state the common difference.78. 24, 16, 8, 0, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is –8, because 16 – 24 = –8; 8 –16 = –8; etc.

79. , 13, 26, …

SOLUTION: Find the difference between the terms.

– =

13 – =

26 – 13 = 13 There is not a common different. The sequence is not an arithmetic sequence

80. 7, 6, 5, 4, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is –1, because 6 – 7 = –1; 5 – 6 = –1; 4 – 5 = –1;etc.

81. 10, 12, 15, 18, …

SOLUTION: Find the difference between the terms. 12 – 10 = 2 15 – 12 = 3 18 – 15 = 3 There is no common difference. The sequence is not an arithmetic sequence.

82. −15, −11, −7, −3, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is 4, because –11 – (–15) = 4; –7 – (–11) = 4; –3 – (–7) = 4;etc.

83. −0.3, 0.2, 0.7, 1.2, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is 0.5, because 0.2 – (–0.3) = 0.5; 0.7 – 0.2 = 0.5; 1.2 – 0.7 = 0.5; etc.

Simplify.

84. t(t5)(t

7)

SOLUTION:

85. n3(n

2)(−2n

3)

SOLUTION:

86. (5t5v

2)(10t

3v

4)

SOLUTION:

87. (−8u4z

5)(5uz

4)

SOLUTION:

88. [(3)2]3

SOLUTION:

89. [(2)3]2

SOLUTION:

90. (2m4k

3)2(−3mk

2)3

SOLUTION:

91. (6xy2)2(2x

2y

2z

2)3

SOLUTION:

eSolutions Manual - Powered by Cognero Page 1

8-1 Adding and Subtracting Polynomials

Determine whether each expression is a polynomial. If it is a polynomial, find the degree and determine whether it is a monomial, binomial, or trinomial.

1. 7ab + 6b2 – 2a

3

SOLUTION:

A polynomial is a monomial or the sum of monomials. 7ab + 6b2 – 2a

3 is the sum of 3 monomials, so it is a

polynomial. The degree of a polynomial is the greatest degree of any term in the polynomial.The degree of each term is 2, 2, and

3, so the degree of 7ab + 6b2 – 2a

3 is 3. The polynomial has three terms, so it is a trinomial.

2. 2y – 5 + 3y2

SOLUTION:

2y – 5 + 3y2

is the sum of monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 1, 0, and

2, so the degree of 2y – 5 + 3y2 is 2. The polynomial has three terms, so it is a trinomial.

3. 3x2

SOLUTION:

A polynomial is a monomial or the sum of monomials, so 3x2 is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of 3x2 is 2. The

polynomial has one term, so it is a monomial.

4.

SOLUTION: A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer exponents. It has

only one term. is a division of two monomials, so it is not a monomial.

5. 5m2p

3 + 6

SOLUTION:

A polynomial is a monomial or the sum of monomials. 5m2p

3 + 6

is the sum of 2 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 5 and 0,

so the degree of 5m2p

3 + 6 is 5. The polynomial has two terms, so it is a binomial.

6. 5q–4

+ 6q

SOLUTION: A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer exponents. It has

only one term. 5q-4 is equal to , which is a division of two monomials, so it is not a monomial.

Write each polynomial in standard form. Identify the leading coefficient.

7. –4d4 + 1 – d

2

SOLUTION: Find the degree of each term.

–4d4→ 4

1 → 0

–d2 → 2

The greatest degree is 4, from the term –4d4, so the leading coefficient of –4d

4 + 1 – d

2 is –4.

Rewrite the polynomial with each monomial in descending order according to degree.

–4d4 – d

2 + 1

8. 2x5 – 12 + 3x

SOLUTION: Find the degree of each term.

2x5 → 5

–12 → 0 3x → 1

The greatest degree is 5, from the term 2x5, so the leading coefficient of 2x

5 – 12 + 3x is 2.

Rewrite the polynomial with each monomial in descending order according to degree.

2x5 + 3x – 12

9. 4z – 2z2 – 5z

4

SOLUTION: Find the degree of each term. 4z → 1

– 2z2 → 2

– 5z4→ 4

The greatest degree is 4, from the term – 5z4, so the leading coefficient of 4z – 2z

2 – 5z

4 is –5.

Rewrite the polynomial with each monomial in descending order according to degree.

–5z4 – 2z

2 + 4z

10. 2a + 4a3 – 5a

2 – 1

SOLUTION: Find the degree of each term. 2a → 1

4a3→ 3

– 5a2 → 2

– 1 → 0

The greatest degree is 3, from the term 4a3, so the leading coefficient of 2a + 4a

3 – 5a

2 – 1 is 4.

Rewrite the polynomial with each monomial in descending order according to degree.

4a3– 5a

2 + 2a – 1

Find each sum or difference.

11. (6x3 − 4) + (−2x

3 + 9)

SOLUTION:

12. (g3 − 2g2 + 5g + 6) − (g2 + 2g)

SOLUTION:

13. (4 + 2a2 − 2a) − (3a

2 − 8a + 7)

SOLUTION:

14. (8y − 4y2) + (3y − 9y

2)

SOLUTION:

15. (−4z3 − 2z + 8) − (4z

3 + 3z2 − 5)

SOLUTION:

16. (−3d2 − 8 + 2d) + (4d − 12 + d2

)

SOLUTION:

17. (y + 5) + (2y + 4y2 – 2)

SOLUTION:

18. (3n3 − 5n + n2

) − (−8n2 + 3n

3)

SOLUTION:

19. CCSS SENSE-MAKING The total number of students T who traveled for spring break consists of two groups: students who flew to their destinations F and students who drove to their destination D. The number (in thousands) of students who flew and the total number of students who flew or drove can be modeled by the following equations,where n is the number of years since 1995. T = 14n + 21

F = 8n + 7 a. Write an equation that models the number of students who drove to their destination for this time period. b. Predict the number of students who will drive to their destination in 2012. c. How many students will drive or fly to their destination in 2015?

SOLUTION: a.

D = 6n + 14 b. n = 2012 – 1995 = 17

The number of students who will drive to their destination in 2012 is 116,000 students. c. n = 2015 – 1995 = 20

The number of students who will drive or fly to their destination in 2015 is 301,000 students.

Determine whether each expression is a polynomial. If it is a polynomial, find the degree and determine whether it is a monomial, binomial, or trinomial.

20.

SOLUTION: A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer exponents. It has

only one term. is a division of two monomials, so it is not a monomial.

21.

SOLUTION: A polynomial is a monomial or the sum of monomials. 21 is a monomial, so it is also a polynomial. The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of 21 is 0. The polynomial has only one term, so it is a monomial.

22. c4 – 2c

2 + 1

SOLUTION:

A polynomial is a monomial or the sum of monomials. c4 – 2c

2 + 1

is the sum of 3 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 4, 2, and

0, so the degree of c4 – 2c2 + 1 is 4. The polynomial has three terms, so it is a trinomial.

23. d + 3dc

SOLUTION: A polynomial is a monomial or the sum of monomials. A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer

exponents. 3dc

has a variable in the exponent, so it is not a monomial.

24. a – a2

SOLUTION:

A polynomial is a monomial or the sum of monomials. a – a2 is the sum of 2 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 1 and 2,

so the degree of a – a2 is 2. The polynomial has two terms, so it is a binomial.

25. 5n3 + nq

3

SOLUTION:

A polynomial is a monomial or the sum of monomials. 5n3 + nq3 is the sum of 2 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 3 and 4,

so the degree of 5n3 + nq3 is 4. The polynomial has two terms, so it is a binomial.

Write each polynomial in standard form. Identify the leading coefficient.

26. 5x2 – 2 + 3x

SOLUTION: Find the degree of each term.

5x2 → 2

– 2 → 0

3x → 1

The greatest degree is 2, from the term 5x2, so the leading coefficient of 5x

2 – 2 + 3x is 5.

Rewrite the polynomial with each monomial in descending order according to degree.

5x2 + 3x – 2

27. 8y + 7y3

SOLUTION: Find the degree of each term.

7y3 → 3

8y → 1

The greatest degree is 3, from the term 7y3, so the leading coefficient of 8y + 7y

3 is 7.

Rewrite the polynomial with each monomial in descending order according to degree.

7y3 + 8y

28. 4 – 3c – 5c2

SOLUTION: Find the degree of each term. 4 → 0 3c → 1

– 5c2 → 2

The greatest degree is 2, from the term – 5c2, so the leading coefficient of 4 – 3c – 5c

2 is –5.

Rewrite the polynomial with each monomial in descending order according to degree.

–5c2 – 3c + 4

29. –y3 + 3y – 3y

2 + 2

SOLUTION: Find the degree of each term.

–y3 → 3

3y → 1

3y2 → 2

2 → 0

The greatest degree is 3, from the term –y3, so the leading coefficient of –y

3 + 3y – 3y

2 + 2 is –1.

Rewrite the polynomial with each monomial in descending order according to degree. –y 3 – 3y 2 + 3y + 2

30. 11t + 2t2 – 3 + t

5

SOLUTION: Find the degree of each term. 11t → 1

2t2 → 2

–3 → 0

t5 → 5

The greatest degree is 5, from the term t5, so the leading coefficient of 11t + 2t2 – 3 + t

5 is 1.

Rewrite the polynomial with each monomial in descending order according to degree.

t5 + 2t

2 + 11t – 3

31. 2 + r – r3

SOLUTION: Find the degree of each term. 2 → 0 r → 1

– r3→ 3

The greatest degree is 3, from the term – r3, so the leading coefficient of 2 + r – r3

is –1. Rewrite the polynomial with each monomial in descending order according to degree.

–r3 + r + 2

32.

SOLUTION: Find the degree of each term.

→ 0

–3x4 → 4

7 → 0

The greatest degree is 4, from the term –3x4, so the leading coefficient of is –3.

Rewrite the polynomial with each monomial in descending order according to degree.

33. –9b2 + 10b – b

6

SOLUTION: Find the degree of each term.

–9b2 → 2

10b → 1

–b6→ 6

The greatest degree is 6, from the term –b6, so the leading coefficient of –9b2 + 10b – b6 is –1. Rewrite the polynomial with each monomial in descending order according to degree.

–b6 – 9b

2 + 10b

Find each sum or difference.

34. (2c2 + 6c + 4) + (5c – 7)

SOLUTION:

35. (2x + 3x2) − (7 − 8x

2)

SOLUTION:

36. (3c3 − c + 11) − (c2 + 2c + 8)

SOLUTION:

37. (z2 + z) + (z

2 − 11)

SOLUTION:

38. (2x − 2y + 1) − (3y + 4x)

SOLUTION:

39. (4a − 5b2 + 3) + (6 − 2a + 3b

2)

SOLUTION:

40. (x2y − 3x

2 + y) + (3y − 2x2y)

SOLUTION:

41. (−8xy + 3x2 − 5y) + (4x

2 − 2y + 6xy)

SOLUTION:

42. (5n − 2p2 + 2np) − (4p

2 + 4n)

SOLUTION:

43. (4rxt − 8r2x + x2

) − (6rx2 + 5rxt − 2x

2)

SOLUTION:

44. PETS From 1999 through 2009, the number of dogs D and the number of cats C (in hundreds) adopted from animal

shelters in the United States are modeled by the equations D = 2n + 3 and C = n + 4, where n is the number of yearssince 1999. a. Write an equation that models the total number T of dogs and cats adopted in hundreds for this time period. b. If this trend continues, how many dogs and cats will be adopted in 2013?

SOLUTION: a.

So, an equation that models the total number of dogs and cats adopted is T = 3n + 7. b. Evaluate the equation for the total number of dogs and cats for n = 2013 – 1999 = 14.

The number of cats and dogs adopted in 2013 will be 49 × 100 or 4900 cats and dogs.

Classify each polynomial according to its degree and number of terms.

45. 4x – 3x2 + 5

SOLUTION:

Find the degree of each term of 4x – 3x2 + 5.

4x → 1

– 3x2 → 2

5 → 0

The greatest degree is 2 and there are 3 terms, so 4x – 3x2 + 5 is a quadratic trinomial.

46. 11z3

SOLUTION: Find the degree of each term. 11z3

→ 3

The greatest degree is 3 and there is one term, so 11z3 is a cubic monomial.

47. 9 + y4

SOLUTION:

Find the degree of each term of 9 + y 4.

9 → 1

y4→ 4

The greatest degree is 4 and there are 2 terms, so 9 + y 4 is a quartic binomial.

48. 3x3 – 7

SOLUTION:

Find the degree of each term of 3x3 – 7.

3x3 → 3

–7 → 0

The greatest degree is 3 and there are 2 terms, so 3x3 – 7 is a cubic binomial.

49. –2x5 – x

2 + 5x – 8

SOLUTION:

Find the degree of each term of –2x5 – x2 + 5x – 8.

–2x5 → 5

–x2 → 2

5x → 1 – 8 → 0

The greatest degree is 5 and there are 4 terms, so –2x5 – x2 + 5x – 8 is a quintic polynomial.

50. 10t – 4t2 + 6t

3

SOLUTION:

Find the degree of each term of 10t – 4t2 + 6t

3.

10t → 1

4t2 → 2

6t3 → 3

The greatest degree is 3 and there are 3 terms, so 10t – 4t2 + 6t

3 is a cubic trinomial.

51. ENROLLMENT In a rapidly growing school system, the numbers (in hundreds) of total students N and

K-5 students P enrolled from 2000 to 2009 are modeled by the equations N = 1.25t2 – t + 7.5 and P =

0.7t2 – 0.95t + 3.8, where t is the number of years since 2000.

a. Write an equation modeling the number of 6-12 students S enrolled for this time period. b. How many 6-12 students were enrolled in the school system in 2007?

SOLUTION: a. To write an equation that represents the number of 6-12 students enrolled, subtract the equations that represent the total number of students and the number of K-5 students.

b. Replace t with 7 in the equation for S to determine the number of students enrolled in 6-12 in 2007.

In 2007 there were 30.3 hundreds or 3030 students enrolled in 6-12.

52. CCSS REASONING The perimeter of the figure shown is represented by the expression 3x2 − 7x + 2. Write a

polynomial that represents the measure of the third side.

SOLUTION:

53. GEOMETRY Consider the rectangle.

a. What does (4x2 + 2x – 1)(2x

2 – x + 3) represent?

b. What does 2(4x2 + 2x – 1) + 2(2x

2 – x + 3) represent?

SOLUTION:

a. (4x2 + 2x – 1)(2x

2 – x + 3) is a multiplication of the length and the width of the rectangle, which is the formula for

the area of a rectangle.

b. 2(4x2 + 2x – 1) + 2(2x

2 – x + 3) is the sum of twice the length and twice the width of a rectangle, which is the

formula for the perimeter of the rectangle

Find each sum or difference.

54. (4x + 2y − 6z) + (5y − 2z + 7x) + (−9z − 2x − 3y)

SOLUTION:

55. (5a2 − 4) + (a

2 − 2a + 12) + (4a2 − 6a + 8)

SOLUTION:

56. (3c2 − 7) + (4c + 7) − (c

2 + 5c − 8)

SOLUTION:

57. (3n3 + 3n − 10) − (4n

2 − 5n) + (4n3 − 3n

2 − 9n + 4)

SOLUTION:

58. FOOTBALL The National Football League is divided into two conferences, the American A and the National N. From 2002 through 2009, the total attendance T (in thousands) for both conferences and for the American Conference games are modeled by the following equations, where x is the number of years since 2002.

T = –0.69x3 + 55.83x

2 + 643.31x + 10,538

A = –3.78x3 + 58.96x

2 + 265.96x + 5257

Determine how many people attended National Conference football games in 2009.

SOLUTION:

Let x = 7 represent 2009, then find how many people attended a National Conference football game in 2009.

In 2009 the number of people who attended National Conference football games was about 8829 thousand. Multiply by 1000 to find the attendance in standard form. 8829 ×1000 = 8,829,000. So, about 8,829,000 people attended National Conference football games in 2009.

59. CAR RENTAL The cost to rent a car for a day is $15 plus $0.15 for each mile driven. a. Write a polynomial that represents the cost of renting a car for m miles. b. If a car is driven 145 miles, how much would it cost to rent? c. If a car is driven 105 miles each day for four days, how much would it cost to rent a car? d. If a car is driven 220 miles each day for seven days, how much would it cost to rent a car?

SOLUTION: a. The cost to rent a car is the daily rate and the mileage cost or 15 + 0.15m. b. Substitute 145 for m to find the daily cost to drive 145 miles.

The cost to rent the car would be $36.75. c. The expression represents the cost per day. For 4 days, multiply the entire expression by 4 to find the cost for a 4 day trip. Substitute 105 for m.

The cost to rent the car would be $123. d. The expression represents the cost per day. For 7 days, multiply the entire expression by 7 to find the cost for a 7 day trip. Substitute 220 for m.

The cost to rent the car would be $336.

60. MULTIPLE REPRESENTATIONS In this problem, you will explore perimeter and area. a. Geometric Draw three rectangles that each have a perimeter of 400 feet. b. Tabular Record the width and length of each rectangle in a table like the one shown below. Find the area of eachrectangle.

c. Graphical On a coordinate system, graph the area of rectangle 4 in terms of the length, x. Use the graph to determine the largest area possible. d. Analytical Determine the length and width that produce the largest area.

SOLUTION: a. For the perimeters to be 400, create lengths and widths that sum to 200. Note that some lengths are already provided in the table in part b.

b. The area is length multiplied by width. Make sure the units are squared for area. The sum of the length and width must be 200, so if the length is x, the width must be 200 – x.

c. The length will be the x-values on the horizontal and the area will be the y-values on the vertical. The length cannot pass 200 since the sum of the length and width is 200. Set the intervals for the x-axis to 25 feet. Extend the table of values to find more points to plot on the graph. It appears that 10,000 is the greatest area, so set the intervals

for the y-axis to 1000 ft2. After graphing, it appears that the highest point on the graph is at an area of 10,000 ft

2.

d. The associated x-value with the maximum area is x = 100, so the length must be 100 and the width must be 200 – 100, or 100. The length and width of the rectangle must be 100 feet each to have the largest area.

61. CCSS CRITIQUE Cheyenne and Sebastian are finding (2x2 − x) − (3x + 3x

2 − 2). Is either of them correct? Explain your reasoning.

SOLUTION:

Neither is correct. Cheyenne, did not distribute the negative to the 2nd and 3rd terms when she found the additive inverse. Sebastian did not distribute the negate to the 3rd terms when he found the additive inverse. To find the additive inverse, all terms should be multiplied by −1.

62. REASONING Determine whether each of the following statements is true or false . Explain your reasoning. a. A binomial can have a degree of zero. b. The order in which polynomials are subtracted does not matter.

SOLUTION: a. If a binomial has two terms that are each a degree of 0, then those terms can be combined and the binomial becomes a monomial. For example, 18 + 7 = 25. If one of the terms of the binomial does not have a degree of 0, then the binomial cannot have a degree of 0, since the degree of a polynomial is the greatest degree of any term in the polynomial. b. Subtraction is not commutative. While 2 + 5 = 5 + 2, 2 – 5 ≠ 5 – 2. This is also true for polynomials. Sample answer: (2x – 3) – (4x – 3) = –2x, but (4x – 3) – (2x – 3) = 2x

63. CHALLENGE Write a polynomial that represents the sum of an odd integer 2n + 1 and the next two consecutive odd integers.

SOLUTION:

64. WRITING IN MATH Why would you add or subtract equations that represent real-world situations? Explain.

SOLUTION:

65. WRITING IN MATH Describe how to add and subtract polynomials using both the vertical and horizontal formats.

SOLUTION: To add polynomials in a horizontal format, you combine like terms. For the vertical format, you write the polynomials in standard form, align like terms in columns, and combine like terms.

To subtract polynomials in a horizontal format you find the additive inverse of the polynomial you are subtracting, andthen combine like terms. For the vertical format, you write the polynomials in standard form, align like terms in columns, and subtract by adding the additive inverse.

66. Three consecutive integers can be represented by x, x + 1, and x + 2. What is the sum of these three integers? A x(x + 1)(x + 2)

B x3 + 3

C 3x + 3 D x + 3

SOLUTION:

The correct choice is C.

67. SHORT RESPONSE What is the perimeter of a square with sides that measure 2x + 3 units?

SOLUTION:

The perimeter of the square is 8x + 12 units.

68. Jim cuts a board in the shape of a regular hexagon and pounds in a nail at each vertex, as shown. How many rubber bands will he need to stretch a rubber band across every possible pair of nails?

F 15 G 14 H 12 J 9

SOLUTION: The first nail would connect to 5 others, the second to 4 others, the third to 3 others, etc.5 + 4 + 3 + 2 + 1 = 15 The correct choice is F.

69. Which ordered pair is in the solution set of the system of inequalities shown in the graph?

A (−3, 0) B (0, −3) C (5, 0) D (0, 5)

SOLUTION: Choice A is outside the shaded area for both inequalities. Choices B and D are inside the shaded area for only one inequality. Choice C is the only point in the solution for both inequalities. So, the correct choice is C.

70. COMPUTERS A computer technician charges by the hour to fix and repair computer equipment. The total cost of the technician for one hour is $75, for two hours is $125, for three hours is $175, for four hours is $225, and so on. Write a recursive formula for the sequence.

SOLUTION: Write out the terms. $75, $125, $175, $225, ... The first term is 75, and 50 is added to form each following term. Therefore, we have a1 = 75, an = an – 1 + 50, n ≥ 2.

Determine whether each sequence is arithmetic, geometric, or neither. Explain.71. 8, –32, 128, –512, ...

SOLUTION: Check for a common difference. –32 – 8 = –40 128 – (–32) = 160 There is no common difference. Check for a common ratio. –32 ÷ 8 = –4 128 ÷ (–32) = –4 Geometric; the common ratio is –4.

72. 25, 8, –9, –26, ...

SOLUTION: Check for a common difference. 8 – 25 = –17 –9 – 8 = –17 Arithmetic; the common difference is –17.

73.

SOLUTION: Check for a common difference.

There is no common difference. Check for a common ratio.

There is no common ratio, so the sequence is not arithmetic or geometric.

74. 43, 52, 61, 70, ...

SOLUTION: Check for a common difference. 52 – 43 = 9 61 – 52 = 9 Arithmetic; the common difference is 9.

75. –27, –16, –5, 6, ...

SOLUTION: Check for a common difference. –16 – (–27) = 11 –5 – (–16) = 11 Arithmetic; the common difference is 11.

76. 200, 100, 50, 25, …

SOLUTION: Check for a common difference. 100 – 200 = –100 50 – 100 = –50 There is no common difference. Check for a common ratio. 100 ÷ 200 = 0.5 50 ÷ 100 = 0.5

Geometric; the common ratio is 0.5 or .

77. JOBS Kimi received an offer for a new job. She wants to compare the offer with her current job. What is total amount of sales that Kimi must get each month to make the same income at either job?

SOLUTION: Let s be Kimi’s monthly sales.

Kimi must sell $80,000 each month to make the same income at either job.

Determine whether each sequence is an arithmetic sequence. If it is, state the common difference.78. 24, 16, 8, 0, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is –8, because 16 – 24 = –8; 8 –16 = –8; etc.

79. , 13, 26, …

SOLUTION: Find the difference between the terms.

– =

13 – =

26 – 13 = 13 There is not a common different. The sequence is not an arithmetic sequence

80. 7, 6, 5, 4, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is –1, because 6 – 7 = –1; 5 – 6 = –1; 4 – 5 = –1;etc.

81. 10, 12, 15, 18, …

SOLUTION: Find the difference between the terms. 12 – 10 = 2 15 – 12 = 3 18 – 15 = 3 There is no common difference. The sequence is not an arithmetic sequence.

82. −15, −11, −7, −3, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is 4, because –11 – (–15) = 4; –7 – (–11) = 4; –3 – (–7) = 4;etc.

83. −0.3, 0.2, 0.7, 1.2, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is 0.5, because 0.2 – (–0.3) = 0.5; 0.7 – 0.2 = 0.5; 1.2 – 0.7 = 0.5; etc.

Simplify.

84. t(t5)(t

7)

SOLUTION:

85. n3(n

2)(−2n

3)

SOLUTION:

86. (5t5v

2)(10t

3v

4)

SOLUTION:

87. (−8u4z

5)(5uz

4)

SOLUTION:

88. [(3)2]3

SOLUTION:

89. [(2)3]2

SOLUTION:

90. (2m4k

3)2(−3mk

2)3

SOLUTION:

91. (6xy2)2(2x

2y

2z

2)3

SOLUTION:

eSolutions Manual - Powered by Cognero Page 2

8-1 Adding and Subtracting Polynomials

Determine whether each expression is a polynomial. If it is a polynomial, find the degree and determine whether it is a monomial, binomial, or trinomial.

1. 7ab + 6b2 – 2a

3

SOLUTION:

A polynomial is a monomial or the sum of monomials. 7ab + 6b2 – 2a

3 is the sum of 3 monomials, so it is a

polynomial. The degree of a polynomial is the greatest degree of any term in the polynomial.The degree of each term is 2, 2, and

3, so the degree of 7ab + 6b2 – 2a

3 is 3. The polynomial has three terms, so it is a trinomial.

2. 2y – 5 + 3y2

SOLUTION:

2y – 5 + 3y2

is the sum of monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 1, 0, and

2, so the degree of 2y – 5 + 3y2 is 2. The polynomial has three terms, so it is a trinomial.

3. 3x2

SOLUTION:

A polynomial is a monomial or the sum of monomials, so 3x2 is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of 3x2 is 2. The

polynomial has one term, so it is a monomial.

4.

SOLUTION: A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer exponents. It has

only one term. is a division of two monomials, so it is not a monomial.

5. 5m2p

3 + 6

SOLUTION:

A polynomial is a monomial or the sum of monomials. 5m2p

3 + 6

is the sum of 2 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 5 and 0,

so the degree of 5m2p

3 + 6 is 5. The polynomial has two terms, so it is a binomial.

6. 5q–4

+ 6q

SOLUTION: A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer exponents. It has

only one term. 5q-4 is equal to , which is a division of two monomials, so it is not a monomial.

Write each polynomial in standard form. Identify the leading coefficient.

7. –4d4 + 1 – d

2

SOLUTION: Find the degree of each term.

–4d4→ 4

1 → 0

–d2 → 2

The greatest degree is 4, from the term –4d4, so the leading coefficient of –4d

4 + 1 – d

2 is –4.

Rewrite the polynomial with each monomial in descending order according to degree.

–4d4 – d

2 + 1

8. 2x5 – 12 + 3x

SOLUTION: Find the degree of each term.

2x5 → 5

–12 → 0 3x → 1

The greatest degree is 5, from the term 2x5, so the leading coefficient of 2x

5 – 12 + 3x is 2.

Rewrite the polynomial with each monomial in descending order according to degree.

2x5 + 3x – 12

9. 4z – 2z2 – 5z

4

SOLUTION: Find the degree of each term. 4z → 1

– 2z2 → 2

– 5z4→ 4

The greatest degree is 4, from the term – 5z4, so the leading coefficient of 4z – 2z

2 – 5z

4 is –5.

Rewrite the polynomial with each monomial in descending order according to degree.

–5z4 – 2z

2 + 4z

10. 2a + 4a3 – 5a

2 – 1

SOLUTION: Find the degree of each term. 2a → 1

4a3→ 3

– 5a2 → 2

– 1 → 0

The greatest degree is 3, from the term 4a3, so the leading coefficient of 2a + 4a

3 – 5a

2 – 1 is 4.

Rewrite the polynomial with each monomial in descending order according to degree.

4a3– 5a

2 + 2a – 1

Find each sum or difference.

11. (6x3 − 4) + (−2x

3 + 9)

SOLUTION:

12. (g3 − 2g2 + 5g + 6) − (g2 + 2g)

SOLUTION:

13. (4 + 2a2 − 2a) − (3a

2 − 8a + 7)

SOLUTION:

14. (8y − 4y2) + (3y − 9y

2)

SOLUTION:

15. (−4z3 − 2z + 8) − (4z

3 + 3z2 − 5)

SOLUTION:

16. (−3d2 − 8 + 2d) + (4d − 12 + d2

)

SOLUTION:

17. (y + 5) + (2y + 4y2 – 2)

SOLUTION:

18. (3n3 − 5n + n2

) − (−8n2 + 3n

3)

SOLUTION:

19. CCSS SENSE-MAKING The total number of students T who traveled for spring break consists of two groups: students who flew to their destinations F and students who drove to their destination D. The number (in thousands) of students who flew and the total number of students who flew or drove can be modeled by the following equations,where n is the number of years since 1995. T = 14n + 21

F = 8n + 7 a. Write an equation that models the number of students who drove to their destination for this time period. b. Predict the number of students who will drive to their destination in 2012. c. How many students will drive or fly to their destination in 2015?

SOLUTION: a.

D = 6n + 14 b. n = 2012 – 1995 = 17

The number of students who will drive to their destination in 2012 is 116,000 students. c. n = 2015 – 1995 = 20

The number of students who will drive or fly to their destination in 2015 is 301,000 students.

Determine whether each expression is a polynomial. If it is a polynomial, find the degree and determine whether it is a monomial, binomial, or trinomial.

20.

SOLUTION: A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer exponents. It has

only one term. is a division of two monomials, so it is not a monomial.

21.

SOLUTION: A polynomial is a monomial or the sum of monomials. 21 is a monomial, so it is also a polynomial. The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of 21 is 0. The polynomial has only one term, so it is a monomial.

22. c4 – 2c

2 + 1

SOLUTION:

A polynomial is a monomial or the sum of monomials. c4 – 2c

2 + 1

is the sum of 3 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 4, 2, and

0, so the degree of c4 – 2c2 + 1 is 4. The polynomial has three terms, so it is a trinomial.

23. d + 3dc

SOLUTION: A polynomial is a monomial or the sum of monomials. A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer

exponents. 3dc

has a variable in the exponent, so it is not a monomial.

24. a – a2

SOLUTION:

A polynomial is a monomial or the sum of monomials. a – a2 is the sum of 2 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 1 and 2,

so the degree of a – a2 is 2. The polynomial has two terms, so it is a binomial.

25. 5n3 + nq

3

SOLUTION:

A polynomial is a monomial or the sum of monomials. 5n3 + nq3 is the sum of 2 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 3 and 4,

so the degree of 5n3 + nq3 is 4. The polynomial has two terms, so it is a binomial.

Write each polynomial in standard form. Identify the leading coefficient.

26. 5x2 – 2 + 3x

SOLUTION: Find the degree of each term.

5x2 → 2

– 2 → 0

3x → 1

The greatest degree is 2, from the term 5x2, so the leading coefficient of 5x

2 – 2 + 3x is 5.

Rewrite the polynomial with each monomial in descending order according to degree.

5x2 + 3x – 2

27. 8y + 7y3

SOLUTION: Find the degree of each term.

7y3 → 3

8y → 1

The greatest degree is 3, from the term 7y3, so the leading coefficient of 8y + 7y

3 is 7.

Rewrite the polynomial with each monomial in descending order according to degree.

7y3 + 8y

28. 4 – 3c – 5c2

SOLUTION: Find the degree of each term. 4 → 0 3c → 1

– 5c2 → 2

The greatest degree is 2, from the term – 5c2, so the leading coefficient of 4 – 3c – 5c

2 is –5.

Rewrite the polynomial with each monomial in descending order according to degree.

–5c2 – 3c + 4

29. –y3 + 3y – 3y

2 + 2

SOLUTION: Find the degree of each term.

–y3 → 3

3y → 1

3y2 → 2

2 → 0

The greatest degree is 3, from the term –y3, so the leading coefficient of –y

3 + 3y – 3y

2 + 2 is –1.

Rewrite the polynomial with each monomial in descending order according to degree. –y 3 – 3y 2 + 3y + 2

30. 11t + 2t2 – 3 + t

5

SOLUTION: Find the degree of each term. 11t → 1

2t2 → 2

–3 → 0

t5 → 5

The greatest degree is 5, from the term t5, so the leading coefficient of 11t + 2t2 – 3 + t

5 is 1.

Rewrite the polynomial with each monomial in descending order according to degree.

t5 + 2t

2 + 11t – 3

31. 2 + r – r3

SOLUTION: Find the degree of each term. 2 → 0 r → 1

– r3→ 3

The greatest degree is 3, from the term – r3, so the leading coefficient of 2 + r – r3

is –1. Rewrite the polynomial with each monomial in descending order according to degree.

–r3 + r + 2

32.

SOLUTION: Find the degree of each term.

→ 0

–3x4 → 4

7 → 0

The greatest degree is 4, from the term –3x4, so the leading coefficient of is –3.

Rewrite the polynomial with each monomial in descending order according to degree.

33. –9b2 + 10b – b

6

SOLUTION: Find the degree of each term.

–9b2 → 2

10b → 1

–b6→ 6

The greatest degree is 6, from the term –b6, so the leading coefficient of –9b2 + 10b – b6 is –1. Rewrite the polynomial with each monomial in descending order according to degree.

–b6 – 9b

2 + 10b

Find each sum or difference.

34. (2c2 + 6c + 4) + (5c – 7)

SOLUTION:

35. (2x + 3x2) − (7 − 8x

2)

SOLUTION:

36. (3c3 − c + 11) − (c2 + 2c + 8)

SOLUTION:

37. (z2 + z) + (z

2 − 11)

SOLUTION:

38. (2x − 2y + 1) − (3y + 4x)

SOLUTION:

39. (4a − 5b2 + 3) + (6 − 2a + 3b

2)

SOLUTION:

40. (x2y − 3x

2 + y) + (3y − 2x2y)

SOLUTION:

41. (−8xy + 3x2 − 5y) + (4x

2 − 2y + 6xy)

SOLUTION:

42. (5n − 2p2 + 2np) − (4p

2 + 4n)

SOLUTION:

43. (4rxt − 8r2x + x2

) − (6rx2 + 5rxt − 2x

2)

SOLUTION:

44. PETS From 1999 through 2009, the number of dogs D and the number of cats C (in hundreds) adopted from animal

shelters in the United States are modeled by the equations D = 2n + 3 and C = n + 4, where n is the number of yearssince 1999. a. Write an equation that models the total number T of dogs and cats adopted in hundreds for this time period. b. If this trend continues, how many dogs and cats will be adopted in 2013?

SOLUTION: a.

So, an equation that models the total number of dogs and cats adopted is T = 3n + 7. b. Evaluate the equation for the total number of dogs and cats for n = 2013 – 1999 = 14.

The number of cats and dogs adopted in 2013 will be 49 × 100 or 4900 cats and dogs.

Classify each polynomial according to its degree and number of terms.

45. 4x – 3x2 + 5

SOLUTION:

Find the degree of each term of 4x – 3x2 + 5.

4x → 1

– 3x2 → 2

5 → 0

The greatest degree is 2 and there are 3 terms, so 4x – 3x2 + 5 is a quadratic trinomial.

46. 11z3

SOLUTION: Find the degree of each term. 11z3

→ 3

The greatest degree is 3 and there is one term, so 11z3 is a cubic monomial.

47. 9 + y4

SOLUTION:

Find the degree of each term of 9 + y 4.

9 → 1

y4→ 4

The greatest degree is 4 and there are 2 terms, so 9 + y 4 is a quartic binomial.

48. 3x3 – 7

SOLUTION:

Find the degree of each term of 3x3 – 7.

3x3 → 3

–7 → 0

The greatest degree is 3 and there are 2 terms, so 3x3 – 7 is a cubic binomial.

49. –2x5 – x

2 + 5x – 8

SOLUTION:

Find the degree of each term of –2x5 – x2 + 5x – 8.

–2x5 → 5

–x2 → 2

5x → 1 – 8 → 0

The greatest degree is 5 and there are 4 terms, so –2x5 – x2 + 5x – 8 is a quintic polynomial.

50. 10t – 4t2 + 6t

3

SOLUTION:

Find the degree of each term of 10t – 4t2 + 6t

3.

10t → 1

4t2 → 2

6t3 → 3

The greatest degree is 3 and there are 3 terms, so 10t – 4t2 + 6t

3 is a cubic trinomial.

51. ENROLLMENT In a rapidly growing school system, the numbers (in hundreds) of total students N and

K-5 students P enrolled from 2000 to 2009 are modeled by the equations N = 1.25t2 – t + 7.5 and P =

0.7t2 – 0.95t + 3.8, where t is the number of years since 2000.

a. Write an equation modeling the number of 6-12 students S enrolled for this time period. b. How many 6-12 students were enrolled in the school system in 2007?

SOLUTION: a. To write an equation that represents the number of 6-12 students enrolled, subtract the equations that represent the total number of students and the number of K-5 students.

b. Replace t with 7 in the equation for S to determine the number of students enrolled in 6-12 in 2007.

In 2007 there were 30.3 hundreds or 3030 students enrolled in 6-12.

52. CCSS REASONING The perimeter of the figure shown is represented by the expression 3x2 − 7x + 2. Write a

polynomial that represents the measure of the third side.

SOLUTION:

53. GEOMETRY Consider the rectangle.

a. What does (4x2 + 2x – 1)(2x

2 – x + 3) represent?

b. What does 2(4x2 + 2x – 1) + 2(2x

2 – x + 3) represent?

SOLUTION:

a. (4x2 + 2x – 1)(2x

2 – x + 3) is a multiplication of the length and the width of the rectangle, which is the formula for

the area of a rectangle.

b. 2(4x2 + 2x – 1) + 2(2x

2 – x + 3) is the sum of twice the length and twice the width of a rectangle, which is the

formula for the perimeter of the rectangle

Find each sum or difference.

54. (4x + 2y − 6z) + (5y − 2z + 7x) + (−9z − 2x − 3y)

SOLUTION:

55. (5a2 − 4) + (a

2 − 2a + 12) + (4a2 − 6a + 8)

SOLUTION:

56. (3c2 − 7) + (4c + 7) − (c

2 + 5c − 8)

SOLUTION:

57. (3n3 + 3n − 10) − (4n

2 − 5n) + (4n3 − 3n

2 − 9n + 4)

SOLUTION:

58. FOOTBALL The National Football League is divided into two conferences, the American A and the National N. From 2002 through 2009, the total attendance T (in thousands) for both conferences and for the American Conference games are modeled by the following equations, where x is the number of years since 2002.

T = –0.69x3 + 55.83x

2 + 643.31x + 10,538

A = –3.78x3 + 58.96x

2 + 265.96x + 5257

Determine how many people attended National Conference football games in 2009.

SOLUTION:

Let x = 7 represent 2009, then find how many people attended a National Conference football game in 2009.

In 2009 the number of people who attended National Conference football games was about 8829 thousand. Multiply by 1000 to find the attendance in standard form. 8829 ×1000 = 8,829,000. So, about 8,829,000 people attended National Conference football games in 2009.

59. CAR RENTAL The cost to rent a car for a day is $15 plus $0.15 for each mile driven. a. Write a polynomial that represents the cost of renting a car for m miles. b. If a car is driven 145 miles, how much would it cost to rent? c. If a car is driven 105 miles each day for four days, how much would it cost to rent a car? d. If a car is driven 220 miles each day for seven days, how much would it cost to rent a car?

SOLUTION: a. The cost to rent a car is the daily rate and the mileage cost or 15 + 0.15m. b. Substitute 145 for m to find the daily cost to drive 145 miles.

The cost to rent the car would be $36.75. c. The expression represents the cost per day. For 4 days, multiply the entire expression by 4 to find the cost for a 4 day trip. Substitute 105 for m.

The cost to rent the car would be $123. d. The expression represents the cost per day. For 7 days, multiply the entire expression by 7 to find the cost for a 7 day trip. Substitute 220 for m.

The cost to rent the car would be $336.

60. MULTIPLE REPRESENTATIONS In this problem, you will explore perimeter and area. a. Geometric Draw three rectangles that each have a perimeter of 400 feet. b. Tabular Record the width and length of each rectangle in a table like the one shown below. Find the area of eachrectangle.

c. Graphical On a coordinate system, graph the area of rectangle 4 in terms of the length, x. Use the graph to determine the largest area possible. d. Analytical Determine the length and width that produce the largest area.

SOLUTION: a. For the perimeters to be 400, create lengths and widths that sum to 200. Note that some lengths are already provided in the table in part b.

b. The area is length multiplied by width. Make sure the units are squared for area. The sum of the length and width must be 200, so if the length is x, the width must be 200 – x.

c. The length will be the x-values on the horizontal and the area will be the y-values on the vertical. The length cannot pass 200 since the sum of the length and width is 200. Set the intervals for the x-axis to 25 feet. Extend the table of values to find more points to plot on the graph. It appears that 10,000 is the greatest area, so set the intervals

for the y-axis to 1000 ft2. After graphing, it appears that the highest point on the graph is at an area of 10,000 ft

2.

d. The associated x-value with the maximum area is x = 100, so the length must be 100 and the width must be 200 – 100, or 100. The length and width of the rectangle must be 100 feet each to have the largest area.

61. CCSS CRITIQUE Cheyenne and Sebastian are finding (2x2 − x) − (3x + 3x

2 − 2). Is either of them correct? Explain your reasoning.

SOLUTION:

Neither is correct. Cheyenne, did not distribute the negative to the 2nd and 3rd terms when she found the additive inverse. Sebastian did not distribute the negate to the 3rd terms when he found the additive inverse. To find the additive inverse, all terms should be multiplied by −1.

62. REASONING Determine whether each of the following statements is true or false . Explain your reasoning. a. A binomial can have a degree of zero. b. The order in which polynomials are subtracted does not matter.

SOLUTION: a. If a binomial has two terms that are each a degree of 0, then those terms can be combined and the binomial becomes a monomial. For example, 18 + 7 = 25. If one of the terms of the binomial does not have a degree of 0, then the binomial cannot have a degree of 0, since the degree of a polynomial is the greatest degree of any term in the polynomial. b. Subtraction is not commutative. While 2 + 5 = 5 + 2, 2 – 5 ≠ 5 – 2. This is also true for polynomials. Sample answer: (2x – 3) – (4x – 3) = –2x, but (4x – 3) – (2x – 3) = 2x

63. CHALLENGE Write a polynomial that represents the sum of an odd integer 2n + 1 and the next two consecutive odd integers.

SOLUTION:

64. WRITING IN MATH Why would you add or subtract equations that represent real-world situations? Explain.

SOLUTION:

65. WRITING IN MATH Describe how to add and subtract polynomials using both the vertical and horizontal formats.

SOLUTION: To add polynomials in a horizontal format, you combine like terms. For the vertical format, you write the polynomials in standard form, align like terms in columns, and combine like terms.

To subtract polynomials in a horizontal format you find the additive inverse of the polynomial you are subtracting, andthen combine like terms. For the vertical format, you write the polynomials in standard form, align like terms in columns, and subtract by adding the additive inverse.

66. Three consecutive integers can be represented by x, x + 1, and x + 2. What is the sum of these three integers? A x(x + 1)(x + 2)

B x3 + 3

C 3x + 3 D x + 3

SOLUTION:

The correct choice is C.

67. SHORT RESPONSE What is the perimeter of a square with sides that measure 2x + 3 units?

SOLUTION:

The perimeter of the square is 8x + 12 units.

68. Jim cuts a board in the shape of a regular hexagon and pounds in a nail at each vertex, as shown. How many rubber bands will he need to stretch a rubber band across every possible pair of nails?

F 15 G 14 H 12 J 9

SOLUTION: The first nail would connect to 5 others, the second to 4 others, the third to 3 others, etc.5 + 4 + 3 + 2 + 1 = 15 The correct choice is F.

69. Which ordered pair is in the solution set of the system of inequalities shown in the graph?

A (−3, 0) B (0, −3) C (5, 0) D (0, 5)

SOLUTION: Choice A is outside the shaded area for both inequalities. Choices B and D are inside the shaded area for only one inequality. Choice C is the only point in the solution for both inequalities. So, the correct choice is C.

70. COMPUTERS A computer technician charges by the hour to fix and repair computer equipment. The total cost of the technician for one hour is $75, for two hours is $125, for three hours is $175, for four hours is $225, and so on. Write a recursive formula for the sequence.

SOLUTION: Write out the terms. $75, $125, $175, $225, ... The first term is 75, and 50 is added to form each following term. Therefore, we have a1 = 75, an = an – 1 + 50, n ≥ 2.

Determine whether each sequence is arithmetic, geometric, or neither. Explain.71. 8, –32, 128, –512, ...

SOLUTION: Check for a common difference. –32 – 8 = –40 128 – (–32) = 160 There is no common difference. Check for a common ratio. –32 ÷ 8 = –4 128 ÷ (–32) = –4 Geometric; the common ratio is –4.

72. 25, 8, –9, –26, ...

SOLUTION: Check for a common difference. 8 – 25 = –17 –9 – 8 = –17 Arithmetic; the common difference is –17.

73.

SOLUTION: Check for a common difference.

There is no common difference. Check for a common ratio.

There is no common ratio, so the sequence is not arithmetic or geometric.

74. 43, 52, 61, 70, ...

SOLUTION: Check for a common difference. 52 – 43 = 9 61 – 52 = 9 Arithmetic; the common difference is 9.

75. –27, –16, –5, 6, ...

SOLUTION: Check for a common difference. –16 – (–27) = 11 –5 – (–16) = 11 Arithmetic; the common difference is 11.

76. 200, 100, 50, 25, …

SOLUTION: Check for a common difference. 100 – 200 = –100 50 – 100 = –50 There is no common difference. Check for a common ratio. 100 ÷ 200 = 0.5 50 ÷ 100 = 0.5

Geometric; the common ratio is 0.5 or .

77. JOBS Kimi received an offer for a new job. She wants to compare the offer with her current job. What is total amount of sales that Kimi must get each month to make the same income at either job?

SOLUTION: Let s be Kimi’s monthly sales.

Kimi must sell $80,000 each month to make the same income at either job.

Determine whether each sequence is an arithmetic sequence. If it is, state the common difference.78. 24, 16, 8, 0, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is –8, because 16 – 24 = –8; 8 –16 = –8; etc.

79. , 13, 26, …

SOLUTION: Find the difference between the terms.

– =

13 – =

26 – 13 = 13 There is not a common different. The sequence is not an arithmetic sequence

80. 7, 6, 5, 4, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is –1, because 6 – 7 = –1; 5 – 6 = –1; 4 – 5 = –1;etc.

81. 10, 12, 15, 18, …

SOLUTION: Find the difference between the terms. 12 – 10 = 2 15 – 12 = 3 18 – 15 = 3 There is no common difference. The sequence is not an arithmetic sequence.

82. −15, −11, −7, −3, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is 4, because –11 – (–15) = 4; –7 – (–11) = 4; –3 – (–7) = 4;etc.

83. −0.3, 0.2, 0.7, 1.2, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is 0.5, because 0.2 – (–0.3) = 0.5; 0.7 – 0.2 = 0.5; 1.2 – 0.7 = 0.5; etc.

Simplify.

84. t(t5)(t

7)

SOLUTION:

85. n3(n

2)(−2n

3)

SOLUTION:

86. (5t5v

2)(10t

3v

4)

SOLUTION:

87. (−8u4z

5)(5uz

4)

SOLUTION:

88. [(3)2]3

SOLUTION:

89. [(2)3]2

SOLUTION:

90. (2m4k

3)2(−3mk

2)3

SOLUTION:

91. (6xy2)2(2x

2y

2z

2)3

SOLUTION:

eSolutions Manual - Powered by Cognero Page 3

8-1 Adding and Subtracting Polynomials

Determine whether each expression is a polynomial. If it is a polynomial, find the degree and determine whether it is a monomial, binomial, or trinomial.

1. 7ab + 6b2 – 2a

3

SOLUTION:

A polynomial is a monomial or the sum of monomials. 7ab + 6b2 – 2a

3 is the sum of 3 monomials, so it is a

polynomial. The degree of a polynomial is the greatest degree of any term in the polynomial.The degree of each term is 2, 2, and

3, so the degree of 7ab + 6b2 – 2a

3 is 3. The polynomial has three terms, so it is a trinomial.

2. 2y – 5 + 3y2

SOLUTION:

2y – 5 + 3y2

is the sum of monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 1, 0, and

2, so the degree of 2y – 5 + 3y2 is 2. The polynomial has three terms, so it is a trinomial.

3. 3x2

SOLUTION:

A polynomial is a monomial or the sum of monomials, so 3x2 is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of 3x2 is 2. The

polynomial has one term, so it is a monomial.

4.

SOLUTION: A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer exponents. It has

only one term. is a division of two monomials, so it is not a monomial.

5. 5m2p

3 + 6

SOLUTION:

A polynomial is a monomial or the sum of monomials. 5m2p

3 + 6

is the sum of 2 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 5 and 0,

so the degree of 5m2p

3 + 6 is 5. The polynomial has two terms, so it is a binomial.

6. 5q–4

+ 6q

SOLUTION: A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer exponents. It has

only one term. 5q-4 is equal to , which is a division of two monomials, so it is not a monomial.

Write each polynomial in standard form. Identify the leading coefficient.

7. –4d4 + 1 – d

2

SOLUTION: Find the degree of each term.

–4d4→ 4

1 → 0

–d2 → 2

The greatest degree is 4, from the term –4d4, so the leading coefficient of –4d

4 + 1 – d

2 is –4.

Rewrite the polynomial with each monomial in descending order according to degree.

–4d4 – d

2 + 1

8. 2x5 – 12 + 3x

SOLUTION: Find the degree of each term.

2x5 → 5

–12 → 0 3x → 1

The greatest degree is 5, from the term 2x5, so the leading coefficient of 2x

5 – 12 + 3x is 2.

Rewrite the polynomial with each monomial in descending order according to degree.

2x5 + 3x – 12

9. 4z – 2z2 – 5z

4

SOLUTION: Find the degree of each term. 4z → 1

– 2z2 → 2

– 5z4→ 4

The greatest degree is 4, from the term – 5z4, so the leading coefficient of 4z – 2z

2 – 5z

4 is –5.

Rewrite the polynomial with each monomial in descending order according to degree.

–5z4 – 2z

2 + 4z

10. 2a + 4a3 – 5a

2 – 1

SOLUTION: Find the degree of each term. 2a → 1

4a3→ 3

– 5a2 → 2

– 1 → 0

The greatest degree is 3, from the term 4a3, so the leading coefficient of 2a + 4a

3 – 5a

2 – 1 is 4.

Rewrite the polynomial with each monomial in descending order according to degree.

4a3– 5a

2 + 2a – 1

Find each sum or difference.

11. (6x3 − 4) + (−2x

3 + 9)

SOLUTION:

12. (g3 − 2g2 + 5g + 6) − (g2 + 2g)

SOLUTION:

13. (4 + 2a2 − 2a) − (3a

2 − 8a + 7)

SOLUTION:

14. (8y − 4y2) + (3y − 9y

2)

SOLUTION:

15. (−4z3 − 2z + 8) − (4z

3 + 3z2 − 5)

SOLUTION:

16. (−3d2 − 8 + 2d) + (4d − 12 + d2

)

SOLUTION:

17. (y + 5) + (2y + 4y2 – 2)

SOLUTION:

18. (3n3 − 5n + n2

) − (−8n2 + 3n

3)

SOLUTION:

19. CCSS SENSE-MAKING The total number of students T who traveled for spring break consists of two groups: students who flew to their destinations F and students who drove to their destination D. The number (in thousands) of students who flew and the total number of students who flew or drove can be modeled by the following equations,where n is the number of years since 1995. T = 14n + 21

F = 8n + 7 a. Write an equation that models the number of students who drove to their destination for this time period. b. Predict the number of students who will drive to their destination in 2012. c. How many students will drive or fly to their destination in 2015?

SOLUTION: a.

D = 6n + 14 b. n = 2012 – 1995 = 17

The number of students who will drive to their destination in 2012 is 116,000 students. c. n = 2015 – 1995 = 20

The number of students who will drive or fly to their destination in 2015 is 301,000 students.

Determine whether each expression is a polynomial. If it is a polynomial, find the degree and determine whether it is a monomial, binomial, or trinomial.

20.

SOLUTION: A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer exponents. It has

only one term. is a division of two monomials, so it is not a monomial.

21.

SOLUTION: A polynomial is a monomial or the sum of monomials. 21 is a monomial, so it is also a polynomial. The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of 21 is 0. The polynomial has only one term, so it is a monomial.

22. c4 – 2c

2 + 1

SOLUTION:

A polynomial is a monomial or the sum of monomials. c4 – 2c

2 + 1

is the sum of 3 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 4, 2, and

0, so the degree of c4 – 2c2 + 1 is 4. The polynomial has three terms, so it is a trinomial.

23. d + 3dc

SOLUTION: A polynomial is a monomial or the sum of monomials. A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer

exponents. 3dc

has a variable in the exponent, so it is not a monomial.

24. a – a2

SOLUTION:

A polynomial is a monomial or the sum of monomials. a – a2 is the sum of 2 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 1 and 2,

so the degree of a – a2 is 2. The polynomial has two terms, so it is a binomial.

25. 5n3 + nq

3

SOLUTION:

A polynomial is a monomial or the sum of monomials. 5n3 + nq3 is the sum of 2 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 3 and 4,

so the degree of 5n3 + nq3 is 4. The polynomial has two terms, so it is a binomial.

Write each polynomial in standard form. Identify the leading coefficient.

26. 5x2 – 2 + 3x

SOLUTION: Find the degree of each term.

5x2 → 2

– 2 → 0

3x → 1

The greatest degree is 2, from the term 5x2, so the leading coefficient of 5x

2 – 2 + 3x is 5.

Rewrite the polynomial with each monomial in descending order according to degree.

5x2 + 3x – 2

27. 8y + 7y3

SOLUTION: Find the degree of each term.

7y3 → 3

8y → 1

The greatest degree is 3, from the term 7y3, so the leading coefficient of 8y + 7y

3 is 7.

Rewrite the polynomial with each monomial in descending order according to degree.

7y3 + 8y

28. 4 – 3c – 5c2

SOLUTION: Find the degree of each term. 4 → 0 3c → 1

– 5c2 → 2

The greatest degree is 2, from the term – 5c2, so the leading coefficient of 4 – 3c – 5c

2 is –5.

Rewrite the polynomial with each monomial in descending order according to degree.

–5c2 – 3c + 4

29. –y3 + 3y – 3y

2 + 2

SOLUTION: Find the degree of each term.

–y3 → 3

3y → 1

3y2 → 2

2 → 0

The greatest degree is 3, from the term –y3, so the leading coefficient of –y

3 + 3y – 3y

2 + 2 is –1.

Rewrite the polynomial with each monomial in descending order according to degree. –y 3 – 3y 2 + 3y + 2

30. 11t + 2t2 – 3 + t

5

SOLUTION: Find the degree of each term. 11t → 1

2t2 → 2

–3 → 0

t5 → 5

The greatest degree is 5, from the term t5, so the leading coefficient of 11t + 2t2 – 3 + t

5 is 1.

Rewrite the polynomial with each monomial in descending order according to degree.

t5 + 2t

2 + 11t – 3

31. 2 + r – r3

SOLUTION: Find the degree of each term. 2 → 0 r → 1

– r3→ 3

The greatest degree is 3, from the term – r3, so the leading coefficient of 2 + r – r3

is –1. Rewrite the polynomial with each monomial in descending order according to degree.

–r3 + r + 2

32.

SOLUTION: Find the degree of each term.

→ 0

–3x4 → 4

7 → 0

The greatest degree is 4, from the term –3x4, so the leading coefficient of is –3.

Rewrite the polynomial with each monomial in descending order according to degree.

33. –9b2 + 10b – b

6

SOLUTION: Find the degree of each term.

–9b2 → 2

10b → 1

–b6→ 6

The greatest degree is 6, from the term –b6, so the leading coefficient of –9b2 + 10b – b6 is –1. Rewrite the polynomial with each monomial in descending order according to degree.

–b6 – 9b

2 + 10b

Find each sum or difference.

34. (2c2 + 6c + 4) + (5c – 7)

SOLUTION:

35. (2x + 3x2) − (7 − 8x

2)

SOLUTION:

36. (3c3 − c + 11) − (c2 + 2c + 8)

SOLUTION:

37. (z2 + z) + (z

2 − 11)

SOLUTION:

38. (2x − 2y + 1) − (3y + 4x)

SOLUTION:

39. (4a − 5b2 + 3) + (6 − 2a + 3b

2)

SOLUTION:

40. (x2y − 3x

2 + y) + (3y − 2x2y)

SOLUTION:

41. (−8xy + 3x2 − 5y) + (4x

2 − 2y + 6xy)

SOLUTION:

42. (5n − 2p2 + 2np) − (4p

2 + 4n)

SOLUTION:

43. (4rxt − 8r2x + x2

) − (6rx2 + 5rxt − 2x

2)

SOLUTION:

44. PETS From 1999 through 2009, the number of dogs D and the number of cats C (in hundreds) adopted from animal

shelters in the United States are modeled by the equations D = 2n + 3 and C = n + 4, where n is the number of yearssince 1999. a. Write an equation that models the total number T of dogs and cats adopted in hundreds for this time period. b. If this trend continues, how many dogs and cats will be adopted in 2013?

SOLUTION: a.

So, an equation that models the total number of dogs and cats adopted is T = 3n + 7. b. Evaluate the equation for the total number of dogs and cats for n = 2013 – 1999 = 14.

The number of cats and dogs adopted in 2013 will be 49 × 100 or 4900 cats and dogs.

Classify each polynomial according to its degree and number of terms.

45. 4x – 3x2 + 5

SOLUTION:

Find the degree of each term of 4x – 3x2 + 5.

4x → 1

– 3x2 → 2

5 → 0

The greatest degree is 2 and there are 3 terms, so 4x – 3x2 + 5 is a quadratic trinomial.

46. 11z3

SOLUTION: Find the degree of each term. 11z3

→ 3

The greatest degree is 3 and there is one term, so 11z3 is a cubic monomial.

47. 9 + y4

SOLUTION:

Find the degree of each term of 9 + y 4.

9 → 1

y4→ 4

The greatest degree is 4 and there are 2 terms, so 9 + y 4 is a quartic binomial.

48. 3x3 – 7

SOLUTION:

Find the degree of each term of 3x3 – 7.

3x3 → 3

–7 → 0

The greatest degree is 3 and there are 2 terms, so 3x3 – 7 is a cubic binomial.

49. –2x5 – x

2 + 5x – 8

SOLUTION:

Find the degree of each term of –2x5 – x2 + 5x – 8.

–2x5 → 5

–x2 → 2

5x → 1 – 8 → 0

The greatest degree is 5 and there are 4 terms, so –2x5 – x2 + 5x – 8 is a quintic polynomial.

50. 10t – 4t2 + 6t

3

SOLUTION:

Find the degree of each term of 10t – 4t2 + 6t

3.

10t → 1

4t2 → 2

6t3 → 3

The greatest degree is 3 and there are 3 terms, so 10t – 4t2 + 6t

3 is a cubic trinomial.

51. ENROLLMENT In a rapidly growing school system, the numbers (in hundreds) of total students N and

K-5 students P enrolled from 2000 to 2009 are modeled by the equations N = 1.25t2 – t + 7.5 and P =

0.7t2 – 0.95t + 3.8, where t is the number of years since 2000.

a. Write an equation modeling the number of 6-12 students S enrolled for this time period. b. How many 6-12 students were enrolled in the school system in 2007?

SOLUTION: a. To write an equation that represents the number of 6-12 students enrolled, subtract the equations that represent the total number of students and the number of K-5 students.

b. Replace t with 7 in the equation for S to determine the number of students enrolled in 6-12 in 2007.

In 2007 there were 30.3 hundreds or 3030 students enrolled in 6-12.

52. CCSS REASONING The perimeter of the figure shown is represented by the expression 3x2 − 7x + 2. Write a

polynomial that represents the measure of the third side.

SOLUTION:

53. GEOMETRY Consider the rectangle.

a. What does (4x2 + 2x – 1)(2x

2 – x + 3) represent?

b. What does 2(4x2 + 2x – 1) + 2(2x

2 – x + 3) represent?

SOLUTION:

a. (4x2 + 2x – 1)(2x

2 – x + 3) is a multiplication of the length and the width of the rectangle, which is the formula for

the area of a rectangle.

b. 2(4x2 + 2x – 1) + 2(2x

2 – x + 3) is the sum of twice the length and twice the width of a rectangle, which is the

formula for the perimeter of the rectangle

Find each sum or difference.

54. (4x + 2y − 6z) + (5y − 2z + 7x) + (−9z − 2x − 3y)

SOLUTION:

55. (5a2 − 4) + (a

2 − 2a + 12) + (4a2 − 6a + 8)

SOLUTION:

56. (3c2 − 7) + (4c + 7) − (c

2 + 5c − 8)

SOLUTION:

57. (3n3 + 3n − 10) − (4n

2 − 5n) + (4n3 − 3n

2 − 9n + 4)

SOLUTION:

58. FOOTBALL The National Football League is divided into two conferences, the American A and the National N. From 2002 through 2009, the total attendance T (in thousands) for both conferences and for the American Conference games are modeled by the following equations, where x is the number of years since 2002.

T = –0.69x3 + 55.83x

2 + 643.31x + 10,538

A = –3.78x3 + 58.96x

2 + 265.96x + 5257

Determine how many people attended National Conference football games in 2009.

SOLUTION:

Let x = 7 represent 2009, then find how many people attended a National Conference football game in 2009.

In 2009 the number of people who attended National Conference football games was about 8829 thousand. Multiply by 1000 to find the attendance in standard form. 8829 ×1000 = 8,829,000. So, about 8,829,000 people attended National Conference football games in 2009.

59. CAR RENTAL The cost to rent a car for a day is $15 plus $0.15 for each mile driven. a. Write a polynomial that represents the cost of renting a car for m miles. b. If a car is driven 145 miles, how much would it cost to rent? c. If a car is driven 105 miles each day for four days, how much would it cost to rent a car? d. If a car is driven 220 miles each day for seven days, how much would it cost to rent a car?

SOLUTION: a. The cost to rent a car is the daily rate and the mileage cost or 15 + 0.15m. b. Substitute 145 for m to find the daily cost to drive 145 miles.

The cost to rent the car would be $36.75. c. The expression represents the cost per day. For 4 days, multiply the entire expression by 4 to find the cost for a 4 day trip. Substitute 105 for m.

The cost to rent the car would be $123. d. The expression represents the cost per day. For 7 days, multiply the entire expression by 7 to find the cost for a 7 day trip. Substitute 220 for m.

The cost to rent the car would be $336.

60. MULTIPLE REPRESENTATIONS In this problem, you will explore perimeter and area. a. Geometric Draw three rectangles that each have a perimeter of 400 feet. b. Tabular Record the width and length of each rectangle in a table like the one shown below. Find the area of eachrectangle.

c. Graphical On a coordinate system, graph the area of rectangle 4 in terms of the length, x. Use the graph to determine the largest area possible. d. Analytical Determine the length and width that produce the largest area.

SOLUTION: a. For the perimeters to be 400, create lengths and widths that sum to 200. Note that some lengths are already provided in the table in part b.

b. The area is length multiplied by width. Make sure the units are squared for area. The sum of the length and width must be 200, so if the length is x, the width must be 200 – x.

c. The length will be the x-values on the horizontal and the area will be the y-values on the vertical. The length cannot pass 200 since the sum of the length and width is 200. Set the intervals for the x-axis to 25 feet. Extend the table of values to find more points to plot on the graph. It appears that 10,000 is the greatest area, so set the intervals

for the y-axis to 1000 ft2. After graphing, it appears that the highest point on the graph is at an area of 10,000 ft

2.

d. The associated x-value with the maximum area is x = 100, so the length must be 100 and the width must be 200 – 100, or 100. The length and width of the rectangle must be 100 feet each to have the largest area.

61. CCSS CRITIQUE Cheyenne and Sebastian are finding (2x2 − x) − (3x + 3x

2 − 2). Is either of them correct? Explain your reasoning.

SOLUTION:

Neither is correct. Cheyenne, did not distribute the negative to the 2nd and 3rd terms when she found the additive inverse. Sebastian did not distribute the negate to the 3rd terms when he found the additive inverse. To find the additive inverse, all terms should be multiplied by −1.

62. REASONING Determine whether each of the following statements is true or false . Explain your reasoning. a. A binomial can have a degree of zero. b. The order in which polynomials are subtracted does not matter.

SOLUTION: a. If a binomial has two terms that are each a degree of 0, then those terms can be combined and the binomial becomes a monomial. For example, 18 + 7 = 25. If one of the terms of the binomial does not have a degree of 0, then the binomial cannot have a degree of 0, since the degree of a polynomial is the greatest degree of any term in the polynomial. b. Subtraction is not commutative. While 2 + 5 = 5 + 2, 2 – 5 ≠ 5 – 2. This is also true for polynomials. Sample answer: (2x – 3) – (4x – 3) = –2x, but (4x – 3) – (2x – 3) = 2x

63. CHALLENGE Write a polynomial that represents the sum of an odd integer 2n + 1 and the next two consecutive odd integers.

SOLUTION:

64. WRITING IN MATH Why would you add or subtract equations that represent real-world situations? Explain.

SOLUTION:

65. WRITING IN MATH Describe how to add and subtract polynomials using both the vertical and horizontal formats.

SOLUTION: To add polynomials in a horizontal format, you combine like terms. For the vertical format, you write the polynomials in standard form, align like terms in columns, and combine like terms.

To subtract polynomials in a horizontal format you find the additive inverse of the polynomial you are subtracting, andthen combine like terms. For the vertical format, you write the polynomials in standard form, align like terms in columns, and subtract by adding the additive inverse.

66. Three consecutive integers can be represented by x, x + 1, and x + 2. What is the sum of these three integers? A x(x + 1)(x + 2)

B x3 + 3

C 3x + 3 D x + 3

SOLUTION:

The correct choice is C.

67. SHORT RESPONSE What is the perimeter of a square with sides that measure 2x + 3 units?

SOLUTION:

The perimeter of the square is 8x + 12 units.

68. Jim cuts a board in the shape of a regular hexagon and pounds in a nail at each vertex, as shown. How many rubber bands will he need to stretch a rubber band across every possible pair of nails?

F 15 G 14 H 12 J 9

SOLUTION: The first nail would connect to 5 others, the second to 4 others, the third to 3 others, etc.5 + 4 + 3 + 2 + 1 = 15 The correct choice is F.

69. Which ordered pair is in the solution set of the system of inequalities shown in the graph?

A (−3, 0) B (0, −3) C (5, 0) D (0, 5)

SOLUTION: Choice A is outside the shaded area for both inequalities. Choices B and D are inside the shaded area for only one inequality. Choice C is the only point in the solution for both inequalities. So, the correct choice is C.

70. COMPUTERS A computer technician charges by the hour to fix and repair computer equipment. The total cost of the technician for one hour is $75, for two hours is $125, for three hours is $175, for four hours is $225, and so on. Write a recursive formula for the sequence.

SOLUTION: Write out the terms. $75, $125, $175, $225, ... The first term is 75, and 50 is added to form each following term. Therefore, we have a1 = 75, an = an – 1 + 50, n ≥ 2.

Determine whether each sequence is arithmetic, geometric, or neither. Explain.71. 8, –32, 128, –512, ...

SOLUTION: Check for a common difference. –32 – 8 = –40 128 – (–32) = 160 There is no common difference. Check for a common ratio. –32 ÷ 8 = –4 128 ÷ (–32) = –4 Geometric; the common ratio is –4.

72. 25, 8, –9, –26, ...

SOLUTION: Check for a common difference. 8 – 25 = –17 –9 – 8 = –17 Arithmetic; the common difference is –17.

73.

SOLUTION: Check for a common difference.

There is no common difference. Check for a common ratio.

There is no common ratio, so the sequence is not arithmetic or geometric.

74. 43, 52, 61, 70, ...

SOLUTION: Check for a common difference. 52 – 43 = 9 61 – 52 = 9 Arithmetic; the common difference is 9.

75. –27, –16, –5, 6, ...

SOLUTION: Check for a common difference. –16 – (–27) = 11 –5 – (–16) = 11 Arithmetic; the common difference is 11.

76. 200, 100, 50, 25, …

SOLUTION: Check for a common difference. 100 – 200 = –100 50 – 100 = –50 There is no common difference. Check for a common ratio. 100 ÷ 200 = 0.5 50 ÷ 100 = 0.5

Geometric; the common ratio is 0.5 or .

77. JOBS Kimi received an offer for a new job. She wants to compare the offer with her current job. What is total amount of sales that Kimi must get each month to make the same income at either job?

SOLUTION: Let s be Kimi’s monthly sales.

Kimi must sell $80,000 each month to make the same income at either job.

Determine whether each sequence is an arithmetic sequence. If it is, state the common difference.78. 24, 16, 8, 0, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is –8, because 16 – 24 = –8; 8 –16 = –8; etc.

79. , 13, 26, …

SOLUTION: Find the difference between the terms.

– =

13 – =

26 – 13 = 13 There is not a common different. The sequence is not an arithmetic sequence

80. 7, 6, 5, 4, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is –1, because 6 – 7 = –1; 5 – 6 = –1; 4 – 5 = –1;etc.

81. 10, 12, 15, 18, …

SOLUTION: Find the difference between the terms. 12 – 10 = 2 15 – 12 = 3 18 – 15 = 3 There is no common difference. The sequence is not an arithmetic sequence.

82. −15, −11, −7, −3, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is 4, because –11 – (–15) = 4; –7 – (–11) = 4; –3 – (–7) = 4;etc.

83. −0.3, 0.2, 0.7, 1.2, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is 0.5, because 0.2 – (–0.3) = 0.5; 0.7 – 0.2 = 0.5; 1.2 – 0.7 = 0.5; etc.

Simplify.

84. t(t5)(t

7)

SOLUTION:

85. n3(n

2)(−2n

3)

SOLUTION:

86. (5t5v

2)(10t

3v

4)

SOLUTION:

87. (−8u4z

5)(5uz

4)

SOLUTION:

88. [(3)2]3

SOLUTION:

89. [(2)3]2

SOLUTION:

90. (2m4k

3)2(−3mk

2)3

SOLUTION:

91. (6xy2)2(2x

2y

2z

2)3

SOLUTION:

eSolutions Manual - Powered by Cognero Page 4

8-1 Adding and Subtracting Polynomials

Determine whether each expression is a polynomial. If it is a polynomial, find the degree and determine whether it is a monomial, binomial, or trinomial.

1. 7ab + 6b2 – 2a

3

SOLUTION:

A polynomial is a monomial or the sum of monomials. 7ab + 6b2 – 2a

3 is the sum of 3 monomials, so it is a

polynomial. The degree of a polynomial is the greatest degree of any term in the polynomial.The degree of each term is 2, 2, and

3, so the degree of 7ab + 6b2 – 2a

3 is 3. The polynomial has three terms, so it is a trinomial.

2. 2y – 5 + 3y2

SOLUTION:

2y – 5 + 3y2

is the sum of monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 1, 0, and

2, so the degree of 2y – 5 + 3y2 is 2. The polynomial has three terms, so it is a trinomial.

3. 3x2

SOLUTION:

A polynomial is a monomial or the sum of monomials, so 3x2 is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of 3x2 is 2. The

polynomial has one term, so it is a monomial.

4.

SOLUTION: A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer exponents. It has

only one term. is a division of two monomials, so it is not a monomial.

5. 5m2p

3 + 6

SOLUTION:

A polynomial is a monomial or the sum of monomials. 5m2p

3 + 6

is the sum of 2 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 5 and 0,

so the degree of 5m2p

3 + 6 is 5. The polynomial has two terms, so it is a binomial.

6. 5q–4

+ 6q

SOLUTION: A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer exponents. It has

only one term. 5q-4 is equal to , which is a division of two monomials, so it is not a monomial.

Write each polynomial in standard form. Identify the leading coefficient.

7. –4d4 + 1 – d

2

SOLUTION: Find the degree of each term.

–4d4→ 4

1 → 0

–d2 → 2

The greatest degree is 4, from the term –4d4, so the leading coefficient of –4d

4 + 1 – d

2 is –4.

Rewrite the polynomial with each monomial in descending order according to degree.

–4d4 – d

2 + 1

8. 2x5 – 12 + 3x

SOLUTION: Find the degree of each term.

2x5 → 5

–12 → 0 3x → 1

The greatest degree is 5, from the term 2x5, so the leading coefficient of 2x

5 – 12 + 3x is 2.

Rewrite the polynomial with each monomial in descending order according to degree.

2x5 + 3x – 12

9. 4z – 2z2 – 5z

4

SOLUTION: Find the degree of each term. 4z → 1

– 2z2 → 2

– 5z4→ 4

The greatest degree is 4, from the term – 5z4, so the leading coefficient of 4z – 2z

2 – 5z

4 is –5.

Rewrite the polynomial with each monomial in descending order according to degree.

–5z4 – 2z

2 + 4z

10. 2a + 4a3 – 5a

2 – 1

SOLUTION: Find the degree of each term. 2a → 1

4a3→ 3

– 5a2 → 2

– 1 → 0

The greatest degree is 3, from the term 4a3, so the leading coefficient of 2a + 4a

3 – 5a

2 – 1 is 4.

Rewrite the polynomial with each monomial in descending order according to degree.

4a3– 5a

2 + 2a – 1

Find each sum or difference.

11. (6x3 − 4) + (−2x

3 + 9)

SOLUTION:

12. (g3 − 2g2 + 5g + 6) − (g2 + 2g)

SOLUTION:

13. (4 + 2a2 − 2a) − (3a

2 − 8a + 7)

SOLUTION:

14. (8y − 4y2) + (3y − 9y

2)

SOLUTION:

15. (−4z3 − 2z + 8) − (4z

3 + 3z2 − 5)

SOLUTION:

16. (−3d2 − 8 + 2d) + (4d − 12 + d2

)

SOLUTION:

17. (y + 5) + (2y + 4y2 – 2)

SOLUTION:

18. (3n3 − 5n + n2

) − (−8n2 + 3n

3)

SOLUTION:

19. CCSS SENSE-MAKING The total number of students T who traveled for spring break consists of two groups: students who flew to their destinations F and students who drove to their destination D. The number (in thousands) of students who flew and the total number of students who flew or drove can be modeled by the following equations,where n is the number of years since 1995. T = 14n + 21

F = 8n + 7 a. Write an equation that models the number of students who drove to their destination for this time period. b. Predict the number of students who will drive to their destination in 2012. c. How many students will drive or fly to their destination in 2015?

SOLUTION: a.

D = 6n + 14 b. n = 2012 – 1995 = 17

The number of students who will drive to their destination in 2012 is 116,000 students. c. n = 2015 – 1995 = 20

The number of students who will drive or fly to their destination in 2015 is 301,000 students.

Determine whether each expression is a polynomial. If it is a polynomial, find the degree and determine whether it is a monomial, binomial, or trinomial.

20.

SOLUTION: A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer exponents. It has

only one term. is a division of two monomials, so it is not a monomial.

21.

SOLUTION: A polynomial is a monomial or the sum of monomials. 21 is a monomial, so it is also a polynomial. The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of 21 is 0. The polynomial has only one term, so it is a monomial.

22. c4 – 2c

2 + 1

SOLUTION:

A polynomial is a monomial or the sum of monomials. c4 – 2c

2 + 1

is the sum of 3 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 4, 2, and

0, so the degree of c4 – 2c2 + 1 is 4. The polynomial has three terms, so it is a trinomial.

23. d + 3dc

SOLUTION: A polynomial is a monomial or the sum of monomials. A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer

exponents. 3dc

has a variable in the exponent, so it is not a monomial.

24. a – a2

SOLUTION:

A polynomial is a monomial or the sum of monomials. a – a2 is the sum of 2 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 1 and 2,

so the degree of a – a2 is 2. The polynomial has two terms, so it is a binomial.

25. 5n3 + nq

3

SOLUTION:

A polynomial is a monomial or the sum of monomials. 5n3 + nq3 is the sum of 2 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 3 and 4,

so the degree of 5n3 + nq3 is 4. The polynomial has two terms, so it is a binomial.

Write each polynomial in standard form. Identify the leading coefficient.

26. 5x2 – 2 + 3x

SOLUTION: Find the degree of each term.

5x2 → 2

– 2 → 0

3x → 1

The greatest degree is 2, from the term 5x2, so the leading coefficient of 5x

2 – 2 + 3x is 5.

Rewrite the polynomial with each monomial in descending order according to degree.

5x2 + 3x – 2

27. 8y + 7y3

SOLUTION: Find the degree of each term.

7y3 → 3

8y → 1

The greatest degree is 3, from the term 7y3, so the leading coefficient of 8y + 7y

3 is 7.

Rewrite the polynomial with each monomial in descending order according to degree.

7y3 + 8y

28. 4 – 3c – 5c2

SOLUTION: Find the degree of each term. 4 → 0 3c → 1

– 5c2 → 2

The greatest degree is 2, from the term – 5c2, so the leading coefficient of 4 – 3c – 5c

2 is –5.

Rewrite the polynomial with each monomial in descending order according to degree.

–5c2 – 3c + 4

29. –y3 + 3y – 3y

2 + 2

SOLUTION: Find the degree of each term.

–y3 → 3

3y → 1

3y2 → 2

2 → 0

The greatest degree is 3, from the term –y3, so the leading coefficient of –y

3 + 3y – 3y

2 + 2 is –1.

Rewrite the polynomial with each monomial in descending order according to degree. –y 3 – 3y 2 + 3y + 2

30. 11t + 2t2 – 3 + t

5

SOLUTION: Find the degree of each term. 11t → 1

2t2 → 2

–3 → 0

t5 → 5

The greatest degree is 5, from the term t5, so the leading coefficient of 11t + 2t2 – 3 + t

5 is 1.

Rewrite the polynomial with each monomial in descending order according to degree.

t5 + 2t

2 + 11t – 3

31. 2 + r – r3

SOLUTION: Find the degree of each term. 2 → 0 r → 1

– r3→ 3

The greatest degree is 3, from the term – r3, so the leading coefficient of 2 + r – r3

is –1. Rewrite the polynomial with each monomial in descending order according to degree.

–r3 + r + 2

32.

SOLUTION: Find the degree of each term.

→ 0

–3x4 → 4

7 → 0

The greatest degree is 4, from the term –3x4, so the leading coefficient of is –3.

Rewrite the polynomial with each monomial in descending order according to degree.

33. –9b2 + 10b – b

6

SOLUTION: Find the degree of each term.

–9b2 → 2

10b → 1

–b6→ 6

The greatest degree is 6, from the term –b6, so the leading coefficient of –9b2 + 10b – b6 is –1. Rewrite the polynomial with each monomial in descending order according to degree.

–b6 – 9b

2 + 10b

Find each sum or difference.

34. (2c2 + 6c + 4) + (5c – 7)

SOLUTION:

35. (2x + 3x2) − (7 − 8x

2)

SOLUTION:

36. (3c3 − c + 11) − (c2 + 2c + 8)

SOLUTION:

37. (z2 + z) + (z

2 − 11)

SOLUTION:

38. (2x − 2y + 1) − (3y + 4x)

SOLUTION:

39. (4a − 5b2 + 3) + (6 − 2a + 3b

2)

SOLUTION:

40. (x2y − 3x

2 + y) + (3y − 2x2y)

SOLUTION:

41. (−8xy + 3x2 − 5y) + (4x

2 − 2y + 6xy)

SOLUTION:

42. (5n − 2p2 + 2np) − (4p

2 + 4n)

SOLUTION:

43. (4rxt − 8r2x + x2

) − (6rx2 + 5rxt − 2x

2)

SOLUTION:

44. PETS From 1999 through 2009, the number of dogs D and the number of cats C (in hundreds) adopted from animal

shelters in the United States are modeled by the equations D = 2n + 3 and C = n + 4, where n is the number of yearssince 1999. a. Write an equation that models the total number T of dogs and cats adopted in hundreds for this time period. b. If this trend continues, how many dogs and cats will be adopted in 2013?

SOLUTION: a.

So, an equation that models the total number of dogs and cats adopted is T = 3n + 7. b. Evaluate the equation for the total number of dogs and cats for n = 2013 – 1999 = 14.

The number of cats and dogs adopted in 2013 will be 49 × 100 or 4900 cats and dogs.

Classify each polynomial according to its degree and number of terms.

45. 4x – 3x2 + 5

SOLUTION:

Find the degree of each term of 4x – 3x2 + 5.

4x → 1

– 3x2 → 2

5 → 0

The greatest degree is 2 and there are 3 terms, so 4x – 3x2 + 5 is a quadratic trinomial.

46. 11z3

SOLUTION: Find the degree of each term. 11z3

→ 3

The greatest degree is 3 and there is one term, so 11z3 is a cubic monomial.

47. 9 + y4

SOLUTION:

Find the degree of each term of 9 + y 4.

9 → 1

y4→ 4

The greatest degree is 4 and there are 2 terms, so 9 + y 4 is a quartic binomial.

48. 3x3 – 7

SOLUTION:

Find the degree of each term of 3x3 – 7.

3x3 → 3

–7 → 0

The greatest degree is 3 and there are 2 terms, so 3x3 – 7 is a cubic binomial.

49. –2x5 – x

2 + 5x – 8

SOLUTION:

Find the degree of each term of –2x5 – x2 + 5x – 8.

–2x5 → 5

–x2 → 2

5x → 1 – 8 → 0

The greatest degree is 5 and there are 4 terms, so –2x5 – x2 + 5x – 8 is a quintic polynomial.

50. 10t – 4t2 + 6t

3

SOLUTION:

Find the degree of each term of 10t – 4t2 + 6t

3.

10t → 1

4t2 → 2

6t3 → 3

The greatest degree is 3 and there are 3 terms, so 10t – 4t2 + 6t

3 is a cubic trinomial.

51. ENROLLMENT In a rapidly growing school system, the numbers (in hundreds) of total students N and

K-5 students P enrolled from 2000 to 2009 are modeled by the equations N = 1.25t2 – t + 7.5 and P =

0.7t2 – 0.95t + 3.8, where t is the number of years since 2000.

a. Write an equation modeling the number of 6-12 students S enrolled for this time period. b. How many 6-12 students were enrolled in the school system in 2007?

SOLUTION: a. To write an equation that represents the number of 6-12 students enrolled, subtract the equations that represent the total number of students and the number of K-5 students.

b. Replace t with 7 in the equation for S to determine the number of students enrolled in 6-12 in 2007.

In 2007 there were 30.3 hundreds or 3030 students enrolled in 6-12.

52. CCSS REASONING The perimeter of the figure shown is represented by the expression 3x2 − 7x + 2. Write a

polynomial that represents the measure of the third side.

SOLUTION:

53. GEOMETRY Consider the rectangle.

a. What does (4x2 + 2x – 1)(2x

2 – x + 3) represent?

b. What does 2(4x2 + 2x – 1) + 2(2x

2 – x + 3) represent?

SOLUTION:

a. (4x2 + 2x – 1)(2x

2 – x + 3) is a multiplication of the length and the width of the rectangle, which is the formula for

the area of a rectangle.

b. 2(4x2 + 2x – 1) + 2(2x

2 – x + 3) is the sum of twice the length and twice the width of a rectangle, which is the

formula for the perimeter of the rectangle

Find each sum or difference.

54. (4x + 2y − 6z) + (5y − 2z + 7x) + (−9z − 2x − 3y)

SOLUTION:

55. (5a2 − 4) + (a

2 − 2a + 12) + (4a2 − 6a + 8)

SOLUTION:

56. (3c2 − 7) + (4c + 7) − (c

2 + 5c − 8)

SOLUTION:

57. (3n3 + 3n − 10) − (4n

2 − 5n) + (4n3 − 3n

2 − 9n + 4)

SOLUTION:

58. FOOTBALL The National Football League is divided into two conferences, the American A and the National N. From 2002 through 2009, the total attendance T (in thousands) for both conferences and for the American Conference games are modeled by the following equations, where x is the number of years since 2002.

T = –0.69x3 + 55.83x

2 + 643.31x + 10,538

A = –3.78x3 + 58.96x

2 + 265.96x + 5257

Determine how many people attended National Conference football games in 2009.

SOLUTION:

Let x = 7 represent 2009, then find how many people attended a National Conference football game in 2009.

In 2009 the number of people who attended National Conference football games was about 8829 thousand. Multiply by 1000 to find the attendance in standard form. 8829 ×1000 = 8,829,000. So, about 8,829,000 people attended National Conference football games in 2009.

59. CAR RENTAL The cost to rent a car for a day is $15 plus $0.15 for each mile driven. a. Write a polynomial that represents the cost of renting a car for m miles. b. If a car is driven 145 miles, how much would it cost to rent? c. If a car is driven 105 miles each day for four days, how much would it cost to rent a car? d. If a car is driven 220 miles each day for seven days, how much would it cost to rent a car?

SOLUTION: a. The cost to rent a car is the daily rate and the mileage cost or 15 + 0.15m. b. Substitute 145 for m to find the daily cost to drive 145 miles.

The cost to rent the car would be $36.75. c. The expression represents the cost per day. For 4 days, multiply the entire expression by 4 to find the cost for a 4 day trip. Substitute 105 for m.

The cost to rent the car would be $123. d. The expression represents the cost per day. For 7 days, multiply the entire expression by 7 to find the cost for a 7 day trip. Substitute 220 for m.

The cost to rent the car would be $336.

60. MULTIPLE REPRESENTATIONS In this problem, you will explore perimeter and area. a. Geometric Draw three rectangles that each have a perimeter of 400 feet. b. Tabular Record the width and length of each rectangle in a table like the one shown below. Find the area of eachrectangle.

c. Graphical On a coordinate system, graph the area of rectangle 4 in terms of the length, x. Use the graph to determine the largest area possible. d. Analytical Determine the length and width that produce the largest area.

SOLUTION: a. For the perimeters to be 400, create lengths and widths that sum to 200. Note that some lengths are already provided in the table in part b.

b. The area is length multiplied by width. Make sure the units are squared for area. The sum of the length and width must be 200, so if the length is x, the width must be 200 – x.

c. The length will be the x-values on the horizontal and the area will be the y-values on the vertical. The length cannot pass 200 since the sum of the length and width is 200. Set the intervals for the x-axis to 25 feet. Extend the table of values to find more points to plot on the graph. It appears that 10,000 is the greatest area, so set the intervals

for the y-axis to 1000 ft2. After graphing, it appears that the highest point on the graph is at an area of 10,000 ft

2.

d. The associated x-value with the maximum area is x = 100, so the length must be 100 and the width must be 200 – 100, or 100. The length and width of the rectangle must be 100 feet each to have the largest area.

61. CCSS CRITIQUE Cheyenne and Sebastian are finding (2x2 − x) − (3x + 3x

2 − 2). Is either of them correct? Explain your reasoning.

SOLUTION:

Neither is correct. Cheyenne, did not distribute the negative to the 2nd and 3rd terms when she found the additive inverse. Sebastian did not distribute the negate to the 3rd terms when he found the additive inverse. To find the additive inverse, all terms should be multiplied by −1.

62. REASONING Determine whether each of the following statements is true or false . Explain your reasoning. a. A binomial can have a degree of zero. b. The order in which polynomials are subtracted does not matter.

SOLUTION: a. If a binomial has two terms that are each a degree of 0, then those terms can be combined and the binomial becomes a monomial. For example, 18 + 7 = 25. If one of the terms of the binomial does not have a degree of 0, then the binomial cannot have a degree of 0, since the degree of a polynomial is the greatest degree of any term in the polynomial. b. Subtraction is not commutative. While 2 + 5 = 5 + 2, 2 – 5 ≠ 5 – 2. This is also true for polynomials. Sample answer: (2x – 3) – (4x – 3) = –2x, but (4x – 3) – (2x – 3) = 2x

63. CHALLENGE Write a polynomial that represents the sum of an odd integer 2n + 1 and the next two consecutive odd integers.

SOLUTION:

64. WRITING IN MATH Why would you add or subtract equations that represent real-world situations? Explain.

SOLUTION:

65. WRITING IN MATH Describe how to add and subtract polynomials using both the vertical and horizontal formats.

SOLUTION: To add polynomials in a horizontal format, you combine like terms. For the vertical format, you write the polynomials in standard form, align like terms in columns, and combine like terms.

To subtract polynomials in a horizontal format you find the additive inverse of the polynomial you are subtracting, andthen combine like terms. For the vertical format, you write the polynomials in standard form, align like terms in columns, and subtract by adding the additive inverse.

66. Three consecutive integers can be represented by x, x + 1, and x + 2. What is the sum of these three integers? A x(x + 1)(x + 2)

B x3 + 3

C 3x + 3 D x + 3

SOLUTION:

The correct choice is C.

67. SHORT RESPONSE What is the perimeter of a square with sides that measure 2x + 3 units?

SOLUTION:

The perimeter of the square is 8x + 12 units.

68. Jim cuts a board in the shape of a regular hexagon and pounds in a nail at each vertex, as shown. How many rubber bands will he need to stretch a rubber band across every possible pair of nails?

F 15 G 14 H 12 J 9

SOLUTION: The first nail would connect to 5 others, the second to 4 others, the third to 3 others, etc.5 + 4 + 3 + 2 + 1 = 15 The correct choice is F.

69. Which ordered pair is in the solution set of the system of inequalities shown in the graph?

A (−3, 0) B (0, −3) C (5, 0) D (0, 5)

SOLUTION: Choice A is outside the shaded area for both inequalities. Choices B and D are inside the shaded area for only one inequality. Choice C is the only point in the solution for both inequalities. So, the correct choice is C.

70. COMPUTERS A computer technician charges by the hour to fix and repair computer equipment. The total cost of the technician for one hour is $75, for two hours is $125, for three hours is $175, for four hours is $225, and so on. Write a recursive formula for the sequence.

SOLUTION: Write out the terms. $75, $125, $175, $225, ... The first term is 75, and 50 is added to form each following term. Therefore, we have a1 = 75, an = an – 1 + 50, n ≥ 2.

Determine whether each sequence is arithmetic, geometric, or neither. Explain.71. 8, –32, 128, –512, ...

SOLUTION: Check for a common difference. –32 – 8 = –40 128 – (–32) = 160 There is no common difference. Check for a common ratio. –32 ÷ 8 = –4 128 ÷ (–32) = –4 Geometric; the common ratio is –4.

72. 25, 8, –9, –26, ...

SOLUTION: Check for a common difference. 8 – 25 = –17 –9 – 8 = –17 Arithmetic; the common difference is –17.

73.

SOLUTION: Check for a common difference.

There is no common difference. Check for a common ratio.

There is no common ratio, so the sequence is not arithmetic or geometric.

74. 43, 52, 61, 70, ...

SOLUTION: Check for a common difference. 52 – 43 = 9 61 – 52 = 9 Arithmetic; the common difference is 9.

75. –27, –16, –5, 6, ...

SOLUTION: Check for a common difference. –16 – (–27) = 11 –5 – (–16) = 11 Arithmetic; the common difference is 11.

76. 200, 100, 50, 25, …

SOLUTION: Check for a common difference. 100 – 200 = –100 50 – 100 = –50 There is no common difference. Check for a common ratio. 100 ÷ 200 = 0.5 50 ÷ 100 = 0.5

Geometric; the common ratio is 0.5 or .

77. JOBS Kimi received an offer for a new job. She wants to compare the offer with her current job. What is total amount of sales that Kimi must get each month to make the same income at either job?

SOLUTION: Let s be Kimi’s monthly sales.

Kimi must sell $80,000 each month to make the same income at either job.

Determine whether each sequence is an arithmetic sequence. If it is, state the common difference.78. 24, 16, 8, 0, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is –8, because 16 – 24 = –8; 8 –16 = –8; etc.

79. , 13, 26, …

SOLUTION: Find the difference between the terms.

– =

13 – =

26 – 13 = 13 There is not a common different. The sequence is not an arithmetic sequence

80. 7, 6, 5, 4, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is –1, because 6 – 7 = –1; 5 – 6 = –1; 4 – 5 = –1;etc.

81. 10, 12, 15, 18, …

SOLUTION: Find the difference between the terms. 12 – 10 = 2 15 – 12 = 3 18 – 15 = 3 There is no common difference. The sequence is not an arithmetic sequence.

82. −15, −11, −7, −3, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is 4, because –11 – (–15) = 4; –7 – (–11) = 4; –3 – (–7) = 4;etc.

83. −0.3, 0.2, 0.7, 1.2, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is 0.5, because 0.2 – (–0.3) = 0.5; 0.7 – 0.2 = 0.5; 1.2 – 0.7 = 0.5; etc.

Simplify.

84. t(t5)(t

7)

SOLUTION:

85. n3(n

2)(−2n

3)

SOLUTION:

86. (5t5v

2)(10t

3v

4)

SOLUTION:

87. (−8u4z

5)(5uz

4)

SOLUTION:

88. [(3)2]3

SOLUTION:

89. [(2)3]2

SOLUTION:

90. (2m4k

3)2(−3mk

2)3

SOLUTION:

91. (6xy2)2(2x

2y

2z

2)3

SOLUTION:

eSolutions Manual - Powered by Cognero Page 5

8-1 Adding and Subtracting Polynomials

Determine whether each expression is a polynomial. If it is a polynomial, find the degree and determine whether it is a monomial, binomial, or trinomial.

1. 7ab + 6b2 – 2a

3

SOLUTION:

A polynomial is a monomial or the sum of monomials. 7ab + 6b2 – 2a

3 is the sum of 3 monomials, so it is a

polynomial. The degree of a polynomial is the greatest degree of any term in the polynomial.The degree of each term is 2, 2, and

3, so the degree of 7ab + 6b2 – 2a

3 is 3. The polynomial has three terms, so it is a trinomial.

2. 2y – 5 + 3y2

SOLUTION:

2y – 5 + 3y2

is the sum of monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 1, 0, and

2, so the degree of 2y – 5 + 3y2 is 2. The polynomial has three terms, so it is a trinomial.

3. 3x2

SOLUTION:

A polynomial is a monomial or the sum of monomials, so 3x2 is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of 3x2 is 2. The

polynomial has one term, so it is a monomial.

4.

SOLUTION: A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer exponents. It has

only one term. is a division of two monomials, so it is not a monomial.

5. 5m2p

3 + 6

SOLUTION:

A polynomial is a monomial or the sum of monomials. 5m2p

3 + 6

is the sum of 2 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 5 and 0,

so the degree of 5m2p

3 + 6 is 5. The polynomial has two terms, so it is a binomial.

6. 5q–4

+ 6q

SOLUTION: A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer exponents. It has

only one term. 5q-4 is equal to , which is a division of two monomials, so it is not a monomial.

Write each polynomial in standard form. Identify the leading coefficient.

7. –4d4 + 1 – d

2

SOLUTION: Find the degree of each term.

–4d4→ 4

1 → 0

–d2 → 2

The greatest degree is 4, from the term –4d4, so the leading coefficient of –4d

4 + 1 – d

2 is –4.

Rewrite the polynomial with each monomial in descending order according to degree.

–4d4 – d

2 + 1

8. 2x5 – 12 + 3x

SOLUTION: Find the degree of each term.

2x5 → 5

–12 → 0 3x → 1

The greatest degree is 5, from the term 2x5, so the leading coefficient of 2x

5 – 12 + 3x is 2.

Rewrite the polynomial with each monomial in descending order according to degree.

2x5 + 3x – 12

9. 4z – 2z2 – 5z

4

SOLUTION: Find the degree of each term. 4z → 1

– 2z2 → 2

– 5z4→ 4

The greatest degree is 4, from the term – 5z4, so the leading coefficient of 4z – 2z

2 – 5z

4 is –5.

Rewrite the polynomial with each monomial in descending order according to degree.

–5z4 – 2z

2 + 4z

10. 2a + 4a3 – 5a

2 – 1

SOLUTION: Find the degree of each term. 2a → 1

4a3→ 3

– 5a2 → 2

– 1 → 0

The greatest degree is 3, from the term 4a3, so the leading coefficient of 2a + 4a

3 – 5a

2 – 1 is 4.

Rewrite the polynomial with each monomial in descending order according to degree.

4a3– 5a

2 + 2a – 1

Find each sum or difference.

11. (6x3 − 4) + (−2x

3 + 9)

SOLUTION:

12. (g3 − 2g2 + 5g + 6) − (g2 + 2g)

SOLUTION:

13. (4 + 2a2 − 2a) − (3a

2 − 8a + 7)

SOLUTION:

14. (8y − 4y2) + (3y − 9y

2)

SOLUTION:

15. (−4z3 − 2z + 8) − (4z

3 + 3z2 − 5)

SOLUTION:

16. (−3d2 − 8 + 2d) + (4d − 12 + d2

)

SOLUTION:

17. (y + 5) + (2y + 4y2 – 2)

SOLUTION:

18. (3n3 − 5n + n2

) − (−8n2 + 3n

3)

SOLUTION:

19. CCSS SENSE-MAKING The total number of students T who traveled for spring break consists of two groups: students who flew to their destinations F and students who drove to their destination D. The number (in thousands) of students who flew and the total number of students who flew or drove can be modeled by the following equations,where n is the number of years since 1995. T = 14n + 21

F = 8n + 7 a. Write an equation that models the number of students who drove to their destination for this time period. b. Predict the number of students who will drive to their destination in 2012. c. How many students will drive or fly to their destination in 2015?

SOLUTION: a.

D = 6n + 14 b. n = 2012 – 1995 = 17

The number of students who will drive to their destination in 2012 is 116,000 students. c. n = 2015 – 1995 = 20

The number of students who will drive or fly to their destination in 2015 is 301,000 students.

Determine whether each expression is a polynomial. If it is a polynomial, find the degree and determine whether it is a monomial, binomial, or trinomial.

20.

SOLUTION: A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer exponents. It has

only one term. is a division of two monomials, so it is not a monomial.

21.

SOLUTION: A polynomial is a monomial or the sum of monomials. 21 is a monomial, so it is also a polynomial. The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of 21 is 0. The polynomial has only one term, so it is a monomial.

22. c4 – 2c

2 + 1

SOLUTION:

A polynomial is a monomial or the sum of monomials. c4 – 2c

2 + 1

is the sum of 3 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 4, 2, and

0, so the degree of c4 – 2c2 + 1 is 4. The polynomial has three terms, so it is a trinomial.

23. d + 3dc

SOLUTION: A polynomial is a monomial or the sum of monomials. A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer

exponents. 3dc

has a variable in the exponent, so it is not a monomial.

24. a – a2

SOLUTION:

A polynomial is a monomial or the sum of monomials. a – a2 is the sum of 2 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 1 and 2,

so the degree of a – a2 is 2. The polynomial has two terms, so it is a binomial.

25. 5n3 + nq

3

SOLUTION:

A polynomial is a monomial or the sum of monomials. 5n3 + nq3 is the sum of 2 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 3 and 4,

so the degree of 5n3 + nq3 is 4. The polynomial has two terms, so it is a binomial.

Write each polynomial in standard form. Identify the leading coefficient.

26. 5x2 – 2 + 3x

SOLUTION: Find the degree of each term.

5x2 → 2

– 2 → 0

3x → 1

The greatest degree is 2, from the term 5x2, so the leading coefficient of 5x

2 – 2 + 3x is 5.

Rewrite the polynomial with each monomial in descending order according to degree.

5x2 + 3x – 2

27. 8y + 7y3

SOLUTION: Find the degree of each term.

7y3 → 3

8y → 1

The greatest degree is 3, from the term 7y3, so the leading coefficient of 8y + 7y

3 is 7.

Rewrite the polynomial with each monomial in descending order according to degree.

7y3 + 8y

28. 4 – 3c – 5c2

SOLUTION: Find the degree of each term. 4 → 0 3c → 1

– 5c2 → 2

The greatest degree is 2, from the term – 5c2, so the leading coefficient of 4 – 3c – 5c

2 is –5.

Rewrite the polynomial with each monomial in descending order according to degree.

–5c2 – 3c + 4

29. –y3 + 3y – 3y

2 + 2

SOLUTION: Find the degree of each term.

–y3 → 3

3y → 1

3y2 → 2

2 → 0

The greatest degree is 3, from the term –y3, so the leading coefficient of –y

3 + 3y – 3y

2 + 2 is –1.

Rewrite the polynomial with each monomial in descending order according to degree. –y 3 – 3y 2 + 3y + 2

30. 11t + 2t2 – 3 + t

5

SOLUTION: Find the degree of each term. 11t → 1

2t2 → 2

–3 → 0

t5 → 5

The greatest degree is 5, from the term t5, so the leading coefficient of 11t + 2t2 – 3 + t

5 is 1.

Rewrite the polynomial with each monomial in descending order according to degree.

t5 + 2t

2 + 11t – 3

31. 2 + r – r3

SOLUTION: Find the degree of each term. 2 → 0 r → 1

– r3→ 3

The greatest degree is 3, from the term – r3, so the leading coefficient of 2 + r – r3

is –1. Rewrite the polynomial with each monomial in descending order according to degree.

–r3 + r + 2

32.

SOLUTION: Find the degree of each term.

→ 0

–3x4 → 4

7 → 0

The greatest degree is 4, from the term –3x4, so the leading coefficient of is –3.

Rewrite the polynomial with each monomial in descending order according to degree.

33. –9b2 + 10b – b

6

SOLUTION: Find the degree of each term.

–9b2 → 2

10b → 1

–b6→ 6

The greatest degree is 6, from the term –b6, so the leading coefficient of –9b2 + 10b – b6 is –1. Rewrite the polynomial with each monomial in descending order according to degree.

–b6 – 9b

2 + 10b

Find each sum or difference.

34. (2c2 + 6c + 4) + (5c – 7)

SOLUTION:

35. (2x + 3x2) − (7 − 8x

2)

SOLUTION:

36. (3c3 − c + 11) − (c2 + 2c + 8)

SOLUTION:

37. (z2 + z) + (z

2 − 11)

SOLUTION:

38. (2x − 2y + 1) − (3y + 4x)

SOLUTION:

39. (4a − 5b2 + 3) + (6 − 2a + 3b

2)

SOLUTION:

40. (x2y − 3x

2 + y) + (3y − 2x2y)

SOLUTION:

41. (−8xy + 3x2 − 5y) + (4x

2 − 2y + 6xy)

SOLUTION:

42. (5n − 2p2 + 2np) − (4p

2 + 4n)

SOLUTION:

43. (4rxt − 8r2x + x2

) − (6rx2 + 5rxt − 2x

2)

SOLUTION:

44. PETS From 1999 through 2009, the number of dogs D and the number of cats C (in hundreds) adopted from animal

shelters in the United States are modeled by the equations D = 2n + 3 and C = n + 4, where n is the number of yearssince 1999. a. Write an equation that models the total number T of dogs and cats adopted in hundreds for this time period. b. If this trend continues, how many dogs and cats will be adopted in 2013?

SOLUTION: a.

So, an equation that models the total number of dogs and cats adopted is T = 3n + 7. b. Evaluate the equation for the total number of dogs and cats for n = 2013 – 1999 = 14.

The number of cats and dogs adopted in 2013 will be 49 × 100 or 4900 cats and dogs.

Classify each polynomial according to its degree and number of terms.

45. 4x – 3x2 + 5

SOLUTION:

Find the degree of each term of 4x – 3x2 + 5.

4x → 1

– 3x2 → 2

5 → 0

The greatest degree is 2 and there are 3 terms, so 4x – 3x2 + 5 is a quadratic trinomial.

46. 11z3

SOLUTION: Find the degree of each term. 11z3

→ 3

The greatest degree is 3 and there is one term, so 11z3 is a cubic monomial.

47. 9 + y4

SOLUTION:

Find the degree of each term of 9 + y 4.

9 → 1

y4→ 4

The greatest degree is 4 and there are 2 terms, so 9 + y 4 is a quartic binomial.

48. 3x3 – 7

SOLUTION:

Find the degree of each term of 3x3 – 7.

3x3 → 3

–7 → 0

The greatest degree is 3 and there are 2 terms, so 3x3 – 7 is a cubic binomial.

49. –2x5 – x

2 + 5x – 8

SOLUTION:

Find the degree of each term of –2x5 – x2 + 5x – 8.

–2x5 → 5

–x2 → 2

5x → 1 – 8 → 0

The greatest degree is 5 and there are 4 terms, so –2x5 – x2 + 5x – 8 is a quintic polynomial.

50. 10t – 4t2 + 6t

3

SOLUTION:

Find the degree of each term of 10t – 4t2 + 6t

3.

10t → 1

4t2 → 2

6t3 → 3

The greatest degree is 3 and there are 3 terms, so 10t – 4t2 + 6t

3 is a cubic trinomial.

51. ENROLLMENT In a rapidly growing school system, the numbers (in hundreds) of total students N and

K-5 students P enrolled from 2000 to 2009 are modeled by the equations N = 1.25t2 – t + 7.5 and P =

0.7t2 – 0.95t + 3.8, where t is the number of years since 2000.

a. Write an equation modeling the number of 6-12 students S enrolled for this time period. b. How many 6-12 students were enrolled in the school system in 2007?

SOLUTION: a. To write an equation that represents the number of 6-12 students enrolled, subtract the equations that represent the total number of students and the number of K-5 students.

b. Replace t with 7 in the equation for S to determine the number of students enrolled in 6-12 in 2007.

In 2007 there were 30.3 hundreds or 3030 students enrolled in 6-12.

52. CCSS REASONING The perimeter of the figure shown is represented by the expression 3x2 − 7x + 2. Write a

polynomial that represents the measure of the third side.

SOLUTION:

53. GEOMETRY Consider the rectangle.

a. What does (4x2 + 2x – 1)(2x

2 – x + 3) represent?

b. What does 2(4x2 + 2x – 1) + 2(2x

2 – x + 3) represent?

SOLUTION:

a. (4x2 + 2x – 1)(2x

2 – x + 3) is a multiplication of the length and the width of the rectangle, which is the formula for

the area of a rectangle.

b. 2(4x2 + 2x – 1) + 2(2x

2 – x + 3) is the sum of twice the length and twice the width of a rectangle, which is the

formula for the perimeter of the rectangle

Find each sum or difference.

54. (4x + 2y − 6z) + (5y − 2z + 7x) + (−9z − 2x − 3y)

SOLUTION:

55. (5a2 − 4) + (a

2 − 2a + 12) + (4a2 − 6a + 8)

SOLUTION:

56. (3c2 − 7) + (4c + 7) − (c

2 + 5c − 8)

SOLUTION:

57. (3n3 + 3n − 10) − (4n

2 − 5n) + (4n3 − 3n

2 − 9n + 4)

SOLUTION:

58. FOOTBALL The National Football League is divided into two conferences, the American A and the National N. From 2002 through 2009, the total attendance T (in thousands) for both conferences and for the American Conference games are modeled by the following equations, where x is the number of years since 2002.

T = –0.69x3 + 55.83x

2 + 643.31x + 10,538

A = –3.78x3 + 58.96x

2 + 265.96x + 5257

Determine how many people attended National Conference football games in 2009.

SOLUTION:

Let x = 7 represent 2009, then find how many people attended a National Conference football game in 2009.

In 2009 the number of people who attended National Conference football games was about 8829 thousand. Multiply by 1000 to find the attendance in standard form. 8829 ×1000 = 8,829,000. So, about 8,829,000 people attended National Conference football games in 2009.

59. CAR RENTAL The cost to rent a car for a day is $15 plus $0.15 for each mile driven. a. Write a polynomial that represents the cost of renting a car for m miles. b. If a car is driven 145 miles, how much would it cost to rent? c. If a car is driven 105 miles each day for four days, how much would it cost to rent a car? d. If a car is driven 220 miles each day for seven days, how much would it cost to rent a car?

SOLUTION: a. The cost to rent a car is the daily rate and the mileage cost or 15 + 0.15m. b. Substitute 145 for m to find the daily cost to drive 145 miles.

The cost to rent the car would be $36.75. c. The expression represents the cost per day. For 4 days, multiply the entire expression by 4 to find the cost for a 4 day trip. Substitute 105 for m.

The cost to rent the car would be $123. d. The expression represents the cost per day. For 7 days, multiply the entire expression by 7 to find the cost for a 7 day trip. Substitute 220 for m.

The cost to rent the car would be $336.

60. MULTIPLE REPRESENTATIONS In this problem, you will explore perimeter and area. a. Geometric Draw three rectangles that each have a perimeter of 400 feet. b. Tabular Record the width and length of each rectangle in a table like the one shown below. Find the area of eachrectangle.

c. Graphical On a coordinate system, graph the area of rectangle 4 in terms of the length, x. Use the graph to determine the largest area possible. d. Analytical Determine the length and width that produce the largest area.

SOLUTION: a. For the perimeters to be 400, create lengths and widths that sum to 200. Note that some lengths are already provided in the table in part b.

b. The area is length multiplied by width. Make sure the units are squared for area. The sum of the length and width must be 200, so if the length is x, the width must be 200 – x.

c. The length will be the x-values on the horizontal and the area will be the y-values on the vertical. The length cannot pass 200 since the sum of the length and width is 200. Set the intervals for the x-axis to 25 feet. Extend the table of values to find more points to plot on the graph. It appears that 10,000 is the greatest area, so set the intervals

for the y-axis to 1000 ft2. After graphing, it appears that the highest point on the graph is at an area of 10,000 ft

2.

d. The associated x-value with the maximum area is x = 100, so the length must be 100 and the width must be 200 – 100, or 100. The length and width of the rectangle must be 100 feet each to have the largest area.

61. CCSS CRITIQUE Cheyenne and Sebastian are finding (2x2 − x) − (3x + 3x

2 − 2). Is either of them correct? Explain your reasoning.

SOLUTION:

Neither is correct. Cheyenne, did not distribute the negative to the 2nd and 3rd terms when she found the additive inverse. Sebastian did not distribute the negate to the 3rd terms when he found the additive inverse. To find the additive inverse, all terms should be multiplied by −1.

62. REASONING Determine whether each of the following statements is true or false . Explain your reasoning. a. A binomial can have a degree of zero. b. The order in which polynomials are subtracted does not matter.

SOLUTION: a. If a binomial has two terms that are each a degree of 0, then those terms can be combined and the binomial becomes a monomial. For example, 18 + 7 = 25. If one of the terms of the binomial does not have a degree of 0, then the binomial cannot have a degree of 0, since the degree of a polynomial is the greatest degree of any term in the polynomial. b. Subtraction is not commutative. While 2 + 5 = 5 + 2, 2 – 5 ≠ 5 – 2. This is also true for polynomials. Sample answer: (2x – 3) – (4x – 3) = –2x, but (4x – 3) – (2x – 3) = 2x

63. CHALLENGE Write a polynomial that represents the sum of an odd integer 2n + 1 and the next two consecutive odd integers.

SOLUTION:

64. WRITING IN MATH Why would you add or subtract equations that represent real-world situations? Explain.

SOLUTION:

65. WRITING IN MATH Describe how to add and subtract polynomials using both the vertical and horizontal formats.

SOLUTION: To add polynomials in a horizontal format, you combine like terms. For the vertical format, you write the polynomials in standard form, align like terms in columns, and combine like terms.

To subtract polynomials in a horizontal format you find the additive inverse of the polynomial you are subtracting, andthen combine like terms. For the vertical format, you write the polynomials in standard form, align like terms in columns, and subtract by adding the additive inverse.

66. Three consecutive integers can be represented by x, x + 1, and x + 2. What is the sum of these three integers? A x(x + 1)(x + 2)

B x3 + 3

C 3x + 3 D x + 3

SOLUTION:

The correct choice is C.

67. SHORT RESPONSE What is the perimeter of a square with sides that measure 2x + 3 units?

SOLUTION:

The perimeter of the square is 8x + 12 units.

68. Jim cuts a board in the shape of a regular hexagon and pounds in a nail at each vertex, as shown. How many rubber bands will he need to stretch a rubber band across every possible pair of nails?

F 15 G 14 H 12 J 9

SOLUTION: The first nail would connect to 5 others, the second to 4 others, the third to 3 others, etc.5 + 4 + 3 + 2 + 1 = 15 The correct choice is F.

69. Which ordered pair is in the solution set of the system of inequalities shown in the graph?

A (−3, 0) B (0, −3) C (5, 0) D (0, 5)

SOLUTION: Choice A is outside the shaded area for both inequalities. Choices B and D are inside the shaded area for only one inequality. Choice C is the only point in the solution for both inequalities. So, the correct choice is C.

70. COMPUTERS A computer technician charges by the hour to fix and repair computer equipment. The total cost of the technician for one hour is $75, for two hours is $125, for three hours is $175, for four hours is $225, and so on. Write a recursive formula for the sequence.

SOLUTION: Write out the terms. $75, $125, $175, $225, ... The first term is 75, and 50 is added to form each following term. Therefore, we have a1 = 75, an = an – 1 + 50, n ≥ 2.

Determine whether each sequence is arithmetic, geometric, or neither. Explain.71. 8, –32, 128, –512, ...

SOLUTION: Check for a common difference. –32 – 8 = –40 128 – (–32) = 160 There is no common difference. Check for a common ratio. –32 ÷ 8 = –4 128 ÷ (–32) = –4 Geometric; the common ratio is –4.

72. 25, 8, –9, –26, ...

SOLUTION: Check for a common difference. 8 – 25 = –17 –9 – 8 = –17 Arithmetic; the common difference is –17.

73.

SOLUTION: Check for a common difference.

There is no common difference. Check for a common ratio.

There is no common ratio, so the sequence is not arithmetic or geometric.

74. 43, 52, 61, 70, ...

SOLUTION: Check for a common difference. 52 – 43 = 9 61 – 52 = 9 Arithmetic; the common difference is 9.

75. –27, –16, –5, 6, ...

SOLUTION: Check for a common difference. –16 – (–27) = 11 –5 – (–16) = 11 Arithmetic; the common difference is 11.

76. 200, 100, 50, 25, …

SOLUTION: Check for a common difference. 100 – 200 = –100 50 – 100 = –50 There is no common difference. Check for a common ratio. 100 ÷ 200 = 0.5 50 ÷ 100 = 0.5

Geometric; the common ratio is 0.5 or .

77. JOBS Kimi received an offer for a new job. She wants to compare the offer with her current job. What is total amount of sales that Kimi must get each month to make the same income at either job?

SOLUTION: Let s be Kimi’s monthly sales.

Kimi must sell $80,000 each month to make the same income at either job.

Determine whether each sequence is an arithmetic sequence. If it is, state the common difference.78. 24, 16, 8, 0, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is –8, because 16 – 24 = –8; 8 –16 = –8; etc.

79. , 13, 26, …

SOLUTION: Find the difference between the terms.

– =

13 – =

26 – 13 = 13 There is not a common different. The sequence is not an arithmetic sequence

80. 7, 6, 5, 4, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is –1, because 6 – 7 = –1; 5 – 6 = –1; 4 – 5 = –1;etc.

81. 10, 12, 15, 18, …

SOLUTION: Find the difference between the terms. 12 – 10 = 2 15 – 12 = 3 18 – 15 = 3 There is no common difference. The sequence is not an arithmetic sequence.

82. −15, −11, −7, −3, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is 4, because –11 – (–15) = 4; –7 – (–11) = 4; –3 – (–7) = 4;etc.

83. −0.3, 0.2, 0.7, 1.2, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is 0.5, because 0.2 – (–0.3) = 0.5; 0.7 – 0.2 = 0.5; 1.2 – 0.7 = 0.5; etc.

Simplify.

84. t(t5)(t

7)

SOLUTION:

85. n3(n

2)(−2n

3)

SOLUTION:

86. (5t5v

2)(10t

3v

4)

SOLUTION:

87. (−8u4z

5)(5uz

4)

SOLUTION:

88. [(3)2]3

SOLUTION:

89. [(2)3]2

SOLUTION:

90. (2m4k

3)2(−3mk

2)3

SOLUTION:

91. (6xy2)2(2x

2y

2z

2)3

SOLUTION:

eSolutions Manual - Powered by Cognero Page 6

8-1 Adding and Subtracting Polynomials

Determine whether each expression is a polynomial. If it is a polynomial, find the degree and determine whether it is a monomial, binomial, or trinomial.

1. 7ab + 6b2 – 2a

3

SOLUTION:

A polynomial is a monomial or the sum of monomials. 7ab + 6b2 – 2a

3 is the sum of 3 monomials, so it is a

polynomial. The degree of a polynomial is the greatest degree of any term in the polynomial.The degree of each term is 2, 2, and

3, so the degree of 7ab + 6b2 – 2a

3 is 3. The polynomial has three terms, so it is a trinomial.

2. 2y – 5 + 3y2

SOLUTION:

2y – 5 + 3y2

is the sum of monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 1, 0, and

2, so the degree of 2y – 5 + 3y2 is 2. The polynomial has three terms, so it is a trinomial.

3. 3x2

SOLUTION:

A polynomial is a monomial or the sum of monomials, so 3x2 is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of 3x2 is 2. The

polynomial has one term, so it is a monomial.

4.

SOLUTION: A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer exponents. It has

only one term. is a division of two monomials, so it is not a monomial.

5. 5m2p

3 + 6

SOLUTION:

A polynomial is a monomial or the sum of monomials. 5m2p

3 + 6

is the sum of 2 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 5 and 0,

so the degree of 5m2p

3 + 6 is 5. The polynomial has two terms, so it is a binomial.

6. 5q–4

+ 6q

SOLUTION: A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer exponents. It has

only one term. 5q-4 is equal to , which is a division of two monomials, so it is not a monomial.

Write each polynomial in standard form. Identify the leading coefficient.

7. –4d4 + 1 – d

2

SOLUTION: Find the degree of each term.

–4d4→ 4

1 → 0

–d2 → 2

The greatest degree is 4, from the term –4d4, so the leading coefficient of –4d

4 + 1 – d

2 is –4.

Rewrite the polynomial with each monomial in descending order according to degree.

–4d4 – d

2 + 1

8. 2x5 – 12 + 3x

SOLUTION: Find the degree of each term.

2x5 → 5

–12 → 0 3x → 1

The greatest degree is 5, from the term 2x5, so the leading coefficient of 2x

5 – 12 + 3x is 2.

Rewrite the polynomial with each monomial in descending order according to degree.

2x5 + 3x – 12

9. 4z – 2z2 – 5z

4

SOLUTION: Find the degree of each term. 4z → 1

– 2z2 → 2

– 5z4→ 4

The greatest degree is 4, from the term – 5z4, so the leading coefficient of 4z – 2z

2 – 5z

4 is –5.

Rewrite the polynomial with each monomial in descending order according to degree.

–5z4 – 2z

2 + 4z

10. 2a + 4a3 – 5a

2 – 1

SOLUTION: Find the degree of each term. 2a → 1

4a3→ 3

– 5a2 → 2

– 1 → 0

The greatest degree is 3, from the term 4a3, so the leading coefficient of 2a + 4a

3 – 5a

2 – 1 is 4.

Rewrite the polynomial with each monomial in descending order according to degree.

4a3– 5a

2 + 2a – 1

Find each sum or difference.

11. (6x3 − 4) + (−2x

3 + 9)

SOLUTION:

12. (g3 − 2g2 + 5g + 6) − (g2 + 2g)

SOLUTION:

13. (4 + 2a2 − 2a) − (3a

2 − 8a + 7)

SOLUTION:

14. (8y − 4y2) + (3y − 9y

2)

SOLUTION:

15. (−4z3 − 2z + 8) − (4z

3 + 3z2 − 5)

SOLUTION:

16. (−3d2 − 8 + 2d) + (4d − 12 + d2

)

SOLUTION:

17. (y + 5) + (2y + 4y2 – 2)

SOLUTION:

18. (3n3 − 5n + n2

) − (−8n2 + 3n

3)

SOLUTION:

19. CCSS SENSE-MAKING The total number of students T who traveled for spring break consists of two groups: students who flew to their destinations F and students who drove to their destination D. The number (in thousands) of students who flew and the total number of students who flew or drove can be modeled by the following equations,where n is the number of years since 1995. T = 14n + 21

F = 8n + 7 a. Write an equation that models the number of students who drove to their destination for this time period. b. Predict the number of students who will drive to their destination in 2012. c. How many students will drive or fly to their destination in 2015?

SOLUTION: a.

D = 6n + 14 b. n = 2012 – 1995 = 17

The number of students who will drive to their destination in 2012 is 116,000 students. c. n = 2015 – 1995 = 20

The number of students who will drive or fly to their destination in 2015 is 301,000 students.

Determine whether each expression is a polynomial. If it is a polynomial, find the degree and determine whether it is a monomial, binomial, or trinomial.

20.

SOLUTION: A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer exponents. It has

only one term. is a division of two monomials, so it is not a monomial.

21.

SOLUTION: A polynomial is a monomial or the sum of monomials. 21 is a monomial, so it is also a polynomial. The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of 21 is 0. The polynomial has only one term, so it is a monomial.

22. c4 – 2c

2 + 1

SOLUTION:

A polynomial is a monomial or the sum of monomials. c4 – 2c

2 + 1

is the sum of 3 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 4, 2, and

0, so the degree of c4 – 2c2 + 1 is 4. The polynomial has three terms, so it is a trinomial.

23. d + 3dc

SOLUTION: A polynomial is a monomial or the sum of monomials. A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer

exponents. 3dc

has a variable in the exponent, so it is not a monomial.

24. a – a2

SOLUTION:

A polynomial is a monomial or the sum of monomials. a – a2 is the sum of 2 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 1 and 2,

so the degree of a – a2 is 2. The polynomial has two terms, so it is a binomial.

25. 5n3 + nq

3

SOLUTION:

A polynomial is a monomial or the sum of monomials. 5n3 + nq3 is the sum of 2 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 3 and 4,

so the degree of 5n3 + nq3 is 4. The polynomial has two terms, so it is a binomial.

Write each polynomial in standard form. Identify the leading coefficient.

26. 5x2 – 2 + 3x

SOLUTION: Find the degree of each term.

5x2 → 2

– 2 → 0

3x → 1

The greatest degree is 2, from the term 5x2, so the leading coefficient of 5x

2 – 2 + 3x is 5.

Rewrite the polynomial with each monomial in descending order according to degree.

5x2 + 3x – 2

27. 8y + 7y3

SOLUTION: Find the degree of each term.

7y3 → 3

8y → 1

The greatest degree is 3, from the term 7y3, so the leading coefficient of 8y + 7y

3 is 7.

Rewrite the polynomial with each monomial in descending order according to degree.

7y3 + 8y

28. 4 – 3c – 5c2

SOLUTION: Find the degree of each term. 4 → 0 3c → 1

– 5c2 → 2

The greatest degree is 2, from the term – 5c2, so the leading coefficient of 4 – 3c – 5c

2 is –5.

Rewrite the polynomial with each monomial in descending order according to degree.

–5c2 – 3c + 4

29. –y3 + 3y – 3y

2 + 2

SOLUTION: Find the degree of each term.

–y3 → 3

3y → 1

3y2 → 2

2 → 0

The greatest degree is 3, from the term –y3, so the leading coefficient of –y

3 + 3y – 3y

2 + 2 is –1.

Rewrite the polynomial with each monomial in descending order according to degree. –y 3 – 3y 2 + 3y + 2

30. 11t + 2t2 – 3 + t

5

SOLUTION: Find the degree of each term. 11t → 1

2t2 → 2

–3 → 0

t5 → 5

The greatest degree is 5, from the term t5, so the leading coefficient of 11t + 2t2 – 3 + t

5 is 1.

Rewrite the polynomial with each monomial in descending order according to degree.

t5 + 2t

2 + 11t – 3

31. 2 + r – r3

SOLUTION: Find the degree of each term. 2 → 0 r → 1

– r3→ 3

The greatest degree is 3, from the term – r3, so the leading coefficient of 2 + r – r3

is –1. Rewrite the polynomial with each monomial in descending order according to degree.

–r3 + r + 2

32.

SOLUTION: Find the degree of each term.

→ 0

–3x4 → 4

7 → 0

The greatest degree is 4, from the term –3x4, so the leading coefficient of is –3.

Rewrite the polynomial with each monomial in descending order according to degree.

33. –9b2 + 10b – b

6

SOLUTION: Find the degree of each term.

–9b2 → 2

10b → 1

–b6→ 6

The greatest degree is 6, from the term –b6, so the leading coefficient of –9b2 + 10b – b6 is –1. Rewrite the polynomial with each monomial in descending order according to degree.

–b6 – 9b

2 + 10b

Find each sum or difference.

34. (2c2 + 6c + 4) + (5c – 7)

SOLUTION:

35. (2x + 3x2) − (7 − 8x

2)

SOLUTION:

36. (3c3 − c + 11) − (c2 + 2c + 8)

SOLUTION:

37. (z2 + z) + (z

2 − 11)

SOLUTION:

38. (2x − 2y + 1) − (3y + 4x)

SOLUTION:

39. (4a − 5b2 + 3) + (6 − 2a + 3b

2)

SOLUTION:

40. (x2y − 3x

2 + y) + (3y − 2x2y)

SOLUTION:

41. (−8xy + 3x2 − 5y) + (4x

2 − 2y + 6xy)

SOLUTION:

42. (5n − 2p2 + 2np) − (4p

2 + 4n)

SOLUTION:

43. (4rxt − 8r2x + x2

) − (6rx2 + 5rxt − 2x

2)

SOLUTION:

44. PETS From 1999 through 2009, the number of dogs D and the number of cats C (in hundreds) adopted from animal

shelters in the United States are modeled by the equations D = 2n + 3 and C = n + 4, where n is the number of yearssince 1999. a. Write an equation that models the total number T of dogs and cats adopted in hundreds for this time period. b. If this trend continues, how many dogs and cats will be adopted in 2013?

SOLUTION: a.

So, an equation that models the total number of dogs and cats adopted is T = 3n + 7. b. Evaluate the equation for the total number of dogs and cats for n = 2013 – 1999 = 14.

The number of cats and dogs adopted in 2013 will be 49 × 100 or 4900 cats and dogs.

Classify each polynomial according to its degree and number of terms.

45. 4x – 3x2 + 5

SOLUTION:

Find the degree of each term of 4x – 3x2 + 5.

4x → 1

– 3x2 → 2

5 → 0

The greatest degree is 2 and there are 3 terms, so 4x – 3x2 + 5 is a quadratic trinomial.

46. 11z3

SOLUTION: Find the degree of each term. 11z3

→ 3

The greatest degree is 3 and there is one term, so 11z3 is a cubic monomial.

47. 9 + y4

SOLUTION:

Find the degree of each term of 9 + y 4.

9 → 1

y4→ 4

The greatest degree is 4 and there are 2 terms, so 9 + y 4 is a quartic binomial.

48. 3x3 – 7

SOLUTION:

Find the degree of each term of 3x3 – 7.

3x3 → 3

–7 → 0

The greatest degree is 3 and there are 2 terms, so 3x3 – 7 is a cubic binomial.

49. –2x5 – x

2 + 5x – 8

SOLUTION:

Find the degree of each term of –2x5 – x2 + 5x – 8.

–2x5 → 5

–x2 → 2

5x → 1 – 8 → 0

The greatest degree is 5 and there are 4 terms, so –2x5 – x2 + 5x – 8 is a quintic polynomial.

50. 10t – 4t2 + 6t

3

SOLUTION:

Find the degree of each term of 10t – 4t2 + 6t

3.

10t → 1

4t2 → 2

6t3 → 3

The greatest degree is 3 and there are 3 terms, so 10t – 4t2 + 6t

3 is a cubic trinomial.

51. ENROLLMENT In a rapidly growing school system, the numbers (in hundreds) of total students N and

K-5 students P enrolled from 2000 to 2009 are modeled by the equations N = 1.25t2 – t + 7.5 and P =

0.7t2 – 0.95t + 3.8, where t is the number of years since 2000.

a. Write an equation modeling the number of 6-12 students S enrolled for this time period. b. How many 6-12 students were enrolled in the school system in 2007?

SOLUTION: a. To write an equation that represents the number of 6-12 students enrolled, subtract the equations that represent the total number of students and the number of K-5 students.

b. Replace t with 7 in the equation for S to determine the number of students enrolled in 6-12 in 2007.

In 2007 there were 30.3 hundreds or 3030 students enrolled in 6-12.

52. CCSS REASONING The perimeter of the figure shown is represented by the expression 3x2 − 7x + 2. Write a

polynomial that represents the measure of the third side.

SOLUTION:

53. GEOMETRY Consider the rectangle.

a. What does (4x2 + 2x – 1)(2x

2 – x + 3) represent?

b. What does 2(4x2 + 2x – 1) + 2(2x

2 – x + 3) represent?

SOLUTION:

a. (4x2 + 2x – 1)(2x

2 – x + 3) is a multiplication of the length and the width of the rectangle, which is the formula for

the area of a rectangle.

b. 2(4x2 + 2x – 1) + 2(2x

2 – x + 3) is the sum of twice the length and twice the width of a rectangle, which is the

formula for the perimeter of the rectangle

Find each sum or difference.

54. (4x + 2y − 6z) + (5y − 2z + 7x) + (−9z − 2x − 3y)

SOLUTION:

55. (5a2 − 4) + (a

2 − 2a + 12) + (4a2 − 6a + 8)

SOLUTION:

56. (3c2 − 7) + (4c + 7) − (c

2 + 5c − 8)

SOLUTION:

57. (3n3 + 3n − 10) − (4n

2 − 5n) + (4n3 − 3n

2 − 9n + 4)

SOLUTION:

58. FOOTBALL The National Football League is divided into two conferences, the American A and the National N. From 2002 through 2009, the total attendance T (in thousands) for both conferences and for the American Conference games are modeled by the following equations, where x is the number of years since 2002.

T = –0.69x3 + 55.83x

2 + 643.31x + 10,538

A = –3.78x3 + 58.96x

2 + 265.96x + 5257

Determine how many people attended National Conference football games in 2009.

SOLUTION:

Let x = 7 represent 2009, then find how many people attended a National Conference football game in 2009.

In 2009 the number of people who attended National Conference football games was about 8829 thousand. Multiply by 1000 to find the attendance in standard form. 8829 ×1000 = 8,829,000. So, about 8,829,000 people attended National Conference football games in 2009.

59. CAR RENTAL The cost to rent a car for a day is $15 plus $0.15 for each mile driven. a. Write a polynomial that represents the cost of renting a car for m miles. b. If a car is driven 145 miles, how much would it cost to rent? c. If a car is driven 105 miles each day for four days, how much would it cost to rent a car? d. If a car is driven 220 miles each day for seven days, how much would it cost to rent a car?

SOLUTION: a. The cost to rent a car is the daily rate and the mileage cost or 15 + 0.15m. b. Substitute 145 for m to find the daily cost to drive 145 miles.

The cost to rent the car would be $36.75. c. The expression represents the cost per day. For 4 days, multiply the entire expression by 4 to find the cost for a 4 day trip. Substitute 105 for m.

The cost to rent the car would be $123. d. The expression represents the cost per day. For 7 days, multiply the entire expression by 7 to find the cost for a 7 day trip. Substitute 220 for m.

The cost to rent the car would be $336.

60. MULTIPLE REPRESENTATIONS In this problem, you will explore perimeter and area. a. Geometric Draw three rectangles that each have a perimeter of 400 feet. b. Tabular Record the width and length of each rectangle in a table like the one shown below. Find the area of eachrectangle.

c. Graphical On a coordinate system, graph the area of rectangle 4 in terms of the length, x. Use the graph to determine the largest area possible. d. Analytical Determine the length and width that produce the largest area.

SOLUTION: a. For the perimeters to be 400, create lengths and widths that sum to 200. Note that some lengths are already provided in the table in part b.

b. The area is length multiplied by width. Make sure the units are squared for area. The sum of the length and width must be 200, so if the length is x, the width must be 200 – x.

c. The length will be the x-values on the horizontal and the area will be the y-values on the vertical. The length cannot pass 200 since the sum of the length and width is 200. Set the intervals for the x-axis to 25 feet. Extend the table of values to find more points to plot on the graph. It appears that 10,000 is the greatest area, so set the intervals

for the y-axis to 1000 ft2. After graphing, it appears that the highest point on the graph is at an area of 10,000 ft

2.

d. The associated x-value with the maximum area is x = 100, so the length must be 100 and the width must be 200 – 100, or 100. The length and width of the rectangle must be 100 feet each to have the largest area.

61. CCSS CRITIQUE Cheyenne and Sebastian are finding (2x2 − x) − (3x + 3x

2 − 2). Is either of them correct? Explain your reasoning.

SOLUTION:

Neither is correct. Cheyenne, did not distribute the negative to the 2nd and 3rd terms when she found the additive inverse. Sebastian did not distribute the negate to the 3rd terms when he found the additive inverse. To find the additive inverse, all terms should be multiplied by −1.

62. REASONING Determine whether each of the following statements is true or false . Explain your reasoning. a. A binomial can have a degree of zero. b. The order in which polynomials are subtracted does not matter.

SOLUTION: a. If a binomial has two terms that are each a degree of 0, then those terms can be combined and the binomial becomes a monomial. For example, 18 + 7 = 25. If one of the terms of the binomial does not have a degree of 0, then the binomial cannot have a degree of 0, since the degree of a polynomial is the greatest degree of any term in the polynomial. b. Subtraction is not commutative. While 2 + 5 = 5 + 2, 2 – 5 ≠ 5 – 2. This is also true for polynomials. Sample answer: (2x – 3) – (4x – 3) = –2x, but (4x – 3) – (2x – 3) = 2x

63. CHALLENGE Write a polynomial that represents the sum of an odd integer 2n + 1 and the next two consecutive odd integers.

SOLUTION:

64. WRITING IN MATH Why would you add or subtract equations that represent real-world situations? Explain.

SOLUTION:

65. WRITING IN MATH Describe how to add and subtract polynomials using both the vertical and horizontal formats.

SOLUTION: To add polynomials in a horizontal format, you combine like terms. For the vertical format, you write the polynomials in standard form, align like terms in columns, and combine like terms.

To subtract polynomials in a horizontal format you find the additive inverse of the polynomial you are subtracting, andthen combine like terms. For the vertical format, you write the polynomials in standard form, align like terms in columns, and subtract by adding the additive inverse.

66. Three consecutive integers can be represented by x, x + 1, and x + 2. What is the sum of these three integers? A x(x + 1)(x + 2)

B x3 + 3

C 3x + 3 D x + 3

SOLUTION:

The correct choice is C.

67. SHORT RESPONSE What is the perimeter of a square with sides that measure 2x + 3 units?

SOLUTION:

The perimeter of the square is 8x + 12 units.

68. Jim cuts a board in the shape of a regular hexagon and pounds in a nail at each vertex, as shown. How many rubber bands will he need to stretch a rubber band across every possible pair of nails?

F 15 G 14 H 12 J 9

SOLUTION: The first nail would connect to 5 others, the second to 4 others, the third to 3 others, etc.5 + 4 + 3 + 2 + 1 = 15 The correct choice is F.

69. Which ordered pair is in the solution set of the system of inequalities shown in the graph?

A (−3, 0) B (0, −3) C (5, 0) D (0, 5)

SOLUTION: Choice A is outside the shaded area for both inequalities. Choices B and D are inside the shaded area for only one inequality. Choice C is the only point in the solution for both inequalities. So, the correct choice is C.

70. COMPUTERS A computer technician charges by the hour to fix and repair computer equipment. The total cost of the technician for one hour is $75, for two hours is $125, for three hours is $175, for four hours is $225, and so on. Write a recursive formula for the sequence.

SOLUTION: Write out the terms. $75, $125, $175, $225, ... The first term is 75, and 50 is added to form each following term. Therefore, we have a1 = 75, an = an – 1 + 50, n ≥ 2.

Determine whether each sequence is arithmetic, geometric, or neither. Explain.71. 8, –32, 128, –512, ...

SOLUTION: Check for a common difference. –32 – 8 = –40 128 – (–32) = 160 There is no common difference. Check for a common ratio. –32 ÷ 8 = –4 128 ÷ (–32) = –4 Geometric; the common ratio is –4.

72. 25, 8, –9, –26, ...

SOLUTION: Check for a common difference. 8 – 25 = –17 –9 – 8 = –17 Arithmetic; the common difference is –17.

73.

SOLUTION: Check for a common difference.

There is no common difference. Check for a common ratio.

There is no common ratio, so the sequence is not arithmetic or geometric.

74. 43, 52, 61, 70, ...

SOLUTION: Check for a common difference. 52 – 43 = 9 61 – 52 = 9 Arithmetic; the common difference is 9.

75. –27, –16, –5, 6, ...

SOLUTION: Check for a common difference. –16 – (–27) = 11 –5 – (–16) = 11 Arithmetic; the common difference is 11.

76. 200, 100, 50, 25, …

SOLUTION: Check for a common difference. 100 – 200 = –100 50 – 100 = –50 There is no common difference. Check for a common ratio. 100 ÷ 200 = 0.5 50 ÷ 100 = 0.5

Geometric; the common ratio is 0.5 or .

77. JOBS Kimi received an offer for a new job. She wants to compare the offer with her current job. What is total amount of sales that Kimi must get each month to make the same income at either job?

SOLUTION: Let s be Kimi’s monthly sales.

Kimi must sell $80,000 each month to make the same income at either job.

Determine whether each sequence is an arithmetic sequence. If it is, state the common difference.78. 24, 16, 8, 0, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is –8, because 16 – 24 = –8; 8 –16 = –8; etc.

79. , 13, 26, …

SOLUTION: Find the difference between the terms.

– =

13 – =

26 – 13 = 13 There is not a common different. The sequence is not an arithmetic sequence

80. 7, 6, 5, 4, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is –1, because 6 – 7 = –1; 5 – 6 = –1; 4 – 5 = –1;etc.

81. 10, 12, 15, 18, …

SOLUTION: Find the difference between the terms. 12 – 10 = 2 15 – 12 = 3 18 – 15 = 3 There is no common difference. The sequence is not an arithmetic sequence.

82. −15, −11, −7, −3, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is 4, because –11 – (–15) = 4; –7 – (–11) = 4; –3 – (–7) = 4;etc.

83. −0.3, 0.2, 0.7, 1.2, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is 0.5, because 0.2 – (–0.3) = 0.5; 0.7 – 0.2 = 0.5; 1.2 – 0.7 = 0.5; etc.

Simplify.

84. t(t5)(t

7)

SOLUTION:

85. n3(n

2)(−2n

3)

SOLUTION:

86. (5t5v

2)(10t

3v

4)

SOLUTION:

87. (−8u4z

5)(5uz

4)

SOLUTION:

88. [(3)2]3

SOLUTION:

89. [(2)3]2

SOLUTION:

90. (2m4k

3)2(−3mk

2)3

SOLUTION:

91. (6xy2)2(2x

2y

2z

2)3

SOLUTION:

eSolutions Manual - Powered by Cognero Page 7

8-1 Adding and Subtracting Polynomials

Determine whether each expression is a polynomial. If it is a polynomial, find the degree and determine whether it is a monomial, binomial, or trinomial.

1. 7ab + 6b2 – 2a

3

SOLUTION:

A polynomial is a monomial or the sum of monomials. 7ab + 6b2 – 2a

3 is the sum of 3 monomials, so it is a

polynomial. The degree of a polynomial is the greatest degree of any term in the polynomial.The degree of each term is 2, 2, and

3, so the degree of 7ab + 6b2 – 2a

3 is 3. The polynomial has three terms, so it is a trinomial.

2. 2y – 5 + 3y2

SOLUTION:

2y – 5 + 3y2

is the sum of monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 1, 0, and

2, so the degree of 2y – 5 + 3y2 is 2. The polynomial has three terms, so it is a trinomial.

3. 3x2

SOLUTION:

A polynomial is a monomial or the sum of monomials, so 3x2 is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of 3x2 is 2. The

polynomial has one term, so it is a monomial.

4.

SOLUTION: A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer exponents. It has

only one term. is a division of two monomials, so it is not a monomial.

5. 5m2p

3 + 6

SOLUTION:

A polynomial is a monomial or the sum of monomials. 5m2p

3 + 6

is the sum of 2 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 5 and 0,

so the degree of 5m2p

3 + 6 is 5. The polynomial has two terms, so it is a binomial.

6. 5q–4

+ 6q

SOLUTION: A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer exponents. It has

only one term. 5q-4 is equal to , which is a division of two monomials, so it is not a monomial.

Write each polynomial in standard form. Identify the leading coefficient.

7. –4d4 + 1 – d

2

SOLUTION: Find the degree of each term.

–4d4→ 4

1 → 0

–d2 → 2

The greatest degree is 4, from the term –4d4, so the leading coefficient of –4d

4 + 1 – d

2 is –4.

Rewrite the polynomial with each monomial in descending order according to degree.

–4d4 – d

2 + 1

8. 2x5 – 12 + 3x

SOLUTION: Find the degree of each term.

2x5 → 5

–12 → 0 3x → 1

The greatest degree is 5, from the term 2x5, so the leading coefficient of 2x

5 – 12 + 3x is 2.

Rewrite the polynomial with each monomial in descending order according to degree.

2x5 + 3x – 12

9. 4z – 2z2 – 5z

4

SOLUTION: Find the degree of each term. 4z → 1

– 2z2 → 2

– 5z4→ 4

The greatest degree is 4, from the term – 5z4, so the leading coefficient of 4z – 2z

2 – 5z

4 is –5.

Rewrite the polynomial with each monomial in descending order according to degree.

–5z4 – 2z

2 + 4z

10. 2a + 4a3 – 5a

2 – 1

SOLUTION: Find the degree of each term. 2a → 1

4a3→ 3

– 5a2 → 2

– 1 → 0

The greatest degree is 3, from the term 4a3, so the leading coefficient of 2a + 4a

3 – 5a

2 – 1 is 4.

Rewrite the polynomial with each monomial in descending order according to degree.

4a3– 5a

2 + 2a – 1

Find each sum or difference.

11. (6x3 − 4) + (−2x

3 + 9)

SOLUTION:

12. (g3 − 2g2 + 5g + 6) − (g2 + 2g)

SOLUTION:

13. (4 + 2a2 − 2a) − (3a

2 − 8a + 7)

SOLUTION:

14. (8y − 4y2) + (3y − 9y

2)

SOLUTION:

15. (−4z3 − 2z + 8) − (4z

3 + 3z2 − 5)

SOLUTION:

16. (−3d2 − 8 + 2d) + (4d − 12 + d2

)

SOLUTION:

17. (y + 5) + (2y + 4y2 – 2)

SOLUTION:

18. (3n3 − 5n + n2

) − (−8n2 + 3n

3)

SOLUTION:

19. CCSS SENSE-MAKING The total number of students T who traveled for spring break consists of two groups: students who flew to their destinations F and students who drove to their destination D. The number (in thousands) of students who flew and the total number of students who flew or drove can be modeled by the following equations,where n is the number of years since 1995. T = 14n + 21

F = 8n + 7 a. Write an equation that models the number of students who drove to their destination for this time period. b. Predict the number of students who will drive to their destination in 2012. c. How many students will drive or fly to their destination in 2015?

SOLUTION: a.

D = 6n + 14 b. n = 2012 – 1995 = 17

The number of students who will drive to their destination in 2012 is 116,000 students. c. n = 2015 – 1995 = 20

The number of students who will drive or fly to their destination in 2015 is 301,000 students.

Determine whether each expression is a polynomial. If it is a polynomial, find the degree and determine whether it is a monomial, binomial, or trinomial.

20.

SOLUTION: A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer exponents. It has

only one term. is a division of two monomials, so it is not a monomial.

21.

SOLUTION: A polynomial is a monomial or the sum of monomials. 21 is a monomial, so it is also a polynomial. The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of 21 is 0. The polynomial has only one term, so it is a monomial.

22. c4 – 2c

2 + 1

SOLUTION:

A polynomial is a monomial or the sum of monomials. c4 – 2c

2 + 1

is the sum of 3 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 4, 2, and

0, so the degree of c4 – 2c2 + 1 is 4. The polynomial has three terms, so it is a trinomial.

23. d + 3dc

SOLUTION: A polynomial is a monomial or the sum of monomials. A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer

exponents. 3dc

has a variable in the exponent, so it is not a monomial.

24. a – a2

SOLUTION:

A polynomial is a monomial or the sum of monomials. a – a2 is the sum of 2 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 1 and 2,

so the degree of a – a2 is 2. The polynomial has two terms, so it is a binomial.

25. 5n3 + nq

3

SOLUTION:

A polynomial is a monomial or the sum of monomials. 5n3 + nq3 is the sum of 2 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 3 and 4,

so the degree of 5n3 + nq3 is 4. The polynomial has two terms, so it is a binomial.

Write each polynomial in standard form. Identify the leading coefficient.

26. 5x2 – 2 + 3x

SOLUTION: Find the degree of each term.

5x2 → 2

– 2 → 0

3x → 1

The greatest degree is 2, from the term 5x2, so the leading coefficient of 5x

2 – 2 + 3x is 5.

Rewrite the polynomial with each monomial in descending order according to degree.

5x2 + 3x – 2

27. 8y + 7y3

SOLUTION: Find the degree of each term.

7y3 → 3

8y → 1

The greatest degree is 3, from the term 7y3, so the leading coefficient of 8y + 7y

3 is 7.

Rewrite the polynomial with each monomial in descending order according to degree.

7y3 + 8y

28. 4 – 3c – 5c2

SOLUTION: Find the degree of each term. 4 → 0 3c → 1

– 5c2 → 2

The greatest degree is 2, from the term – 5c2, so the leading coefficient of 4 – 3c – 5c

2 is –5.

Rewrite the polynomial with each monomial in descending order according to degree.

–5c2 – 3c + 4

29. –y3 + 3y – 3y

2 + 2

SOLUTION: Find the degree of each term.

–y3 → 3

3y → 1

3y2 → 2

2 → 0

The greatest degree is 3, from the term –y3, so the leading coefficient of –y

3 + 3y – 3y

2 + 2 is –1.

Rewrite the polynomial with each monomial in descending order according to degree. –y 3 – 3y 2 + 3y + 2

30. 11t + 2t2 – 3 + t

5

SOLUTION: Find the degree of each term. 11t → 1

2t2 → 2

–3 → 0

t5 → 5

The greatest degree is 5, from the term t5, so the leading coefficient of 11t + 2t2 – 3 + t

5 is 1.

Rewrite the polynomial with each monomial in descending order according to degree.

t5 + 2t

2 + 11t – 3

31. 2 + r – r3

SOLUTION: Find the degree of each term. 2 → 0 r → 1

– r3→ 3

The greatest degree is 3, from the term – r3, so the leading coefficient of 2 + r – r3

is –1. Rewrite the polynomial with each monomial in descending order according to degree.

–r3 + r + 2

32.

SOLUTION: Find the degree of each term.

→ 0

–3x4 → 4

7 → 0

The greatest degree is 4, from the term –3x4, so the leading coefficient of is –3.

Rewrite the polynomial with each monomial in descending order according to degree.

33. –9b2 + 10b – b

6

SOLUTION: Find the degree of each term.

–9b2 → 2

10b → 1

–b6→ 6

The greatest degree is 6, from the term –b6, so the leading coefficient of –9b2 + 10b – b6 is –1. Rewrite the polynomial with each monomial in descending order according to degree.

–b6 – 9b

2 + 10b

Find each sum or difference.

34. (2c2 + 6c + 4) + (5c – 7)

SOLUTION:

35. (2x + 3x2) − (7 − 8x

2)

SOLUTION:

36. (3c3 − c + 11) − (c2 + 2c + 8)

SOLUTION:

37. (z2 + z) + (z

2 − 11)

SOLUTION:

38. (2x − 2y + 1) − (3y + 4x)

SOLUTION:

39. (4a − 5b2 + 3) + (6 − 2a + 3b

2)

SOLUTION:

40. (x2y − 3x

2 + y) + (3y − 2x2y)

SOLUTION:

41. (−8xy + 3x2 − 5y) + (4x

2 − 2y + 6xy)

SOLUTION:

42. (5n − 2p2 + 2np) − (4p

2 + 4n)

SOLUTION:

43. (4rxt − 8r2x + x2

) − (6rx2 + 5rxt − 2x

2)

SOLUTION:

44. PETS From 1999 through 2009, the number of dogs D and the number of cats C (in hundreds) adopted from animal

shelters in the United States are modeled by the equations D = 2n + 3 and C = n + 4, where n is the number of yearssince 1999. a. Write an equation that models the total number T of dogs and cats adopted in hundreds for this time period. b. If this trend continues, how many dogs and cats will be adopted in 2013?

SOLUTION: a.

So, an equation that models the total number of dogs and cats adopted is T = 3n + 7. b. Evaluate the equation for the total number of dogs and cats for n = 2013 – 1999 = 14.

The number of cats and dogs adopted in 2013 will be 49 × 100 or 4900 cats and dogs.

Classify each polynomial according to its degree and number of terms.

45. 4x – 3x2 + 5

SOLUTION:

Find the degree of each term of 4x – 3x2 + 5.

4x → 1

– 3x2 → 2

5 → 0

The greatest degree is 2 and there are 3 terms, so 4x – 3x2 + 5 is a quadratic trinomial.

46. 11z3

SOLUTION: Find the degree of each term. 11z3

→ 3

The greatest degree is 3 and there is one term, so 11z3 is a cubic monomial.

47. 9 + y4

SOLUTION:

Find the degree of each term of 9 + y 4.

9 → 1

y4→ 4

The greatest degree is 4 and there are 2 terms, so 9 + y 4 is a quartic binomial.

48. 3x3 – 7

SOLUTION:

Find the degree of each term of 3x3 – 7.

3x3 → 3

–7 → 0

The greatest degree is 3 and there are 2 terms, so 3x3 – 7 is a cubic binomial.

49. –2x5 – x

2 + 5x – 8

SOLUTION:

Find the degree of each term of –2x5 – x2 + 5x – 8.

–2x5 → 5

–x2 → 2

5x → 1 – 8 → 0

The greatest degree is 5 and there are 4 terms, so –2x5 – x2 + 5x – 8 is a quintic polynomial.

50. 10t – 4t2 + 6t

3

SOLUTION:

Find the degree of each term of 10t – 4t2 + 6t

3.

10t → 1

4t2 → 2

6t3 → 3

The greatest degree is 3 and there are 3 terms, so 10t – 4t2 + 6t

3 is a cubic trinomial.

51. ENROLLMENT In a rapidly growing school system, the numbers (in hundreds) of total students N and

K-5 students P enrolled from 2000 to 2009 are modeled by the equations N = 1.25t2 – t + 7.5 and P =

0.7t2 – 0.95t + 3.8, where t is the number of years since 2000.

a. Write an equation modeling the number of 6-12 students S enrolled for this time period. b. How many 6-12 students were enrolled in the school system in 2007?

SOLUTION: a. To write an equation that represents the number of 6-12 students enrolled, subtract the equations that represent the total number of students and the number of K-5 students.

b. Replace t with 7 in the equation for S to determine the number of students enrolled in 6-12 in 2007.

In 2007 there were 30.3 hundreds or 3030 students enrolled in 6-12.

52. CCSS REASONING The perimeter of the figure shown is represented by the expression 3x2 − 7x + 2. Write a

polynomial that represents the measure of the third side.

SOLUTION:

53. GEOMETRY Consider the rectangle.

a. What does (4x2 + 2x – 1)(2x

2 – x + 3) represent?

b. What does 2(4x2 + 2x – 1) + 2(2x

2 – x + 3) represent?

SOLUTION:

a. (4x2 + 2x – 1)(2x

2 – x + 3) is a multiplication of the length and the width of the rectangle, which is the formula for

the area of a rectangle.

b. 2(4x2 + 2x – 1) + 2(2x

2 – x + 3) is the sum of twice the length and twice the width of a rectangle, which is the

formula for the perimeter of the rectangle

Find each sum or difference.

54. (4x + 2y − 6z) + (5y − 2z + 7x) + (−9z − 2x − 3y)

SOLUTION:

55. (5a2 − 4) + (a

2 − 2a + 12) + (4a2 − 6a + 8)

SOLUTION:

56. (3c2 − 7) + (4c + 7) − (c

2 + 5c − 8)

SOLUTION:

57. (3n3 + 3n − 10) − (4n

2 − 5n) + (4n3 − 3n

2 − 9n + 4)

SOLUTION:

58. FOOTBALL The National Football League is divided into two conferences, the American A and the National N. From 2002 through 2009, the total attendance T (in thousands) for both conferences and for the American Conference games are modeled by the following equations, where x is the number of years since 2002.

T = –0.69x3 + 55.83x

2 + 643.31x + 10,538

A = –3.78x3 + 58.96x

2 + 265.96x + 5257

Determine how many people attended National Conference football games in 2009.

SOLUTION:

Let x = 7 represent 2009, then find how many people attended a National Conference football game in 2009.

In 2009 the number of people who attended National Conference football games was about 8829 thousand. Multiply by 1000 to find the attendance in standard form. 8829 ×1000 = 8,829,000. So, about 8,829,000 people attended National Conference football games in 2009.

59. CAR RENTAL The cost to rent a car for a day is $15 plus $0.15 for each mile driven. a. Write a polynomial that represents the cost of renting a car for m miles. b. If a car is driven 145 miles, how much would it cost to rent? c. If a car is driven 105 miles each day for four days, how much would it cost to rent a car? d. If a car is driven 220 miles each day for seven days, how much would it cost to rent a car?

SOLUTION: a. The cost to rent a car is the daily rate and the mileage cost or 15 + 0.15m. b. Substitute 145 for m to find the daily cost to drive 145 miles.

The cost to rent the car would be $36.75. c. The expression represents the cost per day. For 4 days, multiply the entire expression by 4 to find the cost for a 4 day trip. Substitute 105 for m.

The cost to rent the car would be $123. d. The expression represents the cost per day. For 7 days, multiply the entire expression by 7 to find the cost for a 7 day trip. Substitute 220 for m.

The cost to rent the car would be $336.

60. MULTIPLE REPRESENTATIONS In this problem, you will explore perimeter and area. a. Geometric Draw three rectangles that each have a perimeter of 400 feet. b. Tabular Record the width and length of each rectangle in a table like the one shown below. Find the area of eachrectangle.

c. Graphical On a coordinate system, graph the area of rectangle 4 in terms of the length, x. Use the graph to determine the largest area possible. d. Analytical Determine the length and width that produce the largest area.

SOLUTION: a. For the perimeters to be 400, create lengths and widths that sum to 200. Note that some lengths are already provided in the table in part b.

b. The area is length multiplied by width. Make sure the units are squared for area. The sum of the length and width must be 200, so if the length is x, the width must be 200 – x.

c. The length will be the x-values on the horizontal and the area will be the y-values on the vertical. The length cannot pass 200 since the sum of the length and width is 200. Set the intervals for the x-axis to 25 feet. Extend the table of values to find more points to plot on the graph. It appears that 10,000 is the greatest area, so set the intervals

for the y-axis to 1000 ft2. After graphing, it appears that the highest point on the graph is at an area of 10,000 ft

2.

d. The associated x-value with the maximum area is x = 100, so the length must be 100 and the width must be 200 – 100, or 100. The length and width of the rectangle must be 100 feet each to have the largest area.

61. CCSS CRITIQUE Cheyenne and Sebastian are finding (2x2 − x) − (3x + 3x

2 − 2). Is either of them correct? Explain your reasoning.

SOLUTION:

Neither is correct. Cheyenne, did not distribute the negative to the 2nd and 3rd terms when she found the additive inverse. Sebastian did not distribute the negate to the 3rd terms when he found the additive inverse. To find the additive inverse, all terms should be multiplied by −1.

62. REASONING Determine whether each of the following statements is true or false . Explain your reasoning. a. A binomial can have a degree of zero. b. The order in which polynomials are subtracted does not matter.

SOLUTION: a. If a binomial has two terms that are each a degree of 0, then those terms can be combined and the binomial becomes a monomial. For example, 18 + 7 = 25. If one of the terms of the binomial does not have a degree of 0, then the binomial cannot have a degree of 0, since the degree of a polynomial is the greatest degree of any term in the polynomial. b. Subtraction is not commutative. While 2 + 5 = 5 + 2, 2 – 5 ≠ 5 – 2. This is also true for polynomials. Sample answer: (2x – 3) – (4x – 3) = –2x, but (4x – 3) – (2x – 3) = 2x

63. CHALLENGE Write a polynomial that represents the sum of an odd integer 2n + 1 and the next two consecutive odd integers.

SOLUTION:

64. WRITING IN MATH Why would you add or subtract equations that represent real-world situations? Explain.

SOLUTION:

65. WRITING IN MATH Describe how to add and subtract polynomials using both the vertical and horizontal formats.

SOLUTION: To add polynomials in a horizontal format, you combine like terms. For the vertical format, you write the polynomials in standard form, align like terms in columns, and combine like terms.

To subtract polynomials in a horizontal format you find the additive inverse of the polynomial you are subtracting, andthen combine like terms. For the vertical format, you write the polynomials in standard form, align like terms in columns, and subtract by adding the additive inverse.

66. Three consecutive integers can be represented by x, x + 1, and x + 2. What is the sum of these three integers? A x(x + 1)(x + 2)

B x3 + 3

C 3x + 3 D x + 3

SOLUTION:

The correct choice is C.

67. SHORT RESPONSE What is the perimeter of a square with sides that measure 2x + 3 units?

SOLUTION:

The perimeter of the square is 8x + 12 units.

68. Jim cuts a board in the shape of a regular hexagon and pounds in a nail at each vertex, as shown. How many rubber bands will he need to stretch a rubber band across every possible pair of nails?

F 15 G 14 H 12 J 9

SOLUTION: The first nail would connect to 5 others, the second to 4 others, the third to 3 others, etc.5 + 4 + 3 + 2 + 1 = 15 The correct choice is F.

69. Which ordered pair is in the solution set of the system of inequalities shown in the graph?

A (−3, 0) B (0, −3) C (5, 0) D (0, 5)

SOLUTION: Choice A is outside the shaded area for both inequalities. Choices B and D are inside the shaded area for only one inequality. Choice C is the only point in the solution for both inequalities. So, the correct choice is C.

70. COMPUTERS A computer technician charges by the hour to fix and repair computer equipment. The total cost of the technician for one hour is $75, for two hours is $125, for three hours is $175, for four hours is $225, and so on. Write a recursive formula for the sequence.

SOLUTION: Write out the terms. $75, $125, $175, $225, ... The first term is 75, and 50 is added to form each following term. Therefore, we have a1 = 75, an = an – 1 + 50, n ≥ 2.

Determine whether each sequence is arithmetic, geometric, or neither. Explain.71. 8, –32, 128, –512, ...

SOLUTION: Check for a common difference. –32 – 8 = –40 128 – (–32) = 160 There is no common difference. Check for a common ratio. –32 ÷ 8 = –4 128 ÷ (–32) = –4 Geometric; the common ratio is –4.

72. 25, 8, –9, –26, ...

SOLUTION: Check for a common difference. 8 – 25 = –17 –9 – 8 = –17 Arithmetic; the common difference is –17.

73.

SOLUTION: Check for a common difference.

There is no common difference. Check for a common ratio.

There is no common ratio, so the sequence is not arithmetic or geometric.

74. 43, 52, 61, 70, ...

SOLUTION: Check for a common difference. 52 – 43 = 9 61 – 52 = 9 Arithmetic; the common difference is 9.

75. –27, –16, –5, 6, ...

SOLUTION: Check for a common difference. –16 – (–27) = 11 –5 – (–16) = 11 Arithmetic; the common difference is 11.

76. 200, 100, 50, 25, …

SOLUTION: Check for a common difference. 100 – 200 = –100 50 – 100 = –50 There is no common difference. Check for a common ratio. 100 ÷ 200 = 0.5 50 ÷ 100 = 0.5

Geometric; the common ratio is 0.5 or .

77. JOBS Kimi received an offer for a new job. She wants to compare the offer with her current job. What is total amount of sales that Kimi must get each month to make the same income at either job?

SOLUTION: Let s be Kimi’s monthly sales.

Kimi must sell $80,000 each month to make the same income at either job.

Determine whether each sequence is an arithmetic sequence. If it is, state the common difference.78. 24, 16, 8, 0, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is –8, because 16 – 24 = –8; 8 –16 = –8; etc.

79. , 13, 26, …

SOLUTION: Find the difference between the terms.

– =

13 – =

26 – 13 = 13 There is not a common different. The sequence is not an arithmetic sequence

80. 7, 6, 5, 4, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is –1, because 6 – 7 = –1; 5 – 6 = –1; 4 – 5 = –1;etc.

81. 10, 12, 15, 18, …

SOLUTION: Find the difference between the terms. 12 – 10 = 2 15 – 12 = 3 18 – 15 = 3 There is no common difference. The sequence is not an arithmetic sequence.

82. −15, −11, −7, −3, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is 4, because –11 – (–15) = 4; –7 – (–11) = 4; –3 – (–7) = 4;etc.

83. −0.3, 0.2, 0.7, 1.2, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is 0.5, because 0.2 – (–0.3) = 0.5; 0.7 – 0.2 = 0.5; 1.2 – 0.7 = 0.5; etc.

Simplify.

84. t(t5)(t

7)

SOLUTION:

85. n3(n

2)(−2n

3)

SOLUTION:

86. (5t5v

2)(10t

3v

4)

SOLUTION:

87. (−8u4z

5)(5uz

4)

SOLUTION:

88. [(3)2]3

SOLUTION:

89. [(2)3]2

SOLUTION:

90. (2m4k

3)2(−3mk

2)3

SOLUTION:

91. (6xy2)2(2x

2y

2z

2)3

SOLUTION:

eSolutions Manual - Powered by Cognero Page 8

8-1 Adding and Subtracting Polynomials

Determine whether each expression is a polynomial. If it is a polynomial, find the degree and determine whether it is a monomial, binomial, or trinomial.

1. 7ab + 6b2 – 2a

3

SOLUTION:

A polynomial is a monomial or the sum of monomials. 7ab + 6b2 – 2a

3 is the sum of 3 monomials, so it is a

polynomial. The degree of a polynomial is the greatest degree of any term in the polynomial.The degree of each term is 2, 2, and

3, so the degree of 7ab + 6b2 – 2a

3 is 3. The polynomial has three terms, so it is a trinomial.

2. 2y – 5 + 3y2

SOLUTION:

2y – 5 + 3y2

is the sum of monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 1, 0, and

2, so the degree of 2y – 5 + 3y2 is 2. The polynomial has three terms, so it is a trinomial.

3. 3x2

SOLUTION:

A polynomial is a monomial or the sum of monomials, so 3x2 is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of 3x2 is 2. The

polynomial has one term, so it is a monomial.

4.

SOLUTION: A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer exponents. It has

only one term. is a division of two monomials, so it is not a monomial.

5. 5m2p

3 + 6

SOLUTION:

A polynomial is a monomial or the sum of monomials. 5m2p

3 + 6

is the sum of 2 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 5 and 0,

so the degree of 5m2p

3 + 6 is 5. The polynomial has two terms, so it is a binomial.

6. 5q–4

+ 6q

SOLUTION: A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer exponents. It has

only one term. 5q-4 is equal to , which is a division of two monomials, so it is not a monomial.

Write each polynomial in standard form. Identify the leading coefficient.

7. –4d4 + 1 – d

2

SOLUTION: Find the degree of each term.

–4d4→ 4

1 → 0

–d2 → 2

The greatest degree is 4, from the term –4d4, so the leading coefficient of –4d

4 + 1 – d

2 is –4.

Rewrite the polynomial with each monomial in descending order according to degree.

–4d4 – d

2 + 1

8. 2x5 – 12 + 3x

SOLUTION: Find the degree of each term.

2x5 → 5

–12 → 0 3x → 1

The greatest degree is 5, from the term 2x5, so the leading coefficient of 2x

5 – 12 + 3x is 2.

Rewrite the polynomial with each monomial in descending order according to degree.

2x5 + 3x – 12

9. 4z – 2z2 – 5z

4

SOLUTION: Find the degree of each term. 4z → 1

– 2z2 → 2

– 5z4→ 4

The greatest degree is 4, from the term – 5z4, so the leading coefficient of 4z – 2z

2 – 5z

4 is –5.

Rewrite the polynomial with each monomial in descending order according to degree.

–5z4 – 2z

2 + 4z

10. 2a + 4a3 – 5a

2 – 1

SOLUTION: Find the degree of each term. 2a → 1

4a3→ 3

– 5a2 → 2

– 1 → 0

The greatest degree is 3, from the term 4a3, so the leading coefficient of 2a + 4a

3 – 5a

2 – 1 is 4.

Rewrite the polynomial with each monomial in descending order according to degree.

4a3– 5a

2 + 2a – 1

Find each sum or difference.

11. (6x3 − 4) + (−2x

3 + 9)

SOLUTION:

12. (g3 − 2g2 + 5g + 6) − (g2 + 2g)

SOLUTION:

13. (4 + 2a2 − 2a) − (3a

2 − 8a + 7)

SOLUTION:

14. (8y − 4y2) + (3y − 9y

2)

SOLUTION:

15. (−4z3 − 2z + 8) − (4z

3 + 3z2 − 5)

SOLUTION:

16. (−3d2 − 8 + 2d) + (4d − 12 + d2

)

SOLUTION:

17. (y + 5) + (2y + 4y2 – 2)

SOLUTION:

18. (3n3 − 5n + n2

) − (−8n2 + 3n

3)

SOLUTION:

19. CCSS SENSE-MAKING The total number of students T who traveled for spring break consists of two groups: students who flew to their destinations F and students who drove to their destination D. The number (in thousands) of students who flew and the total number of students who flew or drove can be modeled by the following equations,where n is the number of years since 1995. T = 14n + 21

F = 8n + 7 a. Write an equation that models the number of students who drove to their destination for this time period. b. Predict the number of students who will drive to their destination in 2012. c. How many students will drive or fly to their destination in 2015?

SOLUTION: a.

D = 6n + 14 b. n = 2012 – 1995 = 17

The number of students who will drive to their destination in 2012 is 116,000 students. c. n = 2015 – 1995 = 20

The number of students who will drive or fly to their destination in 2015 is 301,000 students.

Determine whether each expression is a polynomial. If it is a polynomial, find the degree and determine whether it is a monomial, binomial, or trinomial.

20.

SOLUTION: A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer exponents. It has

only one term. is a division of two monomials, so it is not a monomial.

21.

SOLUTION: A polynomial is a monomial or the sum of monomials. 21 is a monomial, so it is also a polynomial. The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of 21 is 0. The polynomial has only one term, so it is a monomial.

22. c4 – 2c

2 + 1

SOLUTION:

A polynomial is a monomial or the sum of monomials. c4 – 2c

2 + 1

is the sum of 3 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 4, 2, and

0, so the degree of c4 – 2c2 + 1 is 4. The polynomial has three terms, so it is a trinomial.

23. d + 3dc

SOLUTION: A polynomial is a monomial or the sum of monomials. A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer

exponents. 3dc

has a variable in the exponent, so it is not a monomial.

24. a – a2

SOLUTION:

A polynomial is a monomial or the sum of monomials. a – a2 is the sum of 2 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 1 and 2,

so the degree of a – a2 is 2. The polynomial has two terms, so it is a binomial.

25. 5n3 + nq

3

SOLUTION:

A polynomial is a monomial or the sum of monomials. 5n3 + nq3 is the sum of 2 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 3 and 4,

so the degree of 5n3 + nq3 is 4. The polynomial has two terms, so it is a binomial.

Write each polynomial in standard form. Identify the leading coefficient.

26. 5x2 – 2 + 3x

SOLUTION: Find the degree of each term.

5x2 → 2

– 2 → 0

3x → 1

The greatest degree is 2, from the term 5x2, so the leading coefficient of 5x

2 – 2 + 3x is 5.

Rewrite the polynomial with each monomial in descending order according to degree.

5x2 + 3x – 2

27. 8y + 7y3

SOLUTION: Find the degree of each term.

7y3 → 3

8y → 1

The greatest degree is 3, from the term 7y3, so the leading coefficient of 8y + 7y

3 is 7.

Rewrite the polynomial with each monomial in descending order according to degree.

7y3 + 8y

28. 4 – 3c – 5c2

SOLUTION: Find the degree of each term. 4 → 0 3c → 1

– 5c2 → 2

The greatest degree is 2, from the term – 5c2, so the leading coefficient of 4 – 3c – 5c

2 is –5.

Rewrite the polynomial with each monomial in descending order according to degree.

–5c2 – 3c + 4

29. –y3 + 3y – 3y

2 + 2

SOLUTION: Find the degree of each term.

–y3 → 3

3y → 1

3y2 → 2

2 → 0

The greatest degree is 3, from the term –y3, so the leading coefficient of –y

3 + 3y – 3y

2 + 2 is –1.

Rewrite the polynomial with each monomial in descending order according to degree. –y 3 – 3y 2 + 3y + 2

30. 11t + 2t2 – 3 + t

5

SOLUTION: Find the degree of each term. 11t → 1

2t2 → 2

–3 → 0

t5 → 5

The greatest degree is 5, from the term t5, so the leading coefficient of 11t + 2t2 – 3 + t

5 is 1.

Rewrite the polynomial with each monomial in descending order according to degree.

t5 + 2t

2 + 11t – 3

31. 2 + r – r3

SOLUTION: Find the degree of each term. 2 → 0 r → 1

– r3→ 3

The greatest degree is 3, from the term – r3, so the leading coefficient of 2 + r – r3

is –1. Rewrite the polynomial with each monomial in descending order according to degree.

–r3 + r + 2

32.

SOLUTION: Find the degree of each term.

→ 0

–3x4 → 4

7 → 0

The greatest degree is 4, from the term –3x4, so the leading coefficient of is –3.

Rewrite the polynomial with each monomial in descending order according to degree.

33. –9b2 + 10b – b

6

SOLUTION: Find the degree of each term.

–9b2 → 2

10b → 1

–b6→ 6

The greatest degree is 6, from the term –b6, so the leading coefficient of –9b2 + 10b – b6 is –1. Rewrite the polynomial with each monomial in descending order according to degree.

–b6 – 9b

2 + 10b

Find each sum or difference.

34. (2c2 + 6c + 4) + (5c – 7)

SOLUTION:

35. (2x + 3x2) − (7 − 8x

2)

SOLUTION:

36. (3c3 − c + 11) − (c2 + 2c + 8)

SOLUTION:

37. (z2 + z) + (z

2 − 11)

SOLUTION:

38. (2x − 2y + 1) − (3y + 4x)

SOLUTION:

39. (4a − 5b2 + 3) + (6 − 2a + 3b

2)

SOLUTION:

40. (x2y − 3x

2 + y) + (3y − 2x2y)

SOLUTION:

41. (−8xy + 3x2 − 5y) + (4x

2 − 2y + 6xy)

SOLUTION:

42. (5n − 2p2 + 2np) − (4p

2 + 4n)

SOLUTION:

43. (4rxt − 8r2x + x2

) − (6rx2 + 5rxt − 2x

2)

SOLUTION:

44. PETS From 1999 through 2009, the number of dogs D and the number of cats C (in hundreds) adopted from animal

shelters in the United States are modeled by the equations D = 2n + 3 and C = n + 4, where n is the number of yearssince 1999. a. Write an equation that models the total number T of dogs and cats adopted in hundreds for this time period. b. If this trend continues, how many dogs and cats will be adopted in 2013?

SOLUTION: a.

So, an equation that models the total number of dogs and cats adopted is T = 3n + 7. b. Evaluate the equation for the total number of dogs and cats for n = 2013 – 1999 = 14.

The number of cats and dogs adopted in 2013 will be 49 × 100 or 4900 cats and dogs.

Classify each polynomial according to its degree and number of terms.

45. 4x – 3x2 + 5

SOLUTION:

Find the degree of each term of 4x – 3x2 + 5.

4x → 1

– 3x2 → 2

5 → 0

The greatest degree is 2 and there are 3 terms, so 4x – 3x2 + 5 is a quadratic trinomial.

46. 11z3

SOLUTION: Find the degree of each term. 11z3

→ 3

The greatest degree is 3 and there is one term, so 11z3 is a cubic monomial.

47. 9 + y4

SOLUTION:

Find the degree of each term of 9 + y 4.

9 → 1

y4→ 4

The greatest degree is 4 and there are 2 terms, so 9 + y 4 is a quartic binomial.

48. 3x3 – 7

SOLUTION:

Find the degree of each term of 3x3 – 7.

3x3 → 3

–7 → 0

The greatest degree is 3 and there are 2 terms, so 3x3 – 7 is a cubic binomial.

49. –2x5 – x

2 + 5x – 8

SOLUTION:

Find the degree of each term of –2x5 – x2 + 5x – 8.

–2x5 → 5

–x2 → 2

5x → 1 – 8 → 0

The greatest degree is 5 and there are 4 terms, so –2x5 – x2 + 5x – 8 is a quintic polynomial.

50. 10t – 4t2 + 6t

3

SOLUTION:

Find the degree of each term of 10t – 4t2 + 6t

3.

10t → 1

4t2 → 2

6t3 → 3

The greatest degree is 3 and there are 3 terms, so 10t – 4t2 + 6t

3 is a cubic trinomial.

51. ENROLLMENT In a rapidly growing school system, the numbers (in hundreds) of total students N and

K-5 students P enrolled from 2000 to 2009 are modeled by the equations N = 1.25t2 – t + 7.5 and P =

0.7t2 – 0.95t + 3.8, where t is the number of years since 2000.

a. Write an equation modeling the number of 6-12 students S enrolled for this time period. b. How many 6-12 students were enrolled in the school system in 2007?

SOLUTION: a. To write an equation that represents the number of 6-12 students enrolled, subtract the equations that represent the total number of students and the number of K-5 students.

b. Replace t with 7 in the equation for S to determine the number of students enrolled in 6-12 in 2007.

In 2007 there were 30.3 hundreds or 3030 students enrolled in 6-12.

52. CCSS REASONING The perimeter of the figure shown is represented by the expression 3x2 − 7x + 2. Write a

polynomial that represents the measure of the third side.

SOLUTION:

53. GEOMETRY Consider the rectangle.

a. What does (4x2 + 2x – 1)(2x

2 – x + 3) represent?

b. What does 2(4x2 + 2x – 1) + 2(2x

2 – x + 3) represent?

SOLUTION:

a. (4x2 + 2x – 1)(2x

2 – x + 3) is a multiplication of the length and the width of the rectangle, which is the formula for

the area of a rectangle.

b. 2(4x2 + 2x – 1) + 2(2x

2 – x + 3) is the sum of twice the length and twice the width of a rectangle, which is the

formula for the perimeter of the rectangle

Find each sum or difference.

54. (4x + 2y − 6z) + (5y − 2z + 7x) + (−9z − 2x − 3y)

SOLUTION:

55. (5a2 − 4) + (a

2 − 2a + 12) + (4a2 − 6a + 8)

SOLUTION:

56. (3c2 − 7) + (4c + 7) − (c

2 + 5c − 8)

SOLUTION:

57. (3n3 + 3n − 10) − (4n

2 − 5n) + (4n3 − 3n

2 − 9n + 4)

SOLUTION:

58. FOOTBALL The National Football League is divided into two conferences, the American A and the National N. From 2002 through 2009, the total attendance T (in thousands) for both conferences and for the American Conference games are modeled by the following equations, where x is the number of years since 2002.

T = –0.69x3 + 55.83x

2 + 643.31x + 10,538

A = –3.78x3 + 58.96x

2 + 265.96x + 5257

Determine how many people attended National Conference football games in 2009.

SOLUTION:

Let x = 7 represent 2009, then find how many people attended a National Conference football game in 2009.

In 2009 the number of people who attended National Conference football games was about 8829 thousand. Multiply by 1000 to find the attendance in standard form. 8829 ×1000 = 8,829,000. So, about 8,829,000 people attended National Conference football games in 2009.

59. CAR RENTAL The cost to rent a car for a day is $15 plus $0.15 for each mile driven. a. Write a polynomial that represents the cost of renting a car for m miles. b. If a car is driven 145 miles, how much would it cost to rent? c. If a car is driven 105 miles each day for four days, how much would it cost to rent a car? d. If a car is driven 220 miles each day for seven days, how much would it cost to rent a car?

SOLUTION: a. The cost to rent a car is the daily rate and the mileage cost or 15 + 0.15m. b. Substitute 145 for m to find the daily cost to drive 145 miles.

The cost to rent the car would be $36.75. c. The expression represents the cost per day. For 4 days, multiply the entire expression by 4 to find the cost for a 4 day trip. Substitute 105 for m.

The cost to rent the car would be $123. d. The expression represents the cost per day. For 7 days, multiply the entire expression by 7 to find the cost for a 7 day trip. Substitute 220 for m.

The cost to rent the car would be $336.

60. MULTIPLE REPRESENTATIONS In this problem, you will explore perimeter and area. a. Geometric Draw three rectangles that each have a perimeter of 400 feet. b. Tabular Record the width and length of each rectangle in a table like the one shown below. Find the area of eachrectangle.

c. Graphical On a coordinate system, graph the area of rectangle 4 in terms of the length, x. Use the graph to determine the largest area possible. d. Analytical Determine the length and width that produce the largest area.

SOLUTION: a. For the perimeters to be 400, create lengths and widths that sum to 200. Note that some lengths are already provided in the table in part b.

b. The area is length multiplied by width. Make sure the units are squared for area. The sum of the length and width must be 200, so if the length is x, the width must be 200 – x.

c. The length will be the x-values on the horizontal and the area will be the y-values on the vertical. The length cannot pass 200 since the sum of the length and width is 200. Set the intervals for the x-axis to 25 feet. Extend the table of values to find more points to plot on the graph. It appears that 10,000 is the greatest area, so set the intervals

for the y-axis to 1000 ft2. After graphing, it appears that the highest point on the graph is at an area of 10,000 ft

2.

d. The associated x-value with the maximum area is x = 100, so the length must be 100 and the width must be 200 – 100, or 100. The length and width of the rectangle must be 100 feet each to have the largest area.

61. CCSS CRITIQUE Cheyenne and Sebastian are finding (2x2 − x) − (3x + 3x

2 − 2). Is either of them correct? Explain your reasoning.

SOLUTION:

Neither is correct. Cheyenne, did not distribute the negative to the 2nd and 3rd terms when she found the additive inverse. Sebastian did not distribute the negate to the 3rd terms when he found the additive inverse. To find the additive inverse, all terms should be multiplied by −1.

62. REASONING Determine whether each of the following statements is true or false . Explain your reasoning. a. A binomial can have a degree of zero. b. The order in which polynomials are subtracted does not matter.

SOLUTION: a. If a binomial has two terms that are each a degree of 0, then those terms can be combined and the binomial becomes a monomial. For example, 18 + 7 = 25. If one of the terms of the binomial does not have a degree of 0, then the binomial cannot have a degree of 0, since the degree of a polynomial is the greatest degree of any term in the polynomial. b. Subtraction is not commutative. While 2 + 5 = 5 + 2, 2 – 5 ≠ 5 – 2. This is also true for polynomials. Sample answer: (2x – 3) – (4x – 3) = –2x, but (4x – 3) – (2x – 3) = 2x

63. CHALLENGE Write a polynomial that represents the sum of an odd integer 2n + 1 and the next two consecutive odd integers.

SOLUTION:

64. WRITING IN MATH Why would you add or subtract equations that represent real-world situations? Explain.

SOLUTION:

65. WRITING IN MATH Describe how to add and subtract polynomials using both the vertical and horizontal formats.

SOLUTION: To add polynomials in a horizontal format, you combine like terms. For the vertical format, you write the polynomials in standard form, align like terms in columns, and combine like terms.

To subtract polynomials in a horizontal format you find the additive inverse of the polynomial you are subtracting, andthen combine like terms. For the vertical format, you write the polynomials in standard form, align like terms in columns, and subtract by adding the additive inverse.

66. Three consecutive integers can be represented by x, x + 1, and x + 2. What is the sum of these three integers? A x(x + 1)(x + 2)

B x3 + 3

C 3x + 3 D x + 3

SOLUTION:

The correct choice is C.

67. SHORT RESPONSE What is the perimeter of a square with sides that measure 2x + 3 units?

SOLUTION:

The perimeter of the square is 8x + 12 units.

68. Jim cuts a board in the shape of a regular hexagon and pounds in a nail at each vertex, as shown. How many rubber bands will he need to stretch a rubber band across every possible pair of nails?

F 15 G 14 H 12 J 9

SOLUTION: The first nail would connect to 5 others, the second to 4 others, the third to 3 others, etc.5 + 4 + 3 + 2 + 1 = 15 The correct choice is F.

69. Which ordered pair is in the solution set of the system of inequalities shown in the graph?

A (−3, 0) B (0, −3) C (5, 0) D (0, 5)

SOLUTION: Choice A is outside the shaded area for both inequalities. Choices B and D are inside the shaded area for only one inequality. Choice C is the only point in the solution for both inequalities. So, the correct choice is C.

70. COMPUTERS A computer technician charges by the hour to fix and repair computer equipment. The total cost of the technician for one hour is $75, for two hours is $125, for three hours is $175, for four hours is $225, and so on. Write a recursive formula for the sequence.

SOLUTION: Write out the terms. $75, $125, $175, $225, ... The first term is 75, and 50 is added to form each following term. Therefore, we have a1 = 75, an = an – 1 + 50, n ≥ 2.

Determine whether each sequence is arithmetic, geometric, or neither. Explain.71. 8, –32, 128, –512, ...

SOLUTION: Check for a common difference. –32 – 8 = –40 128 – (–32) = 160 There is no common difference. Check for a common ratio. –32 ÷ 8 = –4 128 ÷ (–32) = –4 Geometric; the common ratio is –4.

72. 25, 8, –9, –26, ...

SOLUTION: Check for a common difference. 8 – 25 = –17 –9 – 8 = –17 Arithmetic; the common difference is –17.

73.

SOLUTION: Check for a common difference.

There is no common difference. Check for a common ratio.

There is no common ratio, so the sequence is not arithmetic or geometric.

74. 43, 52, 61, 70, ...

SOLUTION: Check for a common difference. 52 – 43 = 9 61 – 52 = 9 Arithmetic; the common difference is 9.

75. –27, –16, –5, 6, ...

SOLUTION: Check for a common difference. –16 – (–27) = 11 –5 – (–16) = 11 Arithmetic; the common difference is 11.

76. 200, 100, 50, 25, …

SOLUTION: Check for a common difference. 100 – 200 = –100 50 – 100 = –50 There is no common difference. Check for a common ratio. 100 ÷ 200 = 0.5 50 ÷ 100 = 0.5

Geometric; the common ratio is 0.5 or .

77. JOBS Kimi received an offer for a new job. She wants to compare the offer with her current job. What is total amount of sales that Kimi must get each month to make the same income at either job?

SOLUTION: Let s be Kimi’s monthly sales.

Kimi must sell $80,000 each month to make the same income at either job.

Determine whether each sequence is an arithmetic sequence. If it is, state the common difference.78. 24, 16, 8, 0, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is –8, because 16 – 24 = –8; 8 –16 = –8; etc.

79. , 13, 26, …

SOLUTION: Find the difference between the terms.

– =

13 – =

26 – 13 = 13 There is not a common different. The sequence is not an arithmetic sequence

80. 7, 6, 5, 4, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is –1, because 6 – 7 = –1; 5 – 6 = –1; 4 – 5 = –1;etc.

81. 10, 12, 15, 18, …

SOLUTION: Find the difference between the terms. 12 – 10 = 2 15 – 12 = 3 18 – 15 = 3 There is no common difference. The sequence is not an arithmetic sequence.

82. −15, −11, −7, −3, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is 4, because –11 – (–15) = 4; –7 – (–11) = 4; –3 – (–7) = 4;etc.

83. −0.3, 0.2, 0.7, 1.2, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is 0.5, because 0.2 – (–0.3) = 0.5; 0.7 – 0.2 = 0.5; 1.2 – 0.7 = 0.5; etc.

Simplify.

84. t(t5)(t

7)

SOLUTION:

85. n3(n

2)(−2n

3)

SOLUTION:

86. (5t5v

2)(10t

3v

4)

SOLUTION:

87. (−8u4z

5)(5uz

4)

SOLUTION:

88. [(3)2]3

SOLUTION:

89. [(2)3]2

SOLUTION:

90. (2m4k

3)2(−3mk

2)3

SOLUTION:

91. (6xy2)2(2x

2y

2z

2)3

SOLUTION:

eSolutions Manual - Powered by Cognero Page 9

8-1 Adding and Subtracting Polynomials

Determine whether each expression is a polynomial. If it is a polynomial, find the degree and determine whether it is a monomial, binomial, or trinomial.

1. 7ab + 6b2 – 2a

3

SOLUTION:

A polynomial is a monomial or the sum of monomials. 7ab + 6b2 – 2a

3 is the sum of 3 monomials, so it is a

polynomial. The degree of a polynomial is the greatest degree of any term in the polynomial.The degree of each term is 2, 2, and

3, so the degree of 7ab + 6b2 – 2a

3 is 3. The polynomial has three terms, so it is a trinomial.

2. 2y – 5 + 3y2

SOLUTION:

2y – 5 + 3y2

is the sum of monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 1, 0, and

2, so the degree of 2y – 5 + 3y2 is 2. The polynomial has three terms, so it is a trinomial.

3. 3x2

SOLUTION:

A polynomial is a monomial or the sum of monomials, so 3x2 is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of 3x2 is 2. The

polynomial has one term, so it is a monomial.

4.

SOLUTION: A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer exponents. It has

only one term. is a division of two monomials, so it is not a monomial.

5. 5m2p

3 + 6

SOLUTION:

A polynomial is a monomial or the sum of monomials. 5m2p

3 + 6

is the sum of 2 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 5 and 0,

so the degree of 5m2p

3 + 6 is 5. The polynomial has two terms, so it is a binomial.

6. 5q–4

+ 6q

SOLUTION: A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer exponents. It has

only one term. 5q-4 is equal to , which is a division of two monomials, so it is not a monomial.

Write each polynomial in standard form. Identify the leading coefficient.

7. –4d4 + 1 – d

2

SOLUTION: Find the degree of each term.

–4d4→ 4

1 → 0

–d2 → 2

The greatest degree is 4, from the term –4d4, so the leading coefficient of –4d

4 + 1 – d

2 is –4.

Rewrite the polynomial with each monomial in descending order according to degree.

–4d4 – d

2 + 1

8. 2x5 – 12 + 3x

SOLUTION: Find the degree of each term.

2x5 → 5

–12 → 0 3x → 1

The greatest degree is 5, from the term 2x5, so the leading coefficient of 2x

5 – 12 + 3x is 2.

Rewrite the polynomial with each monomial in descending order according to degree.

2x5 + 3x – 12

9. 4z – 2z2 – 5z

4

SOLUTION: Find the degree of each term. 4z → 1

– 2z2 → 2

– 5z4→ 4

The greatest degree is 4, from the term – 5z4, so the leading coefficient of 4z – 2z

2 – 5z

4 is –5.

Rewrite the polynomial with each monomial in descending order according to degree.

–5z4 – 2z

2 + 4z

10. 2a + 4a3 – 5a

2 – 1

SOLUTION: Find the degree of each term. 2a → 1

4a3→ 3

– 5a2 → 2

– 1 → 0

The greatest degree is 3, from the term 4a3, so the leading coefficient of 2a + 4a

3 – 5a

2 – 1 is 4.

Rewrite the polynomial with each monomial in descending order according to degree.

4a3– 5a

2 + 2a – 1

Find each sum or difference.

11. (6x3 − 4) + (−2x

3 + 9)

SOLUTION:

12. (g3 − 2g2 + 5g + 6) − (g2 + 2g)

SOLUTION:

13. (4 + 2a2 − 2a) − (3a

2 − 8a + 7)

SOLUTION:

14. (8y − 4y2) + (3y − 9y

2)

SOLUTION:

15. (−4z3 − 2z + 8) − (4z

3 + 3z2 − 5)

SOLUTION:

16. (−3d2 − 8 + 2d) + (4d − 12 + d2

)

SOLUTION:

17. (y + 5) + (2y + 4y2 – 2)

SOLUTION:

18. (3n3 − 5n + n2

) − (−8n2 + 3n

3)

SOLUTION:

19. CCSS SENSE-MAKING The total number of students T who traveled for spring break consists of two groups: students who flew to their destinations F and students who drove to their destination D. The number (in thousands) of students who flew and the total number of students who flew or drove can be modeled by the following equations,where n is the number of years since 1995. T = 14n + 21

F = 8n + 7 a. Write an equation that models the number of students who drove to their destination for this time period. b. Predict the number of students who will drive to their destination in 2012. c. How many students will drive or fly to their destination in 2015?

SOLUTION: a.

D = 6n + 14 b. n = 2012 – 1995 = 17

The number of students who will drive to their destination in 2012 is 116,000 students. c. n = 2015 – 1995 = 20

The number of students who will drive or fly to their destination in 2015 is 301,000 students.

Determine whether each expression is a polynomial. If it is a polynomial, find the degree and determine whether it is a monomial, binomial, or trinomial.

20.

SOLUTION: A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer exponents. It has

only one term. is a division of two monomials, so it is not a monomial.

21.

SOLUTION: A polynomial is a monomial or the sum of monomials. 21 is a monomial, so it is also a polynomial. The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of 21 is 0. The polynomial has only one term, so it is a monomial.

22. c4 – 2c

2 + 1

SOLUTION:

A polynomial is a monomial or the sum of monomials. c4 – 2c

2 + 1

is the sum of 3 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 4, 2, and

0, so the degree of c4 – 2c2 + 1 is 4. The polynomial has three terms, so it is a trinomial.

23. d + 3dc

SOLUTION: A polynomial is a monomial or the sum of monomials. A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer

exponents. 3dc

has a variable in the exponent, so it is not a monomial.

24. a – a2

SOLUTION:

A polynomial is a monomial or the sum of monomials. a – a2 is the sum of 2 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 1 and 2,

so the degree of a – a2 is 2. The polynomial has two terms, so it is a binomial.

25. 5n3 + nq

3

SOLUTION:

A polynomial is a monomial or the sum of monomials. 5n3 + nq3 is the sum of 2 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 3 and 4,

so the degree of 5n3 + nq3 is 4. The polynomial has two terms, so it is a binomial.

Write each polynomial in standard form. Identify the leading coefficient.

26. 5x2 – 2 + 3x

SOLUTION: Find the degree of each term.

5x2 → 2

– 2 → 0

3x → 1

The greatest degree is 2, from the term 5x2, so the leading coefficient of 5x

2 – 2 + 3x is 5.

Rewrite the polynomial with each monomial in descending order according to degree.

5x2 + 3x – 2

27. 8y + 7y3

SOLUTION: Find the degree of each term.

7y3 → 3

8y → 1

The greatest degree is 3, from the term 7y3, so the leading coefficient of 8y + 7y

3 is 7.

Rewrite the polynomial with each monomial in descending order according to degree.

7y3 + 8y

28. 4 – 3c – 5c2

SOLUTION: Find the degree of each term. 4 → 0 3c → 1

– 5c2 → 2

The greatest degree is 2, from the term – 5c2, so the leading coefficient of 4 – 3c – 5c

2 is –5.

Rewrite the polynomial with each monomial in descending order according to degree.

–5c2 – 3c + 4

29. –y3 + 3y – 3y

2 + 2

SOLUTION: Find the degree of each term.

–y3 → 3

3y → 1

3y2 → 2

2 → 0

The greatest degree is 3, from the term –y3, so the leading coefficient of –y

3 + 3y – 3y

2 + 2 is –1.

Rewrite the polynomial with each monomial in descending order according to degree. –y 3 – 3y 2 + 3y + 2

30. 11t + 2t2 – 3 + t

5

SOLUTION: Find the degree of each term. 11t → 1

2t2 → 2

–3 → 0

t5 → 5

The greatest degree is 5, from the term t5, so the leading coefficient of 11t + 2t2 – 3 + t

5 is 1.

Rewrite the polynomial with each monomial in descending order according to degree.

t5 + 2t

2 + 11t – 3

31. 2 + r – r3

SOLUTION: Find the degree of each term. 2 → 0 r → 1

– r3→ 3

The greatest degree is 3, from the term – r3, so the leading coefficient of 2 + r – r3

is –1. Rewrite the polynomial with each monomial in descending order according to degree.

–r3 + r + 2

32.

SOLUTION: Find the degree of each term.

→ 0

–3x4 → 4

7 → 0

The greatest degree is 4, from the term –3x4, so the leading coefficient of is –3.

Rewrite the polynomial with each monomial in descending order according to degree.

33. –9b2 + 10b – b

6

SOLUTION: Find the degree of each term.

–9b2 → 2

10b → 1

–b6→ 6

The greatest degree is 6, from the term –b6, so the leading coefficient of –9b2 + 10b – b6 is –1. Rewrite the polynomial with each monomial in descending order according to degree.

–b6 – 9b

2 + 10b

Find each sum or difference.

34. (2c2 + 6c + 4) + (5c – 7)

SOLUTION:

35. (2x + 3x2) − (7 − 8x

2)

SOLUTION:

36. (3c3 − c + 11) − (c2 + 2c + 8)

SOLUTION:

37. (z2 + z) + (z

2 − 11)

SOLUTION:

38. (2x − 2y + 1) − (3y + 4x)

SOLUTION:

39. (4a − 5b2 + 3) + (6 − 2a + 3b

2)

SOLUTION:

40. (x2y − 3x

2 + y) + (3y − 2x2y)

SOLUTION:

41. (−8xy + 3x2 − 5y) + (4x

2 − 2y + 6xy)

SOLUTION:

42. (5n − 2p2 + 2np) − (4p

2 + 4n)

SOLUTION:

43. (4rxt − 8r2x + x2

) − (6rx2 + 5rxt − 2x

2)

SOLUTION:

44. PETS From 1999 through 2009, the number of dogs D and the number of cats C (in hundreds) adopted from animal

shelters in the United States are modeled by the equations D = 2n + 3 and C = n + 4, where n is the number of yearssince 1999. a. Write an equation that models the total number T of dogs and cats adopted in hundreds for this time period. b. If this trend continues, how many dogs and cats will be adopted in 2013?

SOLUTION: a.

So, an equation that models the total number of dogs and cats adopted is T = 3n + 7. b. Evaluate the equation for the total number of dogs and cats for n = 2013 – 1999 = 14.

The number of cats and dogs adopted in 2013 will be 49 × 100 or 4900 cats and dogs.

Classify each polynomial according to its degree and number of terms.

45. 4x – 3x2 + 5

SOLUTION:

Find the degree of each term of 4x – 3x2 + 5.

4x → 1

– 3x2 → 2

5 → 0

The greatest degree is 2 and there are 3 terms, so 4x – 3x2 + 5 is a quadratic trinomial.

46. 11z3

SOLUTION: Find the degree of each term. 11z3

→ 3

The greatest degree is 3 and there is one term, so 11z3 is a cubic monomial.

47. 9 + y4

SOLUTION:

Find the degree of each term of 9 + y 4.

9 → 1

y4→ 4

The greatest degree is 4 and there are 2 terms, so 9 + y 4 is a quartic binomial.

48. 3x3 – 7

SOLUTION:

Find the degree of each term of 3x3 – 7.

3x3 → 3

–7 → 0

The greatest degree is 3 and there are 2 terms, so 3x3 – 7 is a cubic binomial.

49. –2x5 – x

2 + 5x – 8

SOLUTION:

Find the degree of each term of –2x5 – x2 + 5x – 8.

–2x5 → 5

–x2 → 2

5x → 1 – 8 → 0

The greatest degree is 5 and there are 4 terms, so –2x5 – x2 + 5x – 8 is a quintic polynomial.

50. 10t – 4t2 + 6t

3

SOLUTION:

Find the degree of each term of 10t – 4t2 + 6t

3.

10t → 1

4t2 → 2

6t3 → 3

The greatest degree is 3 and there are 3 terms, so 10t – 4t2 + 6t

3 is a cubic trinomial.

51. ENROLLMENT In a rapidly growing school system, the numbers (in hundreds) of total students N and

K-5 students P enrolled from 2000 to 2009 are modeled by the equations N = 1.25t2 – t + 7.5 and P =

0.7t2 – 0.95t + 3.8, where t is the number of years since 2000.

a. Write an equation modeling the number of 6-12 students S enrolled for this time period. b. How many 6-12 students were enrolled in the school system in 2007?

SOLUTION: a. To write an equation that represents the number of 6-12 students enrolled, subtract the equations that represent the total number of students and the number of K-5 students.

b. Replace t with 7 in the equation for S to determine the number of students enrolled in 6-12 in 2007.

In 2007 there were 30.3 hundreds or 3030 students enrolled in 6-12.

52. CCSS REASONING The perimeter of the figure shown is represented by the expression 3x2 − 7x + 2. Write a

polynomial that represents the measure of the third side.

SOLUTION:

53. GEOMETRY Consider the rectangle.

a. What does (4x2 + 2x – 1)(2x

2 – x + 3) represent?

b. What does 2(4x2 + 2x – 1) + 2(2x

2 – x + 3) represent?

SOLUTION:

a. (4x2 + 2x – 1)(2x

2 – x + 3) is a multiplication of the length and the width of the rectangle, which is the formula for

the area of a rectangle.

b. 2(4x2 + 2x – 1) + 2(2x

2 – x + 3) is the sum of twice the length and twice the width of a rectangle, which is the

formula for the perimeter of the rectangle

Find each sum or difference.

54. (4x + 2y − 6z) + (5y − 2z + 7x) + (−9z − 2x − 3y)

SOLUTION:

55. (5a2 − 4) + (a

2 − 2a + 12) + (4a2 − 6a + 8)

SOLUTION:

56. (3c2 − 7) + (4c + 7) − (c

2 + 5c − 8)

SOLUTION:

57. (3n3 + 3n − 10) − (4n

2 − 5n) + (4n3 − 3n

2 − 9n + 4)

SOLUTION:

58. FOOTBALL The National Football League is divided into two conferences, the American A and the National N. From 2002 through 2009, the total attendance T (in thousands) for both conferences and for the American Conference games are modeled by the following equations, where x is the number of years since 2002.

T = –0.69x3 + 55.83x

2 + 643.31x + 10,538

A = –3.78x3 + 58.96x

2 + 265.96x + 5257

Determine how many people attended National Conference football games in 2009.

SOLUTION:

Let x = 7 represent 2009, then find how many people attended a National Conference football game in 2009.

In 2009 the number of people who attended National Conference football games was about 8829 thousand. Multiply by 1000 to find the attendance in standard form. 8829 ×1000 = 8,829,000. So, about 8,829,000 people attended National Conference football games in 2009.

59. CAR RENTAL The cost to rent a car for a day is $15 plus $0.15 for each mile driven. a. Write a polynomial that represents the cost of renting a car for m miles. b. If a car is driven 145 miles, how much would it cost to rent? c. If a car is driven 105 miles each day for four days, how much would it cost to rent a car? d. If a car is driven 220 miles each day for seven days, how much would it cost to rent a car?

SOLUTION: a. The cost to rent a car is the daily rate and the mileage cost or 15 + 0.15m. b. Substitute 145 for m to find the daily cost to drive 145 miles.

The cost to rent the car would be $36.75. c. The expression represents the cost per day. For 4 days, multiply the entire expression by 4 to find the cost for a 4 day trip. Substitute 105 for m.

The cost to rent the car would be $123. d. The expression represents the cost per day. For 7 days, multiply the entire expression by 7 to find the cost for a 7 day trip. Substitute 220 for m.

The cost to rent the car would be $336.

60. MULTIPLE REPRESENTATIONS In this problem, you will explore perimeter and area. a. Geometric Draw three rectangles that each have a perimeter of 400 feet. b. Tabular Record the width and length of each rectangle in a table like the one shown below. Find the area of eachrectangle.

c. Graphical On a coordinate system, graph the area of rectangle 4 in terms of the length, x. Use the graph to determine the largest area possible. d. Analytical Determine the length and width that produce the largest area.

SOLUTION: a. For the perimeters to be 400, create lengths and widths that sum to 200. Note that some lengths are already provided in the table in part b.

b. The area is length multiplied by width. Make sure the units are squared for area. The sum of the length and width must be 200, so if the length is x, the width must be 200 – x.

c. The length will be the x-values on the horizontal and the area will be the y-values on the vertical. The length cannot pass 200 since the sum of the length and width is 200. Set the intervals for the x-axis to 25 feet. Extend the table of values to find more points to plot on the graph. It appears that 10,000 is the greatest area, so set the intervals

for the y-axis to 1000 ft2. After graphing, it appears that the highest point on the graph is at an area of 10,000 ft

2.

d. The associated x-value with the maximum area is x = 100, so the length must be 100 and the width must be 200 – 100, or 100. The length and width of the rectangle must be 100 feet each to have the largest area.

61. CCSS CRITIQUE Cheyenne and Sebastian are finding (2x2 − x) − (3x + 3x

2 − 2). Is either of them correct? Explain your reasoning.

SOLUTION:

Neither is correct. Cheyenne, did not distribute the negative to the 2nd and 3rd terms when she found the additive inverse. Sebastian did not distribute the negate to the 3rd terms when he found the additive inverse. To find the additive inverse, all terms should be multiplied by −1.

62. REASONING Determine whether each of the following statements is true or false . Explain your reasoning. a. A binomial can have a degree of zero. b. The order in which polynomials are subtracted does not matter.

SOLUTION: a. If a binomial has two terms that are each a degree of 0, then those terms can be combined and the binomial becomes a monomial. For example, 18 + 7 = 25. If one of the terms of the binomial does not have a degree of 0, then the binomial cannot have a degree of 0, since the degree of a polynomial is the greatest degree of any term in the polynomial. b. Subtraction is not commutative. While 2 + 5 = 5 + 2, 2 – 5 ≠ 5 – 2. This is also true for polynomials. Sample answer: (2x – 3) – (4x – 3) = –2x, but (4x – 3) – (2x – 3) = 2x

63. CHALLENGE Write a polynomial that represents the sum of an odd integer 2n + 1 and the next two consecutive odd integers.

SOLUTION:

64. WRITING IN MATH Why would you add or subtract equations that represent real-world situations? Explain.

SOLUTION:

65. WRITING IN MATH Describe how to add and subtract polynomials using both the vertical and horizontal formats.

SOLUTION: To add polynomials in a horizontal format, you combine like terms. For the vertical format, you write the polynomials in standard form, align like terms in columns, and combine like terms.

To subtract polynomials in a horizontal format you find the additive inverse of the polynomial you are subtracting, andthen combine like terms. For the vertical format, you write the polynomials in standard form, align like terms in columns, and subtract by adding the additive inverse.

66. Three consecutive integers can be represented by x, x + 1, and x + 2. What is the sum of these three integers? A x(x + 1)(x + 2)

B x3 + 3

C 3x + 3 D x + 3

SOLUTION:

The correct choice is C.

67. SHORT RESPONSE What is the perimeter of a square with sides that measure 2x + 3 units?

SOLUTION:

The perimeter of the square is 8x + 12 units.

68. Jim cuts a board in the shape of a regular hexagon and pounds in a nail at each vertex, as shown. How many rubber bands will he need to stretch a rubber band across every possible pair of nails?

F 15 G 14 H 12 J 9

SOLUTION: The first nail would connect to 5 others, the second to 4 others, the third to 3 others, etc.5 + 4 + 3 + 2 + 1 = 15 The correct choice is F.

69. Which ordered pair is in the solution set of the system of inequalities shown in the graph?

A (−3, 0) B (0, −3) C (5, 0) D (0, 5)

SOLUTION: Choice A is outside the shaded area for both inequalities. Choices B and D are inside the shaded area for only one inequality. Choice C is the only point in the solution for both inequalities. So, the correct choice is C.

70. COMPUTERS A computer technician charges by the hour to fix and repair computer equipment. The total cost of the technician for one hour is $75, for two hours is $125, for three hours is $175, for four hours is $225, and so on. Write a recursive formula for the sequence.

SOLUTION: Write out the terms. $75, $125, $175, $225, ... The first term is 75, and 50 is added to form each following term. Therefore, we have a1 = 75, an = an – 1 + 50, n ≥ 2.

Determine whether each sequence is arithmetic, geometric, or neither. Explain.71. 8, –32, 128, –512, ...

SOLUTION: Check for a common difference. –32 – 8 = –40 128 – (–32) = 160 There is no common difference. Check for a common ratio. –32 ÷ 8 = –4 128 ÷ (–32) = –4 Geometric; the common ratio is –4.

72. 25, 8, –9, –26, ...

SOLUTION: Check for a common difference. 8 – 25 = –17 –9 – 8 = –17 Arithmetic; the common difference is –17.

73.

SOLUTION: Check for a common difference.

There is no common difference. Check for a common ratio.

There is no common ratio, so the sequence is not arithmetic or geometric.

74. 43, 52, 61, 70, ...

SOLUTION: Check for a common difference. 52 – 43 = 9 61 – 52 = 9 Arithmetic; the common difference is 9.

75. –27, –16, –5, 6, ...

SOLUTION: Check for a common difference. –16 – (–27) = 11 –5 – (–16) = 11 Arithmetic; the common difference is 11.

76. 200, 100, 50, 25, …

SOLUTION: Check for a common difference. 100 – 200 = –100 50 – 100 = –50 There is no common difference. Check for a common ratio. 100 ÷ 200 = 0.5 50 ÷ 100 = 0.5

Geometric; the common ratio is 0.5 or .

77. JOBS Kimi received an offer for a new job. She wants to compare the offer with her current job. What is total amount of sales that Kimi must get each month to make the same income at either job?

SOLUTION: Let s be Kimi’s monthly sales.

Kimi must sell $80,000 each month to make the same income at either job.

Determine whether each sequence is an arithmetic sequence. If it is, state the common difference.78. 24, 16, 8, 0, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is –8, because 16 – 24 = –8; 8 –16 = –8; etc.

79. , 13, 26, …

SOLUTION: Find the difference between the terms.

– =

13 – =

26 – 13 = 13 There is not a common different. The sequence is not an arithmetic sequence

80. 7, 6, 5, 4, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is –1, because 6 – 7 = –1; 5 – 6 = –1; 4 – 5 = –1;etc.

81. 10, 12, 15, 18, …

SOLUTION: Find the difference between the terms. 12 – 10 = 2 15 – 12 = 3 18 – 15 = 3 There is no common difference. The sequence is not an arithmetic sequence.

82. −15, −11, −7, −3, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is 4, because –11 – (–15) = 4; –7 – (–11) = 4; –3 – (–7) = 4;etc.

83. −0.3, 0.2, 0.7, 1.2, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is 0.5, because 0.2 – (–0.3) = 0.5; 0.7 – 0.2 = 0.5; 1.2 – 0.7 = 0.5; etc.

Simplify.

84. t(t5)(t

7)

SOLUTION:

85. n3(n

2)(−2n

3)

SOLUTION:

86. (5t5v

2)(10t

3v

4)

SOLUTION:

87. (−8u4z

5)(5uz

4)

SOLUTION:

88. [(3)2]3

SOLUTION:

89. [(2)3]2

SOLUTION:

90. (2m4k

3)2(−3mk

2)3

SOLUTION:

91. (6xy2)2(2x

2y

2z

2)3

SOLUTION:

eSolutions Manual - Powered by Cognero Page 10

8-1 Adding and Subtracting Polynomials

Determine whether each expression is a polynomial. If it is a polynomial, find the degree and determine whether it is a monomial, binomial, or trinomial.

1. 7ab + 6b2 – 2a

3

SOLUTION:

A polynomial is a monomial or the sum of monomials. 7ab + 6b2 – 2a

3 is the sum of 3 monomials, so it is a

polynomial. The degree of a polynomial is the greatest degree of any term in the polynomial.The degree of each term is 2, 2, and

3, so the degree of 7ab + 6b2 – 2a

3 is 3. The polynomial has three terms, so it is a trinomial.

2. 2y – 5 + 3y2

SOLUTION:

2y – 5 + 3y2

is the sum of monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 1, 0, and

2, so the degree of 2y – 5 + 3y2 is 2. The polynomial has three terms, so it is a trinomial.

3. 3x2

SOLUTION:

A polynomial is a monomial or the sum of monomials, so 3x2 is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of 3x2 is 2. The

polynomial has one term, so it is a monomial.

4.

SOLUTION: A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer exponents. It has

only one term. is a division of two monomials, so it is not a monomial.

5. 5m2p

3 + 6

SOLUTION:

A polynomial is a monomial or the sum of monomials. 5m2p

3 + 6

is the sum of 2 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 5 and 0,

so the degree of 5m2p

3 + 6 is 5. The polynomial has two terms, so it is a binomial.

6. 5q–4

+ 6q

SOLUTION: A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer exponents. It has

only one term. 5q-4 is equal to , which is a division of two monomials, so it is not a monomial.

Write each polynomial in standard form. Identify the leading coefficient.

7. –4d4 + 1 – d

2

SOLUTION: Find the degree of each term.

–4d4→ 4

1 → 0

–d2 → 2

The greatest degree is 4, from the term –4d4, so the leading coefficient of –4d

4 + 1 – d

2 is –4.

Rewrite the polynomial with each monomial in descending order according to degree.

–4d4 – d

2 + 1

8. 2x5 – 12 + 3x

SOLUTION: Find the degree of each term.

2x5 → 5

–12 → 0 3x → 1

The greatest degree is 5, from the term 2x5, so the leading coefficient of 2x

5 – 12 + 3x is 2.

Rewrite the polynomial with each monomial in descending order according to degree.

2x5 + 3x – 12

9. 4z – 2z2 – 5z

4

SOLUTION: Find the degree of each term. 4z → 1

– 2z2 → 2

– 5z4→ 4

The greatest degree is 4, from the term – 5z4, so the leading coefficient of 4z – 2z

2 – 5z

4 is –5.

Rewrite the polynomial with each monomial in descending order according to degree.

–5z4 – 2z

2 + 4z

10. 2a + 4a3 – 5a

2 – 1

SOLUTION: Find the degree of each term. 2a → 1

4a3→ 3

– 5a2 → 2

– 1 → 0

The greatest degree is 3, from the term 4a3, so the leading coefficient of 2a + 4a

3 – 5a

2 – 1 is 4.

Rewrite the polynomial with each monomial in descending order according to degree.

4a3– 5a

2 + 2a – 1

Find each sum or difference.

11. (6x3 − 4) + (−2x

3 + 9)

SOLUTION:

12. (g3 − 2g2 + 5g + 6) − (g2 + 2g)

SOLUTION:

13. (4 + 2a2 − 2a) − (3a

2 − 8a + 7)

SOLUTION:

14. (8y − 4y2) + (3y − 9y

2)

SOLUTION:

15. (−4z3 − 2z + 8) − (4z

3 + 3z2 − 5)

SOLUTION:

16. (−3d2 − 8 + 2d) + (4d − 12 + d2

)

SOLUTION:

17. (y + 5) + (2y + 4y2 – 2)

SOLUTION:

18. (3n3 − 5n + n2

) − (−8n2 + 3n

3)

SOLUTION:

19. CCSS SENSE-MAKING The total number of students T who traveled for spring break consists of two groups: students who flew to their destinations F and students who drove to their destination D. The number (in thousands) of students who flew and the total number of students who flew or drove can be modeled by the following equations,where n is the number of years since 1995. T = 14n + 21

F = 8n + 7 a. Write an equation that models the number of students who drove to their destination for this time period. b. Predict the number of students who will drive to their destination in 2012. c. How many students will drive or fly to their destination in 2015?

SOLUTION: a.

D = 6n + 14 b. n = 2012 – 1995 = 17

The number of students who will drive to their destination in 2012 is 116,000 students. c. n = 2015 – 1995 = 20

The number of students who will drive or fly to their destination in 2015 is 301,000 students.

Determine whether each expression is a polynomial. If it is a polynomial, find the degree and determine whether it is a monomial, binomial, or trinomial.

20.

SOLUTION: A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer exponents. It has

only one term. is a division of two monomials, so it is not a monomial.

21.

SOLUTION: A polynomial is a monomial or the sum of monomials. 21 is a monomial, so it is also a polynomial. The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of 21 is 0. The polynomial has only one term, so it is a monomial.

22. c4 – 2c

2 + 1

SOLUTION:

A polynomial is a monomial or the sum of monomials. c4 – 2c

2 + 1

is the sum of 3 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 4, 2, and

0, so the degree of c4 – 2c2 + 1 is 4. The polynomial has three terms, so it is a trinomial.

23. d + 3dc

SOLUTION: A polynomial is a monomial or the sum of monomials. A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer

exponents. 3dc

has a variable in the exponent, so it is not a monomial.

24. a – a2

SOLUTION:

A polynomial is a monomial or the sum of monomials. a – a2 is the sum of 2 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 1 and 2,

so the degree of a – a2 is 2. The polynomial has two terms, so it is a binomial.

25. 5n3 + nq

3

SOLUTION:

A polynomial is a monomial or the sum of monomials. 5n3 + nq3 is the sum of 2 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 3 and 4,

so the degree of 5n3 + nq3 is 4. The polynomial has two terms, so it is a binomial.

Write each polynomial in standard form. Identify the leading coefficient.

26. 5x2 – 2 + 3x

SOLUTION: Find the degree of each term.

5x2 → 2

– 2 → 0

3x → 1

The greatest degree is 2, from the term 5x2, so the leading coefficient of 5x

2 – 2 + 3x is 5.

Rewrite the polynomial with each monomial in descending order according to degree.

5x2 + 3x – 2

27. 8y + 7y3

SOLUTION: Find the degree of each term.

7y3 → 3

8y → 1

The greatest degree is 3, from the term 7y3, so the leading coefficient of 8y + 7y

3 is 7.

Rewrite the polynomial with each monomial in descending order according to degree.

7y3 + 8y

28. 4 – 3c – 5c2

SOLUTION: Find the degree of each term. 4 → 0 3c → 1

– 5c2 → 2

The greatest degree is 2, from the term – 5c2, so the leading coefficient of 4 – 3c – 5c

2 is –5.

Rewrite the polynomial with each monomial in descending order according to degree.

–5c2 – 3c + 4

29. –y3 + 3y – 3y

2 + 2

SOLUTION: Find the degree of each term.

–y3 → 3

3y → 1

3y2 → 2

2 → 0

The greatest degree is 3, from the term –y3, so the leading coefficient of –y

3 + 3y – 3y

2 + 2 is –1.

Rewrite the polynomial with each monomial in descending order according to degree. –y 3 – 3y 2 + 3y + 2

30. 11t + 2t2 – 3 + t

5

SOLUTION: Find the degree of each term. 11t → 1

2t2 → 2

–3 → 0

t5 → 5

The greatest degree is 5, from the term t5, so the leading coefficient of 11t + 2t2 – 3 + t

5 is 1.

Rewrite the polynomial with each monomial in descending order according to degree.

t5 + 2t

2 + 11t – 3

31. 2 + r – r3

SOLUTION: Find the degree of each term. 2 → 0 r → 1

– r3→ 3

The greatest degree is 3, from the term – r3, so the leading coefficient of 2 + r – r3

is –1. Rewrite the polynomial with each monomial in descending order according to degree.

–r3 + r + 2

32.

SOLUTION: Find the degree of each term.

→ 0

–3x4 → 4

7 → 0

The greatest degree is 4, from the term –3x4, so the leading coefficient of is –3.

Rewrite the polynomial with each monomial in descending order according to degree.

33. –9b2 + 10b – b

6

SOLUTION: Find the degree of each term.

–9b2 → 2

10b → 1

–b6→ 6

The greatest degree is 6, from the term –b6, so the leading coefficient of –9b2 + 10b – b6 is –1. Rewrite the polynomial with each monomial in descending order according to degree.

–b6 – 9b

2 + 10b

Find each sum or difference.

34. (2c2 + 6c + 4) + (5c – 7)

SOLUTION:

35. (2x + 3x2) − (7 − 8x

2)

SOLUTION:

36. (3c3 − c + 11) − (c2 + 2c + 8)

SOLUTION:

37. (z2 + z) + (z

2 − 11)

SOLUTION:

38. (2x − 2y + 1) − (3y + 4x)

SOLUTION:

39. (4a − 5b2 + 3) + (6 − 2a + 3b

2)

SOLUTION:

40. (x2y − 3x

2 + y) + (3y − 2x2y)

SOLUTION:

41. (−8xy + 3x2 − 5y) + (4x

2 − 2y + 6xy)

SOLUTION:

42. (5n − 2p2 + 2np) − (4p

2 + 4n)

SOLUTION:

43. (4rxt − 8r2x + x2

) − (6rx2 + 5rxt − 2x

2)

SOLUTION:

44. PETS From 1999 through 2009, the number of dogs D and the number of cats C (in hundreds) adopted from animal

shelters in the United States are modeled by the equations D = 2n + 3 and C = n + 4, where n is the number of yearssince 1999. a. Write an equation that models the total number T of dogs and cats adopted in hundreds for this time period. b. If this trend continues, how many dogs and cats will be adopted in 2013?

SOLUTION: a.

So, an equation that models the total number of dogs and cats adopted is T = 3n + 7. b. Evaluate the equation for the total number of dogs and cats for n = 2013 – 1999 = 14.

The number of cats and dogs adopted in 2013 will be 49 × 100 or 4900 cats and dogs.

Classify each polynomial according to its degree and number of terms.

45. 4x – 3x2 + 5

SOLUTION:

Find the degree of each term of 4x – 3x2 + 5.

4x → 1

– 3x2 → 2

5 → 0

The greatest degree is 2 and there are 3 terms, so 4x – 3x2 + 5 is a quadratic trinomial.

46. 11z3

SOLUTION: Find the degree of each term. 11z3

→ 3

The greatest degree is 3 and there is one term, so 11z3 is a cubic monomial.

47. 9 + y4

SOLUTION:

Find the degree of each term of 9 + y 4.

9 → 1

y4→ 4

The greatest degree is 4 and there are 2 terms, so 9 + y 4 is a quartic binomial.

48. 3x3 – 7

SOLUTION:

Find the degree of each term of 3x3 – 7.

3x3 → 3

–7 → 0

The greatest degree is 3 and there are 2 terms, so 3x3 – 7 is a cubic binomial.

49. –2x5 – x

2 + 5x – 8

SOLUTION:

Find the degree of each term of –2x5 – x2 + 5x – 8.

–2x5 → 5

–x2 → 2

5x → 1 – 8 → 0

The greatest degree is 5 and there are 4 terms, so –2x5 – x2 + 5x – 8 is a quintic polynomial.

50. 10t – 4t2 + 6t

3

SOLUTION:

Find the degree of each term of 10t – 4t2 + 6t

3.

10t → 1

4t2 → 2

6t3 → 3

The greatest degree is 3 and there are 3 terms, so 10t – 4t2 + 6t

3 is a cubic trinomial.

51. ENROLLMENT In a rapidly growing school system, the numbers (in hundreds) of total students N and

K-5 students P enrolled from 2000 to 2009 are modeled by the equations N = 1.25t2 – t + 7.5 and P =

0.7t2 – 0.95t + 3.8, where t is the number of years since 2000.

a. Write an equation modeling the number of 6-12 students S enrolled for this time period. b. How many 6-12 students were enrolled in the school system in 2007?

SOLUTION: a. To write an equation that represents the number of 6-12 students enrolled, subtract the equations that represent the total number of students and the number of K-5 students.

b. Replace t with 7 in the equation for S to determine the number of students enrolled in 6-12 in 2007.

In 2007 there were 30.3 hundreds or 3030 students enrolled in 6-12.

52. CCSS REASONING The perimeter of the figure shown is represented by the expression 3x2 − 7x + 2. Write a

polynomial that represents the measure of the third side.

SOLUTION:

53. GEOMETRY Consider the rectangle.

a. What does (4x2 + 2x – 1)(2x

2 – x + 3) represent?

b. What does 2(4x2 + 2x – 1) + 2(2x

2 – x + 3) represent?

SOLUTION:

a. (4x2 + 2x – 1)(2x

2 – x + 3) is a multiplication of the length and the width of the rectangle, which is the formula for

the area of a rectangle.

b. 2(4x2 + 2x – 1) + 2(2x

2 – x + 3) is the sum of twice the length and twice the width of a rectangle, which is the

formula for the perimeter of the rectangle

Find each sum or difference.

54. (4x + 2y − 6z) + (5y − 2z + 7x) + (−9z − 2x − 3y)

SOLUTION:

55. (5a2 − 4) + (a

2 − 2a + 12) + (4a2 − 6a + 8)

SOLUTION:

56. (3c2 − 7) + (4c + 7) − (c

2 + 5c − 8)

SOLUTION:

57. (3n3 + 3n − 10) − (4n

2 − 5n) + (4n3 − 3n

2 − 9n + 4)

SOLUTION:

58. FOOTBALL The National Football League is divided into two conferences, the American A and the National N. From 2002 through 2009, the total attendance T (in thousands) for both conferences and for the American Conference games are modeled by the following equations, where x is the number of years since 2002.

T = –0.69x3 + 55.83x

2 + 643.31x + 10,538

A = –3.78x3 + 58.96x

2 + 265.96x + 5257

Determine how many people attended National Conference football games in 2009.

SOLUTION:

Let x = 7 represent 2009, then find how many people attended a National Conference football game in 2009.

In 2009 the number of people who attended National Conference football games was about 8829 thousand. Multiply by 1000 to find the attendance in standard form. 8829 ×1000 = 8,829,000. So, about 8,829,000 people attended National Conference football games in 2009.

59. CAR RENTAL The cost to rent a car for a day is $15 plus $0.15 for each mile driven. a. Write a polynomial that represents the cost of renting a car for m miles. b. If a car is driven 145 miles, how much would it cost to rent? c. If a car is driven 105 miles each day for four days, how much would it cost to rent a car? d. If a car is driven 220 miles each day for seven days, how much would it cost to rent a car?

SOLUTION: a. The cost to rent a car is the daily rate and the mileage cost or 15 + 0.15m. b. Substitute 145 for m to find the daily cost to drive 145 miles.

The cost to rent the car would be $36.75. c. The expression represents the cost per day. For 4 days, multiply the entire expression by 4 to find the cost for a 4 day trip. Substitute 105 for m.

The cost to rent the car would be $123. d. The expression represents the cost per day. For 7 days, multiply the entire expression by 7 to find the cost for a 7 day trip. Substitute 220 for m.

The cost to rent the car would be $336.

60. MULTIPLE REPRESENTATIONS In this problem, you will explore perimeter and area. a. Geometric Draw three rectangles that each have a perimeter of 400 feet. b. Tabular Record the width and length of each rectangle in a table like the one shown below. Find the area of eachrectangle.

c. Graphical On a coordinate system, graph the area of rectangle 4 in terms of the length, x. Use the graph to determine the largest area possible. d. Analytical Determine the length and width that produce the largest area.

SOLUTION: a. For the perimeters to be 400, create lengths and widths that sum to 200. Note that some lengths are already provided in the table in part b.

b. The area is length multiplied by width. Make sure the units are squared for area. The sum of the length and width must be 200, so if the length is x, the width must be 200 – x.

c. The length will be the x-values on the horizontal and the area will be the y-values on the vertical. The length cannot pass 200 since the sum of the length and width is 200. Set the intervals for the x-axis to 25 feet. Extend the table of values to find more points to plot on the graph. It appears that 10,000 is the greatest area, so set the intervals

for the y-axis to 1000 ft2. After graphing, it appears that the highest point on the graph is at an area of 10,000 ft

2.

d. The associated x-value with the maximum area is x = 100, so the length must be 100 and the width must be 200 – 100, or 100. The length and width of the rectangle must be 100 feet each to have the largest area.

61. CCSS CRITIQUE Cheyenne and Sebastian are finding (2x2 − x) − (3x + 3x

2 − 2). Is either of them correct? Explain your reasoning.

SOLUTION:

Neither is correct. Cheyenne, did not distribute the negative to the 2nd and 3rd terms when she found the additive inverse. Sebastian did not distribute the negate to the 3rd terms when he found the additive inverse. To find the additive inverse, all terms should be multiplied by −1.

62. REASONING Determine whether each of the following statements is true or false . Explain your reasoning. a. A binomial can have a degree of zero. b. The order in which polynomials are subtracted does not matter.

SOLUTION: a. If a binomial has two terms that are each a degree of 0, then those terms can be combined and the binomial becomes a monomial. For example, 18 + 7 = 25. If one of the terms of the binomial does not have a degree of 0, then the binomial cannot have a degree of 0, since the degree of a polynomial is the greatest degree of any term in the polynomial. b. Subtraction is not commutative. While 2 + 5 = 5 + 2, 2 – 5 ≠ 5 – 2. This is also true for polynomials. Sample answer: (2x – 3) – (4x – 3) = –2x, but (4x – 3) – (2x – 3) = 2x

63. CHALLENGE Write a polynomial that represents the sum of an odd integer 2n + 1 and the next two consecutive odd integers.

SOLUTION:

64. WRITING IN MATH Why would you add or subtract equations that represent real-world situations? Explain.

SOLUTION:

65. WRITING IN MATH Describe how to add and subtract polynomials using both the vertical and horizontal formats.

SOLUTION: To add polynomials in a horizontal format, you combine like terms. For the vertical format, you write the polynomials in standard form, align like terms in columns, and combine like terms.

To subtract polynomials in a horizontal format you find the additive inverse of the polynomial you are subtracting, andthen combine like terms. For the vertical format, you write the polynomials in standard form, align like terms in columns, and subtract by adding the additive inverse.

66. Three consecutive integers can be represented by x, x + 1, and x + 2. What is the sum of these three integers? A x(x + 1)(x + 2)

B x3 + 3

C 3x + 3 D x + 3

SOLUTION:

The correct choice is C.

67. SHORT RESPONSE What is the perimeter of a square with sides that measure 2x + 3 units?

SOLUTION:

The perimeter of the square is 8x + 12 units.

68. Jim cuts a board in the shape of a regular hexagon and pounds in a nail at each vertex, as shown. How many rubber bands will he need to stretch a rubber band across every possible pair of nails?

F 15 G 14 H 12 J 9

SOLUTION: The first nail would connect to 5 others, the second to 4 others, the third to 3 others, etc.5 + 4 + 3 + 2 + 1 = 15 The correct choice is F.

69. Which ordered pair is in the solution set of the system of inequalities shown in the graph?

A (−3, 0) B (0, −3) C (5, 0) D (0, 5)

SOLUTION: Choice A is outside the shaded area for both inequalities. Choices B and D are inside the shaded area for only one inequality. Choice C is the only point in the solution for both inequalities. So, the correct choice is C.

70. COMPUTERS A computer technician charges by the hour to fix and repair computer equipment. The total cost of the technician for one hour is $75, for two hours is $125, for three hours is $175, for four hours is $225, and so on. Write a recursive formula for the sequence.

SOLUTION: Write out the terms. $75, $125, $175, $225, ... The first term is 75, and 50 is added to form each following term. Therefore, we have a1 = 75, an = an – 1 + 50, n ≥ 2.

Determine whether each sequence is arithmetic, geometric, or neither. Explain.71. 8, –32, 128, –512, ...

SOLUTION: Check for a common difference. –32 – 8 = –40 128 – (–32) = 160 There is no common difference. Check for a common ratio. –32 ÷ 8 = –4 128 ÷ (–32) = –4 Geometric; the common ratio is –4.

72. 25, 8, –9, –26, ...

SOLUTION: Check for a common difference. 8 – 25 = –17 –9 – 8 = –17 Arithmetic; the common difference is –17.

73.

SOLUTION: Check for a common difference.

There is no common difference. Check for a common ratio.

There is no common ratio, so the sequence is not arithmetic or geometric.

74. 43, 52, 61, 70, ...

SOLUTION: Check for a common difference. 52 – 43 = 9 61 – 52 = 9 Arithmetic; the common difference is 9.

75. –27, –16, –5, 6, ...

SOLUTION: Check for a common difference. –16 – (–27) = 11 –5 – (–16) = 11 Arithmetic; the common difference is 11.

76. 200, 100, 50, 25, …

SOLUTION: Check for a common difference. 100 – 200 = –100 50 – 100 = –50 There is no common difference. Check for a common ratio. 100 ÷ 200 = 0.5 50 ÷ 100 = 0.5

Geometric; the common ratio is 0.5 or .

77. JOBS Kimi received an offer for a new job. She wants to compare the offer with her current job. What is total amount of sales that Kimi must get each month to make the same income at either job?

SOLUTION: Let s be Kimi’s monthly sales.

Kimi must sell $80,000 each month to make the same income at either job.

Determine whether each sequence is an arithmetic sequence. If it is, state the common difference.78. 24, 16, 8, 0, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is –8, because 16 – 24 = –8; 8 –16 = –8; etc.

79. , 13, 26, …

SOLUTION: Find the difference between the terms.

– =

13 – =

26 – 13 = 13 There is not a common different. The sequence is not an arithmetic sequence

80. 7, 6, 5, 4, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is –1, because 6 – 7 = –1; 5 – 6 = –1; 4 – 5 = –1;etc.

81. 10, 12, 15, 18, …

SOLUTION: Find the difference between the terms. 12 – 10 = 2 15 – 12 = 3 18 – 15 = 3 There is no common difference. The sequence is not an arithmetic sequence.

82. −15, −11, −7, −3, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is 4, because –11 – (–15) = 4; –7 – (–11) = 4; –3 – (–7) = 4;etc.

83. −0.3, 0.2, 0.7, 1.2, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is 0.5, because 0.2 – (–0.3) = 0.5; 0.7 – 0.2 = 0.5; 1.2 – 0.7 = 0.5; etc.

Simplify.

84. t(t5)(t

7)

SOLUTION:

85. n3(n

2)(−2n

3)

SOLUTION:

86. (5t5v

2)(10t

3v

4)

SOLUTION:

87. (−8u4z

5)(5uz

4)

SOLUTION:

88. [(3)2]3

SOLUTION:

89. [(2)3]2

SOLUTION:

90. (2m4k

3)2(−3mk

2)3

SOLUTION:

91. (6xy2)2(2x

2y

2z

2)3

SOLUTION:

eSolutions Manual - Powered by Cognero Page 11

8-1 Adding and Subtracting Polynomials

Determine whether each expression is a polynomial. If it is a polynomial, find the degree and determine whether it is a monomial, binomial, or trinomial.

1. 7ab + 6b2 – 2a

3

SOLUTION:

A polynomial is a monomial or the sum of monomials. 7ab + 6b2 – 2a

3 is the sum of 3 monomials, so it is a

polynomial. The degree of a polynomial is the greatest degree of any term in the polynomial.The degree of each term is 2, 2, and

3, so the degree of 7ab + 6b2 – 2a

3 is 3. The polynomial has three terms, so it is a trinomial.

2. 2y – 5 + 3y2

SOLUTION:

2y – 5 + 3y2

is the sum of monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 1, 0, and

2, so the degree of 2y – 5 + 3y2 is 2. The polynomial has three terms, so it is a trinomial.

3. 3x2

SOLUTION:

A polynomial is a monomial or the sum of monomials, so 3x2 is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of 3x2 is 2. The

polynomial has one term, so it is a monomial.

4.

SOLUTION: A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer exponents. It has

only one term. is a division of two monomials, so it is not a monomial.

5. 5m2p

3 + 6

SOLUTION:

A polynomial is a monomial or the sum of monomials. 5m2p

3 + 6

is the sum of 2 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 5 and 0,

so the degree of 5m2p

3 + 6 is 5. The polynomial has two terms, so it is a binomial.

6. 5q–4

+ 6q

SOLUTION: A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer exponents. It has

only one term. 5q-4 is equal to , which is a division of two monomials, so it is not a monomial.

Write each polynomial in standard form. Identify the leading coefficient.

7. –4d4 + 1 – d

2

SOLUTION: Find the degree of each term.

–4d4→ 4

1 → 0

–d2 → 2

The greatest degree is 4, from the term –4d4, so the leading coefficient of –4d

4 + 1 – d

2 is –4.

Rewrite the polynomial with each monomial in descending order according to degree.

–4d4 – d

2 + 1

8. 2x5 – 12 + 3x

SOLUTION: Find the degree of each term.

2x5 → 5

–12 → 0 3x → 1

The greatest degree is 5, from the term 2x5, so the leading coefficient of 2x

5 – 12 + 3x is 2.

Rewrite the polynomial with each monomial in descending order according to degree.

2x5 + 3x – 12

9. 4z – 2z2 – 5z

4

SOLUTION: Find the degree of each term. 4z → 1

– 2z2 → 2

– 5z4→ 4

The greatest degree is 4, from the term – 5z4, so the leading coefficient of 4z – 2z

2 – 5z

4 is –5.

Rewrite the polynomial with each monomial in descending order according to degree.

–5z4 – 2z

2 + 4z

10. 2a + 4a3 – 5a

2 – 1

SOLUTION: Find the degree of each term. 2a → 1

4a3→ 3

– 5a2 → 2

– 1 → 0

The greatest degree is 3, from the term 4a3, so the leading coefficient of 2a + 4a

3 – 5a

2 – 1 is 4.

Rewrite the polynomial with each monomial in descending order according to degree.

4a3– 5a

2 + 2a – 1

Find each sum or difference.

11. (6x3 − 4) + (−2x

3 + 9)

SOLUTION:

12. (g3 − 2g2 + 5g + 6) − (g2 + 2g)

SOLUTION:

13. (4 + 2a2 − 2a) − (3a

2 − 8a + 7)

SOLUTION:

14. (8y − 4y2) + (3y − 9y

2)

SOLUTION:

15. (−4z3 − 2z + 8) − (4z

3 + 3z2 − 5)

SOLUTION:

16. (−3d2 − 8 + 2d) + (4d − 12 + d2

)

SOLUTION:

17. (y + 5) + (2y + 4y2 – 2)

SOLUTION:

18. (3n3 − 5n + n2

) − (−8n2 + 3n

3)

SOLUTION:

19. CCSS SENSE-MAKING The total number of students T who traveled for spring break consists of two groups: students who flew to their destinations F and students who drove to their destination D. The number (in thousands) of students who flew and the total number of students who flew or drove can be modeled by the following equations,where n is the number of years since 1995. T = 14n + 21

F = 8n + 7 a. Write an equation that models the number of students who drove to their destination for this time period. b. Predict the number of students who will drive to their destination in 2012. c. How many students will drive or fly to their destination in 2015?

SOLUTION: a.

D = 6n + 14 b. n = 2012 – 1995 = 17

The number of students who will drive to their destination in 2012 is 116,000 students. c. n = 2015 – 1995 = 20

The number of students who will drive or fly to their destination in 2015 is 301,000 students.

Determine whether each expression is a polynomial. If it is a polynomial, find the degree and determine whether it is a monomial, binomial, or trinomial.

20.

SOLUTION: A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer exponents. It has

only one term. is a division of two monomials, so it is not a monomial.

21.

SOLUTION: A polynomial is a monomial or the sum of monomials. 21 is a monomial, so it is also a polynomial. The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of 21 is 0. The polynomial has only one term, so it is a monomial.

22. c4 – 2c

2 + 1

SOLUTION:

A polynomial is a monomial or the sum of monomials. c4 – 2c

2 + 1

is the sum of 3 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 4, 2, and

0, so the degree of c4 – 2c2 + 1 is 4. The polynomial has three terms, so it is a trinomial.

23. d + 3dc

SOLUTION: A polynomial is a monomial or the sum of monomials. A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer

exponents. 3dc

has a variable in the exponent, so it is not a monomial.

24. a – a2

SOLUTION:

A polynomial is a monomial or the sum of monomials. a – a2 is the sum of 2 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 1 and 2,

so the degree of a – a2 is 2. The polynomial has two terms, so it is a binomial.

25. 5n3 + nq

3

SOLUTION:

A polynomial is a monomial or the sum of monomials. 5n3 + nq3 is the sum of 2 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 3 and 4,

so the degree of 5n3 + nq3 is 4. The polynomial has two terms, so it is a binomial.

Write each polynomial in standard form. Identify the leading coefficient.

26. 5x2 – 2 + 3x

SOLUTION: Find the degree of each term.

5x2 → 2

– 2 → 0

3x → 1

The greatest degree is 2, from the term 5x2, so the leading coefficient of 5x

2 – 2 + 3x is 5.

Rewrite the polynomial with each monomial in descending order according to degree.

5x2 + 3x – 2

27. 8y + 7y3

SOLUTION: Find the degree of each term.

7y3 → 3

8y → 1

The greatest degree is 3, from the term 7y3, so the leading coefficient of 8y + 7y

3 is 7.

Rewrite the polynomial with each monomial in descending order according to degree.

7y3 + 8y

28. 4 – 3c – 5c2

SOLUTION: Find the degree of each term. 4 → 0 3c → 1

– 5c2 → 2

The greatest degree is 2, from the term – 5c2, so the leading coefficient of 4 – 3c – 5c

2 is –5.

Rewrite the polynomial with each monomial in descending order according to degree.

–5c2 – 3c + 4

29. –y3 + 3y – 3y

2 + 2

SOLUTION: Find the degree of each term.

–y3 → 3

3y → 1

3y2 → 2

2 → 0

The greatest degree is 3, from the term –y3, so the leading coefficient of –y

3 + 3y – 3y

2 + 2 is –1.

Rewrite the polynomial with each monomial in descending order according to degree. –y 3 – 3y 2 + 3y + 2

30. 11t + 2t2 – 3 + t

5

SOLUTION: Find the degree of each term. 11t → 1

2t2 → 2

–3 → 0

t5 → 5

The greatest degree is 5, from the term t5, so the leading coefficient of 11t + 2t2 – 3 + t

5 is 1.

Rewrite the polynomial with each monomial in descending order according to degree.

t5 + 2t

2 + 11t – 3

31. 2 + r – r3

SOLUTION: Find the degree of each term. 2 → 0 r → 1

– r3→ 3

The greatest degree is 3, from the term – r3, so the leading coefficient of 2 + r – r3

is –1. Rewrite the polynomial with each monomial in descending order according to degree.

–r3 + r + 2

32.

SOLUTION: Find the degree of each term.

→ 0

–3x4 → 4

7 → 0

The greatest degree is 4, from the term –3x4, so the leading coefficient of is –3.

Rewrite the polynomial with each monomial in descending order according to degree.

33. –9b2 + 10b – b

6

SOLUTION: Find the degree of each term.

–9b2 → 2

10b → 1

–b6→ 6

The greatest degree is 6, from the term –b6, so the leading coefficient of –9b2 + 10b – b6 is –1. Rewrite the polynomial with each monomial in descending order according to degree.

–b6 – 9b

2 + 10b

Find each sum or difference.

34. (2c2 + 6c + 4) + (5c – 7)

SOLUTION:

35. (2x + 3x2) − (7 − 8x

2)

SOLUTION:

36. (3c3 − c + 11) − (c2 + 2c + 8)

SOLUTION:

37. (z2 + z) + (z

2 − 11)

SOLUTION:

38. (2x − 2y + 1) − (3y + 4x)

SOLUTION:

39. (4a − 5b2 + 3) + (6 − 2a + 3b

2)

SOLUTION:

40. (x2y − 3x

2 + y) + (3y − 2x2y)

SOLUTION:

41. (−8xy + 3x2 − 5y) + (4x

2 − 2y + 6xy)

SOLUTION:

42. (5n − 2p2 + 2np) − (4p

2 + 4n)

SOLUTION:

43. (4rxt − 8r2x + x2

) − (6rx2 + 5rxt − 2x

2)

SOLUTION:

44. PETS From 1999 through 2009, the number of dogs D and the number of cats C (in hundreds) adopted from animal

shelters in the United States are modeled by the equations D = 2n + 3 and C = n + 4, where n is the number of yearssince 1999. a. Write an equation that models the total number T of dogs and cats adopted in hundreds for this time period. b. If this trend continues, how many dogs and cats will be adopted in 2013?

SOLUTION: a.

So, an equation that models the total number of dogs and cats adopted is T = 3n + 7. b. Evaluate the equation for the total number of dogs and cats for n = 2013 – 1999 = 14.

The number of cats and dogs adopted in 2013 will be 49 × 100 or 4900 cats and dogs.

Classify each polynomial according to its degree and number of terms.

45. 4x – 3x2 + 5

SOLUTION:

Find the degree of each term of 4x – 3x2 + 5.

4x → 1

– 3x2 → 2

5 → 0

The greatest degree is 2 and there are 3 terms, so 4x – 3x2 + 5 is a quadratic trinomial.

46. 11z3

SOLUTION: Find the degree of each term. 11z3

→ 3

The greatest degree is 3 and there is one term, so 11z3 is a cubic monomial.

47. 9 + y4

SOLUTION:

Find the degree of each term of 9 + y 4.

9 → 1

y4→ 4

The greatest degree is 4 and there are 2 terms, so 9 + y 4 is a quartic binomial.

48. 3x3 – 7

SOLUTION:

Find the degree of each term of 3x3 – 7.

3x3 → 3

–7 → 0

The greatest degree is 3 and there are 2 terms, so 3x3 – 7 is a cubic binomial.

49. –2x5 – x

2 + 5x – 8

SOLUTION:

Find the degree of each term of –2x5 – x2 + 5x – 8.

–2x5 → 5

–x2 → 2

5x → 1 – 8 → 0

The greatest degree is 5 and there are 4 terms, so –2x5 – x2 + 5x – 8 is a quintic polynomial.

50. 10t – 4t2 + 6t

3

SOLUTION:

Find the degree of each term of 10t – 4t2 + 6t

3.

10t → 1

4t2 → 2

6t3 → 3

The greatest degree is 3 and there are 3 terms, so 10t – 4t2 + 6t

3 is a cubic trinomial.

51. ENROLLMENT In a rapidly growing school system, the numbers (in hundreds) of total students N and

K-5 students P enrolled from 2000 to 2009 are modeled by the equations N = 1.25t2 – t + 7.5 and P =

0.7t2 – 0.95t + 3.8, where t is the number of years since 2000.

a. Write an equation modeling the number of 6-12 students S enrolled for this time period. b. How many 6-12 students were enrolled in the school system in 2007?

SOLUTION: a. To write an equation that represents the number of 6-12 students enrolled, subtract the equations that represent the total number of students and the number of K-5 students.

b. Replace t with 7 in the equation for S to determine the number of students enrolled in 6-12 in 2007.

In 2007 there were 30.3 hundreds or 3030 students enrolled in 6-12.

52. CCSS REASONING The perimeter of the figure shown is represented by the expression 3x2 − 7x + 2. Write a

polynomial that represents the measure of the third side.

SOLUTION:

53. GEOMETRY Consider the rectangle.

a. What does (4x2 + 2x – 1)(2x

2 – x + 3) represent?

b. What does 2(4x2 + 2x – 1) + 2(2x

2 – x + 3) represent?

SOLUTION:

a. (4x2 + 2x – 1)(2x

2 – x + 3) is a multiplication of the length and the width of the rectangle, which is the formula for

the area of a rectangle.

b. 2(4x2 + 2x – 1) + 2(2x

2 – x + 3) is the sum of twice the length and twice the width of a rectangle, which is the

formula for the perimeter of the rectangle

Find each sum or difference.

54. (4x + 2y − 6z) + (5y − 2z + 7x) + (−9z − 2x − 3y)

SOLUTION:

55. (5a2 − 4) + (a

2 − 2a + 12) + (4a2 − 6a + 8)

SOLUTION:

56. (3c2 − 7) + (4c + 7) − (c

2 + 5c − 8)

SOLUTION:

57. (3n3 + 3n − 10) − (4n

2 − 5n) + (4n3 − 3n

2 − 9n + 4)

SOLUTION:

58. FOOTBALL The National Football League is divided into two conferences, the American A and the National N. From 2002 through 2009, the total attendance T (in thousands) for both conferences and for the American Conference games are modeled by the following equations, where x is the number of years since 2002.

T = –0.69x3 + 55.83x

2 + 643.31x + 10,538

A = –3.78x3 + 58.96x

2 + 265.96x + 5257

Determine how many people attended National Conference football games in 2009.

SOLUTION:

Let x = 7 represent 2009, then find how many people attended a National Conference football game in 2009.

In 2009 the number of people who attended National Conference football games was about 8829 thousand. Multiply by 1000 to find the attendance in standard form. 8829 ×1000 = 8,829,000. So, about 8,829,000 people attended National Conference football games in 2009.

59. CAR RENTAL The cost to rent a car for a day is $15 plus $0.15 for each mile driven. a. Write a polynomial that represents the cost of renting a car for m miles. b. If a car is driven 145 miles, how much would it cost to rent? c. If a car is driven 105 miles each day for four days, how much would it cost to rent a car? d. If a car is driven 220 miles each day for seven days, how much would it cost to rent a car?

SOLUTION: a. The cost to rent a car is the daily rate and the mileage cost or 15 + 0.15m. b. Substitute 145 for m to find the daily cost to drive 145 miles.

The cost to rent the car would be $36.75. c. The expression represents the cost per day. For 4 days, multiply the entire expression by 4 to find the cost for a 4 day trip. Substitute 105 for m.

The cost to rent the car would be $123. d. The expression represents the cost per day. For 7 days, multiply the entire expression by 7 to find the cost for a 7 day trip. Substitute 220 for m.

The cost to rent the car would be $336.

60. MULTIPLE REPRESENTATIONS In this problem, you will explore perimeter and area. a. Geometric Draw three rectangles that each have a perimeter of 400 feet. b. Tabular Record the width and length of each rectangle in a table like the one shown below. Find the area of eachrectangle.

c. Graphical On a coordinate system, graph the area of rectangle 4 in terms of the length, x. Use the graph to determine the largest area possible. d. Analytical Determine the length and width that produce the largest area.

SOLUTION: a. For the perimeters to be 400, create lengths and widths that sum to 200. Note that some lengths are already provided in the table in part b.

b. The area is length multiplied by width. Make sure the units are squared for area. The sum of the length and width must be 200, so if the length is x, the width must be 200 – x.

c. The length will be the x-values on the horizontal and the area will be the y-values on the vertical. The length cannot pass 200 since the sum of the length and width is 200. Set the intervals for the x-axis to 25 feet. Extend the table of values to find more points to plot on the graph. It appears that 10,000 is the greatest area, so set the intervals

for the y-axis to 1000 ft2. After graphing, it appears that the highest point on the graph is at an area of 10,000 ft

2.

d. The associated x-value with the maximum area is x = 100, so the length must be 100 and the width must be 200 – 100, or 100. The length and width of the rectangle must be 100 feet each to have the largest area.

61. CCSS CRITIQUE Cheyenne and Sebastian are finding (2x2 − x) − (3x + 3x

2 − 2). Is either of them correct? Explain your reasoning.

SOLUTION:

Neither is correct. Cheyenne, did not distribute the negative to the 2nd and 3rd terms when she found the additive inverse. Sebastian did not distribute the negate to the 3rd terms when he found the additive inverse. To find the additive inverse, all terms should be multiplied by −1.

62. REASONING Determine whether each of the following statements is true or false . Explain your reasoning. a. A binomial can have a degree of zero. b. The order in which polynomials are subtracted does not matter.

SOLUTION: a. If a binomial has two terms that are each a degree of 0, then those terms can be combined and the binomial becomes a monomial. For example, 18 + 7 = 25. If one of the terms of the binomial does not have a degree of 0, then the binomial cannot have a degree of 0, since the degree of a polynomial is the greatest degree of any term in the polynomial. b. Subtraction is not commutative. While 2 + 5 = 5 + 2, 2 – 5 ≠ 5 – 2. This is also true for polynomials. Sample answer: (2x – 3) – (4x – 3) = –2x, but (4x – 3) – (2x – 3) = 2x

63. CHALLENGE Write a polynomial that represents the sum of an odd integer 2n + 1 and the next two consecutive odd integers.

SOLUTION:

64. WRITING IN MATH Why would you add or subtract equations that represent real-world situations? Explain.

SOLUTION:

65. WRITING IN MATH Describe how to add and subtract polynomials using both the vertical and horizontal formats.

SOLUTION: To add polynomials in a horizontal format, you combine like terms. For the vertical format, you write the polynomials in standard form, align like terms in columns, and combine like terms.

To subtract polynomials in a horizontal format you find the additive inverse of the polynomial you are subtracting, andthen combine like terms. For the vertical format, you write the polynomials in standard form, align like terms in columns, and subtract by adding the additive inverse.

66. Three consecutive integers can be represented by x, x + 1, and x + 2. What is the sum of these three integers? A x(x + 1)(x + 2)

B x3 + 3

C 3x + 3 D x + 3

SOLUTION:

The correct choice is C.

67. SHORT RESPONSE What is the perimeter of a square with sides that measure 2x + 3 units?

SOLUTION:

The perimeter of the square is 8x + 12 units.

68. Jim cuts a board in the shape of a regular hexagon and pounds in a nail at each vertex, as shown. How many rubber bands will he need to stretch a rubber band across every possible pair of nails?

F 15 G 14 H 12 J 9

SOLUTION: The first nail would connect to 5 others, the second to 4 others, the third to 3 others, etc.5 + 4 + 3 + 2 + 1 = 15 The correct choice is F.

69. Which ordered pair is in the solution set of the system of inequalities shown in the graph?

A (−3, 0) B (0, −3) C (5, 0) D (0, 5)

SOLUTION: Choice A is outside the shaded area for both inequalities. Choices B and D are inside the shaded area for only one inequality. Choice C is the only point in the solution for both inequalities. So, the correct choice is C.

70. COMPUTERS A computer technician charges by the hour to fix and repair computer equipment. The total cost of the technician for one hour is $75, for two hours is $125, for three hours is $175, for four hours is $225, and so on. Write a recursive formula for the sequence.

SOLUTION: Write out the terms. $75, $125, $175, $225, ... The first term is 75, and 50 is added to form each following term. Therefore, we have a1 = 75, an = an – 1 + 50, n ≥ 2.

Determine whether each sequence is arithmetic, geometric, or neither. Explain.71. 8, –32, 128, –512, ...

SOLUTION: Check for a common difference. –32 – 8 = –40 128 – (–32) = 160 There is no common difference. Check for a common ratio. –32 ÷ 8 = –4 128 ÷ (–32) = –4 Geometric; the common ratio is –4.

72. 25, 8, –9, –26, ...

SOLUTION: Check for a common difference. 8 – 25 = –17 –9 – 8 = –17 Arithmetic; the common difference is –17.

73.

SOLUTION: Check for a common difference.

There is no common difference. Check for a common ratio.

There is no common ratio, so the sequence is not arithmetic or geometric.

74. 43, 52, 61, 70, ...

SOLUTION: Check for a common difference. 52 – 43 = 9 61 – 52 = 9 Arithmetic; the common difference is 9.

75. –27, –16, –5, 6, ...

SOLUTION: Check for a common difference. –16 – (–27) = 11 –5 – (–16) = 11 Arithmetic; the common difference is 11.

76. 200, 100, 50, 25, …

SOLUTION: Check for a common difference. 100 – 200 = –100 50 – 100 = –50 There is no common difference. Check for a common ratio. 100 ÷ 200 = 0.5 50 ÷ 100 = 0.5

Geometric; the common ratio is 0.5 or .

77. JOBS Kimi received an offer for a new job. She wants to compare the offer with her current job. What is total amount of sales that Kimi must get each month to make the same income at either job?

SOLUTION: Let s be Kimi’s monthly sales.

Kimi must sell $80,000 each month to make the same income at either job.

Determine whether each sequence is an arithmetic sequence. If it is, state the common difference.78. 24, 16, 8, 0, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is –8, because 16 – 24 = –8; 8 –16 = –8; etc.

79. , 13, 26, …

SOLUTION: Find the difference between the terms.

– =

13 – =

26 – 13 = 13 There is not a common different. The sequence is not an arithmetic sequence

80. 7, 6, 5, 4, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is –1, because 6 – 7 = –1; 5 – 6 = –1; 4 – 5 = –1;etc.

81. 10, 12, 15, 18, …

SOLUTION: Find the difference between the terms. 12 – 10 = 2 15 – 12 = 3 18 – 15 = 3 There is no common difference. The sequence is not an arithmetic sequence.

82. −15, −11, −7, −3, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is 4, because –11 – (–15) = 4; –7 – (–11) = 4; –3 – (–7) = 4;etc.

83. −0.3, 0.2, 0.7, 1.2, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is 0.5, because 0.2 – (–0.3) = 0.5; 0.7 – 0.2 = 0.5; 1.2 – 0.7 = 0.5; etc.

Simplify.

84. t(t5)(t

7)

SOLUTION:

85. n3(n

2)(−2n

3)

SOLUTION:

86. (5t5v

2)(10t

3v

4)

SOLUTION:

87. (−8u4z

5)(5uz

4)

SOLUTION:

88. [(3)2]3

SOLUTION:

89. [(2)3]2

SOLUTION:

90. (2m4k

3)2(−3mk

2)3

SOLUTION:

91. (6xy2)2(2x

2y

2z

2)3

SOLUTION:

eSolutions Manual - Powered by Cognero Page 12

8-1 Adding and Subtracting Polynomials

Determine whether each expression is a polynomial. If it is a polynomial, find the degree and determine whether it is a monomial, binomial, or trinomial.

1. 7ab + 6b2 – 2a

3

SOLUTION:

A polynomial is a monomial or the sum of monomials. 7ab + 6b2 – 2a

3 is the sum of 3 monomials, so it is a

polynomial. The degree of a polynomial is the greatest degree of any term in the polynomial.The degree of each term is 2, 2, and

3, so the degree of 7ab + 6b2 – 2a

3 is 3. The polynomial has three terms, so it is a trinomial.

2. 2y – 5 + 3y2

SOLUTION:

2y – 5 + 3y2

is the sum of monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 1, 0, and

2, so the degree of 2y – 5 + 3y2 is 2. The polynomial has three terms, so it is a trinomial.

3. 3x2

SOLUTION:

A polynomial is a monomial or the sum of monomials, so 3x2 is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of 3x2 is 2. The

polynomial has one term, so it is a monomial.

4.

SOLUTION: A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer exponents. It has

only one term. is a division of two monomials, so it is not a monomial.

5. 5m2p

3 + 6

SOLUTION:

A polynomial is a monomial or the sum of monomials. 5m2p

3 + 6

is the sum of 2 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 5 and 0,

so the degree of 5m2p

3 + 6 is 5. The polynomial has two terms, so it is a binomial.

6. 5q–4

+ 6q

SOLUTION: A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer exponents. It has

only one term. 5q-4 is equal to , which is a division of two monomials, so it is not a monomial.

Write each polynomial in standard form. Identify the leading coefficient.

7. –4d4 + 1 – d

2

SOLUTION: Find the degree of each term.

–4d4→ 4

1 → 0

–d2 → 2

The greatest degree is 4, from the term –4d4, so the leading coefficient of –4d

4 + 1 – d

2 is –4.

Rewrite the polynomial with each monomial in descending order according to degree.

–4d4 – d

2 + 1

8. 2x5 – 12 + 3x

SOLUTION: Find the degree of each term.

2x5 → 5

–12 → 0 3x → 1

The greatest degree is 5, from the term 2x5, so the leading coefficient of 2x

5 – 12 + 3x is 2.

Rewrite the polynomial with each monomial in descending order according to degree.

2x5 + 3x – 12

9. 4z – 2z2 – 5z

4

SOLUTION: Find the degree of each term. 4z → 1

– 2z2 → 2

– 5z4→ 4

The greatest degree is 4, from the term – 5z4, so the leading coefficient of 4z – 2z

2 – 5z

4 is –5.

Rewrite the polynomial with each monomial in descending order according to degree.

–5z4 – 2z

2 + 4z

10. 2a + 4a3 – 5a

2 – 1

SOLUTION: Find the degree of each term. 2a → 1

4a3→ 3

– 5a2 → 2

– 1 → 0

The greatest degree is 3, from the term 4a3, so the leading coefficient of 2a + 4a

3 – 5a

2 – 1 is 4.

Rewrite the polynomial with each monomial in descending order according to degree.

4a3– 5a

2 + 2a – 1

Find each sum or difference.

11. (6x3 − 4) + (−2x

3 + 9)

SOLUTION:

12. (g3 − 2g2 + 5g + 6) − (g2 + 2g)

SOLUTION:

13. (4 + 2a2 − 2a) − (3a

2 − 8a + 7)

SOLUTION:

14. (8y − 4y2) + (3y − 9y

2)

SOLUTION:

15. (−4z3 − 2z + 8) − (4z

3 + 3z2 − 5)

SOLUTION:

16. (−3d2 − 8 + 2d) + (4d − 12 + d2

)

SOLUTION:

17. (y + 5) + (2y + 4y2 – 2)

SOLUTION:

18. (3n3 − 5n + n2

) − (−8n2 + 3n

3)

SOLUTION:

19. CCSS SENSE-MAKING The total number of students T who traveled for spring break consists of two groups: students who flew to their destinations F and students who drove to their destination D. The number (in thousands) of students who flew and the total number of students who flew or drove can be modeled by the following equations,where n is the number of years since 1995. T = 14n + 21

F = 8n + 7 a. Write an equation that models the number of students who drove to their destination for this time period. b. Predict the number of students who will drive to their destination in 2012. c. How many students will drive or fly to their destination in 2015?

SOLUTION: a.

D = 6n + 14 b. n = 2012 – 1995 = 17

The number of students who will drive to their destination in 2012 is 116,000 students. c. n = 2015 – 1995 = 20

The number of students who will drive or fly to their destination in 2015 is 301,000 students.

Determine whether each expression is a polynomial. If it is a polynomial, find the degree and determine whether it is a monomial, binomial, or trinomial.

20.

SOLUTION: A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer exponents. It has

only one term. is a division of two monomials, so it is not a monomial.

21.

SOLUTION: A polynomial is a monomial or the sum of monomials. 21 is a monomial, so it is also a polynomial. The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of 21 is 0. The polynomial has only one term, so it is a monomial.

22. c4 – 2c

2 + 1

SOLUTION:

A polynomial is a monomial or the sum of monomials. c4 – 2c

2 + 1

is the sum of 3 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 4, 2, and

0, so the degree of c4 – 2c2 + 1 is 4. The polynomial has three terms, so it is a trinomial.

23. d + 3dc

SOLUTION: A polynomial is a monomial or the sum of monomials. A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer

exponents. 3dc

has a variable in the exponent, so it is not a monomial.

24. a – a2

SOLUTION:

A polynomial is a monomial or the sum of monomials. a – a2 is the sum of 2 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 1 and 2,

so the degree of a – a2 is 2. The polynomial has two terms, so it is a binomial.

25. 5n3 + nq

3

SOLUTION:

A polynomial is a monomial or the sum of monomials. 5n3 + nq3 is the sum of 2 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 3 and 4,

so the degree of 5n3 + nq3 is 4. The polynomial has two terms, so it is a binomial.

Write each polynomial in standard form. Identify the leading coefficient.

26. 5x2 – 2 + 3x

SOLUTION: Find the degree of each term.

5x2 → 2

– 2 → 0

3x → 1

The greatest degree is 2, from the term 5x2, so the leading coefficient of 5x

2 – 2 + 3x is 5.

Rewrite the polynomial with each monomial in descending order according to degree.

5x2 + 3x – 2

27. 8y + 7y3

SOLUTION: Find the degree of each term.

7y3 → 3

8y → 1

The greatest degree is 3, from the term 7y3, so the leading coefficient of 8y + 7y

3 is 7.

Rewrite the polynomial with each monomial in descending order according to degree.

7y3 + 8y

28. 4 – 3c – 5c2

SOLUTION: Find the degree of each term. 4 → 0 3c → 1

– 5c2 → 2

The greatest degree is 2, from the term – 5c2, so the leading coefficient of 4 – 3c – 5c

2 is –5.

Rewrite the polynomial with each monomial in descending order according to degree.

–5c2 – 3c + 4

29. –y3 + 3y – 3y

2 + 2

SOLUTION: Find the degree of each term.

–y3 → 3

3y → 1

3y2 → 2

2 → 0

The greatest degree is 3, from the term –y3, so the leading coefficient of –y

3 + 3y – 3y

2 + 2 is –1.

Rewrite the polynomial with each monomial in descending order according to degree. –y 3 – 3y 2 + 3y + 2

30. 11t + 2t2 – 3 + t

5

SOLUTION: Find the degree of each term. 11t → 1

2t2 → 2

–3 → 0

t5 → 5

The greatest degree is 5, from the term t5, so the leading coefficient of 11t + 2t2 – 3 + t

5 is 1.

Rewrite the polynomial with each monomial in descending order according to degree.

t5 + 2t

2 + 11t – 3

31. 2 + r – r3

SOLUTION: Find the degree of each term. 2 → 0 r → 1

– r3→ 3

The greatest degree is 3, from the term – r3, so the leading coefficient of 2 + r – r3

is –1. Rewrite the polynomial with each monomial in descending order according to degree.

–r3 + r + 2

32.

SOLUTION: Find the degree of each term.

→ 0

–3x4 → 4

7 → 0

The greatest degree is 4, from the term –3x4, so the leading coefficient of is –3.

Rewrite the polynomial with each monomial in descending order according to degree.

33. –9b2 + 10b – b

6

SOLUTION: Find the degree of each term.

–9b2 → 2

10b → 1

–b6→ 6

The greatest degree is 6, from the term –b6, so the leading coefficient of –9b2 + 10b – b6 is –1. Rewrite the polynomial with each monomial in descending order according to degree.

–b6 – 9b

2 + 10b

Find each sum or difference.

34. (2c2 + 6c + 4) + (5c – 7)

SOLUTION:

35. (2x + 3x2) − (7 − 8x

2)

SOLUTION:

36. (3c3 − c + 11) − (c2 + 2c + 8)

SOLUTION:

37. (z2 + z) + (z

2 − 11)

SOLUTION:

38. (2x − 2y + 1) − (3y + 4x)

SOLUTION:

39. (4a − 5b2 + 3) + (6 − 2a + 3b

2)

SOLUTION:

40. (x2y − 3x

2 + y) + (3y − 2x2y)

SOLUTION:

41. (−8xy + 3x2 − 5y) + (4x

2 − 2y + 6xy)

SOLUTION:

42. (5n − 2p2 + 2np) − (4p

2 + 4n)

SOLUTION:

43. (4rxt − 8r2x + x2

) − (6rx2 + 5rxt − 2x

2)

SOLUTION:

44. PETS From 1999 through 2009, the number of dogs D and the number of cats C (in hundreds) adopted from animal

shelters in the United States are modeled by the equations D = 2n + 3 and C = n + 4, where n is the number of yearssince 1999. a. Write an equation that models the total number T of dogs and cats adopted in hundreds for this time period. b. If this trend continues, how many dogs and cats will be adopted in 2013?

SOLUTION: a.

So, an equation that models the total number of dogs and cats adopted is T = 3n + 7. b. Evaluate the equation for the total number of dogs and cats for n = 2013 – 1999 = 14.

The number of cats and dogs adopted in 2013 will be 49 × 100 or 4900 cats and dogs.

Classify each polynomial according to its degree and number of terms.

45. 4x – 3x2 + 5

SOLUTION:

Find the degree of each term of 4x – 3x2 + 5.

4x → 1

– 3x2 → 2

5 → 0

The greatest degree is 2 and there are 3 terms, so 4x – 3x2 + 5 is a quadratic trinomial.

46. 11z3

SOLUTION: Find the degree of each term. 11z3

→ 3

The greatest degree is 3 and there is one term, so 11z3 is a cubic monomial.

47. 9 + y4

SOLUTION:

Find the degree of each term of 9 + y 4.

9 → 1

y4→ 4

The greatest degree is 4 and there are 2 terms, so 9 + y 4 is a quartic binomial.

48. 3x3 – 7

SOLUTION:

Find the degree of each term of 3x3 – 7.

3x3 → 3

–7 → 0

The greatest degree is 3 and there are 2 terms, so 3x3 – 7 is a cubic binomial.

49. –2x5 – x

2 + 5x – 8

SOLUTION:

Find the degree of each term of –2x5 – x2 + 5x – 8.

–2x5 → 5

–x2 → 2

5x → 1 – 8 → 0

The greatest degree is 5 and there are 4 terms, so –2x5 – x2 + 5x – 8 is a quintic polynomial.

50. 10t – 4t2 + 6t

3

SOLUTION:

Find the degree of each term of 10t – 4t2 + 6t

3.

10t → 1

4t2 → 2

6t3 → 3

The greatest degree is 3 and there are 3 terms, so 10t – 4t2 + 6t

3 is a cubic trinomial.

51. ENROLLMENT In a rapidly growing school system, the numbers (in hundreds) of total students N and

K-5 students P enrolled from 2000 to 2009 are modeled by the equations N = 1.25t2 – t + 7.5 and P =

0.7t2 – 0.95t + 3.8, where t is the number of years since 2000.

a. Write an equation modeling the number of 6-12 students S enrolled for this time period. b. How many 6-12 students were enrolled in the school system in 2007?

SOLUTION: a. To write an equation that represents the number of 6-12 students enrolled, subtract the equations that represent the total number of students and the number of K-5 students.

b. Replace t with 7 in the equation for S to determine the number of students enrolled in 6-12 in 2007.

In 2007 there were 30.3 hundreds or 3030 students enrolled in 6-12.

52. CCSS REASONING The perimeter of the figure shown is represented by the expression 3x2 − 7x + 2. Write a

polynomial that represents the measure of the third side.

SOLUTION:

53. GEOMETRY Consider the rectangle.

a. What does (4x2 + 2x – 1)(2x

2 – x + 3) represent?

b. What does 2(4x2 + 2x – 1) + 2(2x

2 – x + 3) represent?

SOLUTION:

a. (4x2 + 2x – 1)(2x

2 – x + 3) is a multiplication of the length and the width of the rectangle, which is the formula for

the area of a rectangle.

b. 2(4x2 + 2x – 1) + 2(2x

2 – x + 3) is the sum of twice the length and twice the width of a rectangle, which is the

formula for the perimeter of the rectangle

Find each sum or difference.

54. (4x + 2y − 6z) + (5y − 2z + 7x) + (−9z − 2x − 3y)

SOLUTION:

55. (5a2 − 4) + (a

2 − 2a + 12) + (4a2 − 6a + 8)

SOLUTION:

56. (3c2 − 7) + (4c + 7) − (c

2 + 5c − 8)

SOLUTION:

57. (3n3 + 3n − 10) − (4n

2 − 5n) + (4n3 − 3n

2 − 9n + 4)

SOLUTION:

58. FOOTBALL The National Football League is divided into two conferences, the American A and the National N. From 2002 through 2009, the total attendance T (in thousands) for both conferences and for the American Conference games are modeled by the following equations, where x is the number of years since 2002.

T = –0.69x3 + 55.83x

2 + 643.31x + 10,538

A = –3.78x3 + 58.96x

2 + 265.96x + 5257

Determine how many people attended National Conference football games in 2009.

SOLUTION:

Let x = 7 represent 2009, then find how many people attended a National Conference football game in 2009.

In 2009 the number of people who attended National Conference football games was about 8829 thousand. Multiply by 1000 to find the attendance in standard form. 8829 ×1000 = 8,829,000. So, about 8,829,000 people attended National Conference football games in 2009.

59. CAR RENTAL The cost to rent a car for a day is $15 plus $0.15 for each mile driven. a. Write a polynomial that represents the cost of renting a car for m miles. b. If a car is driven 145 miles, how much would it cost to rent? c. If a car is driven 105 miles each day for four days, how much would it cost to rent a car? d. If a car is driven 220 miles each day for seven days, how much would it cost to rent a car?

SOLUTION: a. The cost to rent a car is the daily rate and the mileage cost or 15 + 0.15m. b. Substitute 145 for m to find the daily cost to drive 145 miles.

The cost to rent the car would be $36.75. c. The expression represents the cost per day. For 4 days, multiply the entire expression by 4 to find the cost for a 4 day trip. Substitute 105 for m.

The cost to rent the car would be $123. d. The expression represents the cost per day. For 7 days, multiply the entire expression by 7 to find the cost for a 7 day trip. Substitute 220 for m.

The cost to rent the car would be $336.

60. MULTIPLE REPRESENTATIONS In this problem, you will explore perimeter and area. a. Geometric Draw three rectangles that each have a perimeter of 400 feet. b. Tabular Record the width and length of each rectangle in a table like the one shown below. Find the area of eachrectangle.

c. Graphical On a coordinate system, graph the area of rectangle 4 in terms of the length, x. Use the graph to determine the largest area possible. d. Analytical Determine the length and width that produce the largest area.

SOLUTION: a. For the perimeters to be 400, create lengths and widths that sum to 200. Note that some lengths are already provided in the table in part b.

b. The area is length multiplied by width. Make sure the units are squared for area. The sum of the length and width must be 200, so if the length is x, the width must be 200 – x.

c. The length will be the x-values on the horizontal and the area will be the y-values on the vertical. The length cannot pass 200 since the sum of the length and width is 200. Set the intervals for the x-axis to 25 feet. Extend the table of values to find more points to plot on the graph. It appears that 10,000 is the greatest area, so set the intervals

for the y-axis to 1000 ft2. After graphing, it appears that the highest point on the graph is at an area of 10,000 ft

2.

d. The associated x-value with the maximum area is x = 100, so the length must be 100 and the width must be 200 – 100, or 100. The length and width of the rectangle must be 100 feet each to have the largest area.

61. CCSS CRITIQUE Cheyenne and Sebastian are finding (2x2 − x) − (3x + 3x

2 − 2). Is either of them correct? Explain your reasoning.

SOLUTION:

Neither is correct. Cheyenne, did not distribute the negative to the 2nd and 3rd terms when she found the additive inverse. Sebastian did not distribute the negate to the 3rd terms when he found the additive inverse. To find the additive inverse, all terms should be multiplied by −1.

62. REASONING Determine whether each of the following statements is true or false . Explain your reasoning. a. A binomial can have a degree of zero. b. The order in which polynomials are subtracted does not matter.

SOLUTION: a. If a binomial has two terms that are each a degree of 0, then those terms can be combined and the binomial becomes a monomial. For example, 18 + 7 = 25. If one of the terms of the binomial does not have a degree of 0, then the binomial cannot have a degree of 0, since the degree of a polynomial is the greatest degree of any term in the polynomial. b. Subtraction is not commutative. While 2 + 5 = 5 + 2, 2 – 5 ≠ 5 – 2. This is also true for polynomials. Sample answer: (2x – 3) – (4x – 3) = –2x, but (4x – 3) – (2x – 3) = 2x

63. CHALLENGE Write a polynomial that represents the sum of an odd integer 2n + 1 and the next two consecutive odd integers.

SOLUTION:

64. WRITING IN MATH Why would you add or subtract equations that represent real-world situations? Explain.

SOLUTION:

65. WRITING IN MATH Describe how to add and subtract polynomials using both the vertical and horizontal formats.

SOLUTION: To add polynomials in a horizontal format, you combine like terms. For the vertical format, you write the polynomials in standard form, align like terms in columns, and combine like terms.

To subtract polynomials in a horizontal format you find the additive inverse of the polynomial you are subtracting, andthen combine like terms. For the vertical format, you write the polynomials in standard form, align like terms in columns, and subtract by adding the additive inverse.

66. Three consecutive integers can be represented by x, x + 1, and x + 2. What is the sum of these three integers? A x(x + 1)(x + 2)

B x3 + 3

C 3x + 3 D x + 3

SOLUTION:

The correct choice is C.

67. SHORT RESPONSE What is the perimeter of a square with sides that measure 2x + 3 units?

SOLUTION:

The perimeter of the square is 8x + 12 units.

68. Jim cuts a board in the shape of a regular hexagon and pounds in a nail at each vertex, as shown. How many rubber bands will he need to stretch a rubber band across every possible pair of nails?

F 15 G 14 H 12 J 9

SOLUTION: The first nail would connect to 5 others, the second to 4 others, the third to 3 others, etc.5 + 4 + 3 + 2 + 1 = 15 The correct choice is F.

69. Which ordered pair is in the solution set of the system of inequalities shown in the graph?

A (−3, 0) B (0, −3) C (5, 0) D (0, 5)

SOLUTION: Choice A is outside the shaded area for both inequalities. Choices B and D are inside the shaded area for only one inequality. Choice C is the only point in the solution for both inequalities. So, the correct choice is C.

70. COMPUTERS A computer technician charges by the hour to fix and repair computer equipment. The total cost of the technician for one hour is $75, for two hours is $125, for three hours is $175, for four hours is $225, and so on. Write a recursive formula for the sequence.

SOLUTION: Write out the terms. $75, $125, $175, $225, ... The first term is 75, and 50 is added to form each following term. Therefore, we have a1 = 75, an = an – 1 + 50, n ≥ 2.

Determine whether each sequence is arithmetic, geometric, or neither. Explain.71. 8, –32, 128, –512, ...

SOLUTION: Check for a common difference. –32 – 8 = –40 128 – (–32) = 160 There is no common difference. Check for a common ratio. –32 ÷ 8 = –4 128 ÷ (–32) = –4 Geometric; the common ratio is –4.

72. 25, 8, –9, –26, ...

SOLUTION: Check for a common difference. 8 – 25 = –17 –9 – 8 = –17 Arithmetic; the common difference is –17.

73.

SOLUTION: Check for a common difference.

There is no common difference. Check for a common ratio.

There is no common ratio, so the sequence is not arithmetic or geometric.

74. 43, 52, 61, 70, ...

SOLUTION: Check for a common difference. 52 – 43 = 9 61 – 52 = 9 Arithmetic; the common difference is 9.

75. –27, –16, –5, 6, ...

SOLUTION: Check for a common difference. –16 – (–27) = 11 –5 – (–16) = 11 Arithmetic; the common difference is 11.

76. 200, 100, 50, 25, …

SOLUTION: Check for a common difference. 100 – 200 = –100 50 – 100 = –50 There is no common difference. Check for a common ratio. 100 ÷ 200 = 0.5 50 ÷ 100 = 0.5

Geometric; the common ratio is 0.5 or .

77. JOBS Kimi received an offer for a new job. She wants to compare the offer with her current job. What is total amount of sales that Kimi must get each month to make the same income at either job?

SOLUTION: Let s be Kimi’s monthly sales.

Kimi must sell $80,000 each month to make the same income at either job.

Determine whether each sequence is an arithmetic sequence. If it is, state the common difference.78. 24, 16, 8, 0, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is –8, because 16 – 24 = –8; 8 –16 = –8; etc.

79. , 13, 26, …

SOLUTION: Find the difference between the terms.

– =

13 – =

26 – 13 = 13 There is not a common different. The sequence is not an arithmetic sequence

80. 7, 6, 5, 4, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is –1, because 6 – 7 = –1; 5 – 6 = –1; 4 – 5 = –1;etc.

81. 10, 12, 15, 18, …

SOLUTION: Find the difference between the terms. 12 – 10 = 2 15 – 12 = 3 18 – 15 = 3 There is no common difference. The sequence is not an arithmetic sequence.

82. −15, −11, −7, −3, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is 4, because –11 – (–15) = 4; –7 – (–11) = 4; –3 – (–7) = 4;etc.

83. −0.3, 0.2, 0.7, 1.2, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is 0.5, because 0.2 – (–0.3) = 0.5; 0.7 – 0.2 = 0.5; 1.2 – 0.7 = 0.5; etc.

Simplify.

84. t(t5)(t

7)

SOLUTION:

85. n3(n

2)(−2n

3)

SOLUTION:

86. (5t5v

2)(10t

3v

4)

SOLUTION:

87. (−8u4z

5)(5uz

4)

SOLUTION:

88. [(3)2]3

SOLUTION:

89. [(2)3]2

SOLUTION:

90. (2m4k

3)2(−3mk

2)3

SOLUTION:

91. (6xy2)2(2x

2y

2z

2)3

SOLUTION:

eSolutions Manual - Powered by Cognero Page 13

8-1 Adding and Subtracting Polynomials

Determine whether each expression is a polynomial. If it is a polynomial, find the degree and determine whether it is a monomial, binomial, or trinomial.

1. 7ab + 6b2 – 2a

3

SOLUTION:

A polynomial is a monomial or the sum of monomials. 7ab + 6b2 – 2a

3 is the sum of 3 monomials, so it is a

polynomial. The degree of a polynomial is the greatest degree of any term in the polynomial.The degree of each term is 2, 2, and

3, so the degree of 7ab + 6b2 – 2a

3 is 3. The polynomial has three terms, so it is a trinomial.

2. 2y – 5 + 3y2

SOLUTION:

2y – 5 + 3y2

is the sum of monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 1, 0, and

2, so the degree of 2y – 5 + 3y2 is 2. The polynomial has three terms, so it is a trinomial.

3. 3x2

SOLUTION:

A polynomial is a monomial or the sum of monomials, so 3x2 is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of 3x2 is 2. The

polynomial has one term, so it is a monomial.

4.

SOLUTION: A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer exponents. It has

only one term. is a division of two monomials, so it is not a monomial.

5. 5m2p

3 + 6

SOLUTION:

A polynomial is a monomial or the sum of monomials. 5m2p

3 + 6

is the sum of 2 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 5 and 0,

so the degree of 5m2p

3 + 6 is 5. The polynomial has two terms, so it is a binomial.

6. 5q–4

+ 6q

SOLUTION: A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer exponents. It has

only one term. 5q-4 is equal to , which is a division of two monomials, so it is not a monomial.

Write each polynomial in standard form. Identify the leading coefficient.

7. –4d4 + 1 – d

2

SOLUTION: Find the degree of each term.

–4d4→ 4

1 → 0

–d2 → 2

The greatest degree is 4, from the term –4d4, so the leading coefficient of –4d

4 + 1 – d

2 is –4.

Rewrite the polynomial with each monomial in descending order according to degree.

–4d4 – d

2 + 1

8. 2x5 – 12 + 3x

SOLUTION: Find the degree of each term.

2x5 → 5

–12 → 0 3x → 1

The greatest degree is 5, from the term 2x5, so the leading coefficient of 2x

5 – 12 + 3x is 2.

Rewrite the polynomial with each monomial in descending order according to degree.

2x5 + 3x – 12

9. 4z – 2z2 – 5z

4

SOLUTION: Find the degree of each term. 4z → 1

– 2z2 → 2

– 5z4→ 4

The greatest degree is 4, from the term – 5z4, so the leading coefficient of 4z – 2z

2 – 5z

4 is –5.

Rewrite the polynomial with each monomial in descending order according to degree.

–5z4 – 2z

2 + 4z

10. 2a + 4a3 – 5a

2 – 1

SOLUTION: Find the degree of each term. 2a → 1

4a3→ 3

– 5a2 → 2

– 1 → 0

The greatest degree is 3, from the term 4a3, so the leading coefficient of 2a + 4a

3 – 5a

2 – 1 is 4.

Rewrite the polynomial with each monomial in descending order according to degree.

4a3– 5a

2 + 2a – 1

Find each sum or difference.

11. (6x3 − 4) + (−2x

3 + 9)

SOLUTION:

12. (g3 − 2g2 + 5g + 6) − (g2 + 2g)

SOLUTION:

13. (4 + 2a2 − 2a) − (3a

2 − 8a + 7)

SOLUTION:

14. (8y − 4y2) + (3y − 9y

2)

SOLUTION:

15. (−4z3 − 2z + 8) − (4z

3 + 3z2 − 5)

SOLUTION:

16. (−3d2 − 8 + 2d) + (4d − 12 + d2

)

SOLUTION:

17. (y + 5) + (2y + 4y2 – 2)

SOLUTION:

18. (3n3 − 5n + n2

) − (−8n2 + 3n

3)

SOLUTION:

19. CCSS SENSE-MAKING The total number of students T who traveled for spring break consists of two groups: students who flew to their destinations F and students who drove to their destination D. The number (in thousands) of students who flew and the total number of students who flew or drove can be modeled by the following equations,where n is the number of years since 1995. T = 14n + 21

F = 8n + 7 a. Write an equation that models the number of students who drove to their destination for this time period. b. Predict the number of students who will drive to their destination in 2012. c. How many students will drive or fly to their destination in 2015?

SOLUTION: a.

D = 6n + 14 b. n = 2012 – 1995 = 17

The number of students who will drive to their destination in 2012 is 116,000 students. c. n = 2015 – 1995 = 20

The number of students who will drive or fly to their destination in 2015 is 301,000 students.

Determine whether each expression is a polynomial. If it is a polynomial, find the degree and determine whether it is a monomial, binomial, or trinomial.

20.

SOLUTION: A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer exponents. It has

only one term. is a division of two monomials, so it is not a monomial.

21.

SOLUTION: A polynomial is a monomial or the sum of monomials. 21 is a monomial, so it is also a polynomial. The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of 21 is 0. The polynomial has only one term, so it is a monomial.

22. c4 – 2c

2 + 1

SOLUTION:

A polynomial is a monomial or the sum of monomials. c4 – 2c

2 + 1

is the sum of 3 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 4, 2, and

0, so the degree of c4 – 2c2 + 1 is 4. The polynomial has three terms, so it is a trinomial.

23. d + 3dc

SOLUTION: A polynomial is a monomial or the sum of monomials. A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer

exponents. 3dc

has a variable in the exponent, so it is not a monomial.

24. a – a2

SOLUTION:

A polynomial is a monomial or the sum of monomials. a – a2 is the sum of 2 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 1 and 2,

so the degree of a – a2 is 2. The polynomial has two terms, so it is a binomial.

25. 5n3 + nq

3

SOLUTION:

A polynomial is a monomial or the sum of monomials. 5n3 + nq3 is the sum of 2 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 3 and 4,

so the degree of 5n3 + nq3 is 4. The polynomial has two terms, so it is a binomial.

Write each polynomial in standard form. Identify the leading coefficient.

26. 5x2 – 2 + 3x

SOLUTION: Find the degree of each term.

5x2 → 2

– 2 → 0

3x → 1

The greatest degree is 2, from the term 5x2, so the leading coefficient of 5x

2 – 2 + 3x is 5.

Rewrite the polynomial with each monomial in descending order according to degree.

5x2 + 3x – 2

27. 8y + 7y3

SOLUTION: Find the degree of each term.

7y3 → 3

8y → 1

The greatest degree is 3, from the term 7y3, so the leading coefficient of 8y + 7y

3 is 7.

Rewrite the polynomial with each monomial in descending order according to degree.

7y3 + 8y

28. 4 – 3c – 5c2

SOLUTION: Find the degree of each term. 4 → 0 3c → 1

– 5c2 → 2

The greatest degree is 2, from the term – 5c2, so the leading coefficient of 4 – 3c – 5c

2 is –5.

Rewrite the polynomial with each monomial in descending order according to degree.

–5c2 – 3c + 4

29. –y3 + 3y – 3y

2 + 2

SOLUTION: Find the degree of each term.

–y3 → 3

3y → 1

3y2 → 2

2 → 0

The greatest degree is 3, from the term –y3, so the leading coefficient of –y

3 + 3y – 3y

2 + 2 is –1.

Rewrite the polynomial with each monomial in descending order according to degree. –y 3 – 3y 2 + 3y + 2

30. 11t + 2t2 – 3 + t

5

SOLUTION: Find the degree of each term. 11t → 1

2t2 → 2

–3 → 0

t5 → 5

The greatest degree is 5, from the term t5, so the leading coefficient of 11t + 2t2 – 3 + t

5 is 1.

Rewrite the polynomial with each monomial in descending order according to degree.

t5 + 2t

2 + 11t – 3

31. 2 + r – r3

SOLUTION: Find the degree of each term. 2 → 0 r → 1

– r3→ 3

The greatest degree is 3, from the term – r3, so the leading coefficient of 2 + r – r3

is –1. Rewrite the polynomial with each monomial in descending order according to degree.

–r3 + r + 2

32.

SOLUTION: Find the degree of each term.

→ 0

–3x4 → 4

7 → 0

The greatest degree is 4, from the term –3x4, so the leading coefficient of is –3.

Rewrite the polynomial with each monomial in descending order according to degree.

33. –9b2 + 10b – b

6

SOLUTION: Find the degree of each term.

–9b2 → 2

10b → 1

–b6→ 6

The greatest degree is 6, from the term –b6, so the leading coefficient of –9b2 + 10b – b6 is –1. Rewrite the polynomial with each monomial in descending order according to degree.

–b6 – 9b

2 + 10b

Find each sum or difference.

34. (2c2 + 6c + 4) + (5c – 7)

SOLUTION:

35. (2x + 3x2) − (7 − 8x

2)

SOLUTION:

36. (3c3 − c + 11) − (c2 + 2c + 8)

SOLUTION:

37. (z2 + z) + (z

2 − 11)

SOLUTION:

38. (2x − 2y + 1) − (3y + 4x)

SOLUTION:

39. (4a − 5b2 + 3) + (6 − 2a + 3b

2)

SOLUTION:

40. (x2y − 3x

2 + y) + (3y − 2x2y)

SOLUTION:

41. (−8xy + 3x2 − 5y) + (4x

2 − 2y + 6xy)

SOLUTION:

42. (5n − 2p2 + 2np) − (4p

2 + 4n)

SOLUTION:

43. (4rxt − 8r2x + x2

) − (6rx2 + 5rxt − 2x

2)

SOLUTION:

44. PETS From 1999 through 2009, the number of dogs D and the number of cats C (in hundreds) adopted from animal

shelters in the United States are modeled by the equations D = 2n + 3 and C = n + 4, where n is the number of yearssince 1999. a. Write an equation that models the total number T of dogs and cats adopted in hundreds for this time period. b. If this trend continues, how many dogs and cats will be adopted in 2013?

SOLUTION: a.

So, an equation that models the total number of dogs and cats adopted is T = 3n + 7. b. Evaluate the equation for the total number of dogs and cats for n = 2013 – 1999 = 14.

The number of cats and dogs adopted in 2013 will be 49 × 100 or 4900 cats and dogs.

Classify each polynomial according to its degree and number of terms.

45. 4x – 3x2 + 5

SOLUTION:

Find the degree of each term of 4x – 3x2 + 5.

4x → 1

– 3x2 → 2

5 → 0

The greatest degree is 2 and there are 3 terms, so 4x – 3x2 + 5 is a quadratic trinomial.

46. 11z3

SOLUTION: Find the degree of each term. 11z3

→ 3

The greatest degree is 3 and there is one term, so 11z3 is a cubic monomial.

47. 9 + y4

SOLUTION:

Find the degree of each term of 9 + y 4.

9 → 1

y4→ 4

The greatest degree is 4 and there are 2 terms, so 9 + y 4 is a quartic binomial.

48. 3x3 – 7

SOLUTION:

Find the degree of each term of 3x3 – 7.

3x3 → 3

–7 → 0

The greatest degree is 3 and there are 2 terms, so 3x3 – 7 is a cubic binomial.

49. –2x5 – x

2 + 5x – 8

SOLUTION:

Find the degree of each term of –2x5 – x2 + 5x – 8.

–2x5 → 5

–x2 → 2

5x → 1 – 8 → 0

The greatest degree is 5 and there are 4 terms, so –2x5 – x2 + 5x – 8 is a quintic polynomial.

50. 10t – 4t2 + 6t

3

SOLUTION:

Find the degree of each term of 10t – 4t2 + 6t

3.

10t → 1

4t2 → 2

6t3 → 3

The greatest degree is 3 and there are 3 terms, so 10t – 4t2 + 6t

3 is a cubic trinomial.

51. ENROLLMENT In a rapidly growing school system, the numbers (in hundreds) of total students N and

K-5 students P enrolled from 2000 to 2009 are modeled by the equations N = 1.25t2 – t + 7.5 and P =

0.7t2 – 0.95t + 3.8, where t is the number of years since 2000.

a. Write an equation modeling the number of 6-12 students S enrolled for this time period. b. How many 6-12 students were enrolled in the school system in 2007?

SOLUTION: a. To write an equation that represents the number of 6-12 students enrolled, subtract the equations that represent the total number of students and the number of K-5 students.

b. Replace t with 7 in the equation for S to determine the number of students enrolled in 6-12 in 2007.

In 2007 there were 30.3 hundreds or 3030 students enrolled in 6-12.

52. CCSS REASONING The perimeter of the figure shown is represented by the expression 3x2 − 7x + 2. Write a

polynomial that represents the measure of the third side.

SOLUTION:

53. GEOMETRY Consider the rectangle.

a. What does (4x2 + 2x – 1)(2x

2 – x + 3) represent?

b. What does 2(4x2 + 2x – 1) + 2(2x

2 – x + 3) represent?

SOLUTION:

a. (4x2 + 2x – 1)(2x

2 – x + 3) is a multiplication of the length and the width of the rectangle, which is the formula for

the area of a rectangle.

b. 2(4x2 + 2x – 1) + 2(2x

2 – x + 3) is the sum of twice the length and twice the width of a rectangle, which is the

formula for the perimeter of the rectangle

Find each sum or difference.

54. (4x + 2y − 6z) + (5y − 2z + 7x) + (−9z − 2x − 3y)

SOLUTION:

55. (5a2 − 4) + (a

2 − 2a + 12) + (4a2 − 6a + 8)

SOLUTION:

56. (3c2 − 7) + (4c + 7) − (c

2 + 5c − 8)

SOLUTION:

57. (3n3 + 3n − 10) − (4n

2 − 5n) + (4n3 − 3n

2 − 9n + 4)

SOLUTION:

58. FOOTBALL The National Football League is divided into two conferences, the American A and the National N. From 2002 through 2009, the total attendance T (in thousands) for both conferences and for the American Conference games are modeled by the following equations, where x is the number of years since 2002.

T = –0.69x3 + 55.83x

2 + 643.31x + 10,538

A = –3.78x3 + 58.96x

2 + 265.96x + 5257

Determine how many people attended National Conference football games in 2009.

SOLUTION:

Let x = 7 represent 2009, then find how many people attended a National Conference football game in 2009.

In 2009 the number of people who attended National Conference football games was about 8829 thousand. Multiply by 1000 to find the attendance in standard form. 8829 ×1000 = 8,829,000. So, about 8,829,000 people attended National Conference football games in 2009.

59. CAR RENTAL The cost to rent a car for a day is $15 plus $0.15 for each mile driven. a. Write a polynomial that represents the cost of renting a car for m miles. b. If a car is driven 145 miles, how much would it cost to rent? c. If a car is driven 105 miles each day for four days, how much would it cost to rent a car? d. If a car is driven 220 miles each day for seven days, how much would it cost to rent a car?

SOLUTION: a. The cost to rent a car is the daily rate and the mileage cost or 15 + 0.15m. b. Substitute 145 for m to find the daily cost to drive 145 miles.

The cost to rent the car would be $36.75. c. The expression represents the cost per day. For 4 days, multiply the entire expression by 4 to find the cost for a 4 day trip. Substitute 105 for m.

The cost to rent the car would be $123. d. The expression represents the cost per day. For 7 days, multiply the entire expression by 7 to find the cost for a 7 day trip. Substitute 220 for m.

The cost to rent the car would be $336.

60. MULTIPLE REPRESENTATIONS In this problem, you will explore perimeter and area. a. Geometric Draw three rectangles that each have a perimeter of 400 feet. b. Tabular Record the width and length of each rectangle in a table like the one shown below. Find the area of eachrectangle.

c. Graphical On a coordinate system, graph the area of rectangle 4 in terms of the length, x. Use the graph to determine the largest area possible. d. Analytical Determine the length and width that produce the largest area.

SOLUTION: a. For the perimeters to be 400, create lengths and widths that sum to 200. Note that some lengths are already provided in the table in part b.

b. The area is length multiplied by width. Make sure the units are squared for area. The sum of the length and width must be 200, so if the length is x, the width must be 200 – x.

c. The length will be the x-values on the horizontal and the area will be the y-values on the vertical. The length cannot pass 200 since the sum of the length and width is 200. Set the intervals for the x-axis to 25 feet. Extend the table of values to find more points to plot on the graph. It appears that 10,000 is the greatest area, so set the intervals

for the y-axis to 1000 ft2. After graphing, it appears that the highest point on the graph is at an area of 10,000 ft

2.

d. The associated x-value with the maximum area is x = 100, so the length must be 100 and the width must be 200 – 100, or 100. The length and width of the rectangle must be 100 feet each to have the largest area.

61. CCSS CRITIQUE Cheyenne and Sebastian are finding (2x2 − x) − (3x + 3x

2 − 2). Is either of them correct? Explain your reasoning.

SOLUTION:

Neither is correct. Cheyenne, did not distribute the negative to the 2nd and 3rd terms when she found the additive inverse. Sebastian did not distribute the negate to the 3rd terms when he found the additive inverse. To find the additive inverse, all terms should be multiplied by −1.

62. REASONING Determine whether each of the following statements is true or false . Explain your reasoning. a. A binomial can have a degree of zero. b. The order in which polynomials are subtracted does not matter.

SOLUTION: a. If a binomial has two terms that are each a degree of 0, then those terms can be combined and the binomial becomes a monomial. For example, 18 + 7 = 25. If one of the terms of the binomial does not have a degree of 0, then the binomial cannot have a degree of 0, since the degree of a polynomial is the greatest degree of any term in the polynomial. b. Subtraction is not commutative. While 2 + 5 = 5 + 2, 2 – 5 ≠ 5 – 2. This is also true for polynomials. Sample answer: (2x – 3) – (4x – 3) = –2x, but (4x – 3) – (2x – 3) = 2x

63. CHALLENGE Write a polynomial that represents the sum of an odd integer 2n + 1 and the next two consecutive odd integers.

SOLUTION:

64. WRITING IN MATH Why would you add or subtract equations that represent real-world situations? Explain.

SOLUTION:

65. WRITING IN MATH Describe how to add and subtract polynomials using both the vertical and horizontal formats.

SOLUTION: To add polynomials in a horizontal format, you combine like terms. For the vertical format, you write the polynomials in standard form, align like terms in columns, and combine like terms.

To subtract polynomials in a horizontal format you find the additive inverse of the polynomial you are subtracting, andthen combine like terms. For the vertical format, you write the polynomials in standard form, align like terms in columns, and subtract by adding the additive inverse.

66. Three consecutive integers can be represented by x, x + 1, and x + 2. What is the sum of these three integers? A x(x + 1)(x + 2)

B x3 + 3

C 3x + 3 D x + 3

SOLUTION:

The correct choice is C.

67. SHORT RESPONSE What is the perimeter of a square with sides that measure 2x + 3 units?

SOLUTION:

The perimeter of the square is 8x + 12 units.

68. Jim cuts a board in the shape of a regular hexagon and pounds in a nail at each vertex, as shown. How many rubber bands will he need to stretch a rubber band across every possible pair of nails?

F 15 G 14 H 12 J 9

SOLUTION: The first nail would connect to 5 others, the second to 4 others, the third to 3 others, etc.5 + 4 + 3 + 2 + 1 = 15 The correct choice is F.

69. Which ordered pair is in the solution set of the system of inequalities shown in the graph?

A (−3, 0) B (0, −3) C (5, 0) D (0, 5)

SOLUTION: Choice A is outside the shaded area for both inequalities. Choices B and D are inside the shaded area for only one inequality. Choice C is the only point in the solution for both inequalities. So, the correct choice is C.

70. COMPUTERS A computer technician charges by the hour to fix and repair computer equipment. The total cost of the technician for one hour is $75, for two hours is $125, for three hours is $175, for four hours is $225, and so on. Write a recursive formula for the sequence.

SOLUTION: Write out the terms. $75, $125, $175, $225, ... The first term is 75, and 50 is added to form each following term. Therefore, we have a1 = 75, an = an – 1 + 50, n ≥ 2.

Determine whether each sequence is arithmetic, geometric, or neither. Explain.71. 8, –32, 128, –512, ...

SOLUTION: Check for a common difference. –32 – 8 = –40 128 – (–32) = 160 There is no common difference. Check for a common ratio. –32 ÷ 8 = –4 128 ÷ (–32) = –4 Geometric; the common ratio is –4.

72. 25, 8, –9, –26, ...

SOLUTION: Check for a common difference. 8 – 25 = –17 –9 – 8 = –17 Arithmetic; the common difference is –17.

73.

SOLUTION: Check for a common difference.

There is no common difference. Check for a common ratio.

There is no common ratio, so the sequence is not arithmetic or geometric.

74. 43, 52, 61, 70, ...

SOLUTION: Check for a common difference. 52 – 43 = 9 61 – 52 = 9 Arithmetic; the common difference is 9.

75. –27, –16, –5, 6, ...

SOLUTION: Check for a common difference. –16 – (–27) = 11 –5 – (–16) = 11 Arithmetic; the common difference is 11.

76. 200, 100, 50, 25, …

SOLUTION: Check for a common difference. 100 – 200 = –100 50 – 100 = –50 There is no common difference. Check for a common ratio. 100 ÷ 200 = 0.5 50 ÷ 100 = 0.5

Geometric; the common ratio is 0.5 or .

77. JOBS Kimi received an offer for a new job. She wants to compare the offer with her current job. What is total amount of sales that Kimi must get each month to make the same income at either job?

SOLUTION: Let s be Kimi’s monthly sales.

Kimi must sell $80,000 each month to make the same income at either job.

Determine whether each sequence is an arithmetic sequence. If it is, state the common difference.78. 24, 16, 8, 0, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is –8, because 16 – 24 = –8; 8 –16 = –8; etc.

79. , 13, 26, …

SOLUTION: Find the difference between the terms.

– =

13 – =

26 – 13 = 13 There is not a common different. The sequence is not an arithmetic sequence

80. 7, 6, 5, 4, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is –1, because 6 – 7 = –1; 5 – 6 = –1; 4 – 5 = –1;etc.

81. 10, 12, 15, 18, …

SOLUTION: Find the difference between the terms. 12 – 10 = 2 15 – 12 = 3 18 – 15 = 3 There is no common difference. The sequence is not an arithmetic sequence.

82. −15, −11, −7, −3, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is 4, because –11 – (–15) = 4; –7 – (–11) = 4; –3 – (–7) = 4;etc.

83. −0.3, 0.2, 0.7, 1.2, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is 0.5, because 0.2 – (–0.3) = 0.5; 0.7 – 0.2 = 0.5; 1.2 – 0.7 = 0.5; etc.

Simplify.

84. t(t5)(t

7)

SOLUTION:

85. n3(n

2)(−2n

3)

SOLUTION:

86. (5t5v

2)(10t

3v

4)

SOLUTION:

87. (−8u4z

5)(5uz

4)

SOLUTION:

88. [(3)2]3

SOLUTION:

89. [(2)3]2

SOLUTION:

90. (2m4k

3)2(−3mk

2)3

SOLUTION:

91. (6xy2)2(2x

2y

2z

2)3

SOLUTION:

eSolutions Manual - Powered by Cognero Page 14

8-1 Adding and Subtracting Polynomials

Determine whether each expression is a polynomial. If it is a polynomial, find the degree and determine whether it is a monomial, binomial, or trinomial.

1. 7ab + 6b2 – 2a

3

SOLUTION:

A polynomial is a monomial or the sum of monomials. 7ab + 6b2 – 2a

3 is the sum of 3 monomials, so it is a

polynomial. The degree of a polynomial is the greatest degree of any term in the polynomial.The degree of each term is 2, 2, and

3, so the degree of 7ab + 6b2 – 2a

3 is 3. The polynomial has three terms, so it is a trinomial.

2. 2y – 5 + 3y2

SOLUTION:

2y – 5 + 3y2

is the sum of monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 1, 0, and

2, so the degree of 2y – 5 + 3y2 is 2. The polynomial has three terms, so it is a trinomial.

3. 3x2

SOLUTION:

A polynomial is a monomial or the sum of monomials, so 3x2 is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of 3x2 is 2. The

polynomial has one term, so it is a monomial.

4.

SOLUTION: A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer exponents. It has

only one term. is a division of two monomials, so it is not a monomial.

5. 5m2p

3 + 6

SOLUTION:

A polynomial is a monomial or the sum of monomials. 5m2p

3 + 6

is the sum of 2 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 5 and 0,

so the degree of 5m2p

3 + 6 is 5. The polynomial has two terms, so it is a binomial.

6. 5q–4

+ 6q

SOLUTION: A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer exponents. It has

only one term. 5q-4 is equal to , which is a division of two monomials, so it is not a monomial.

Write each polynomial in standard form. Identify the leading coefficient.

7. –4d4 + 1 – d

2

SOLUTION: Find the degree of each term.

–4d4→ 4

1 → 0

–d2 → 2

The greatest degree is 4, from the term –4d4, so the leading coefficient of –4d

4 + 1 – d

2 is –4.

Rewrite the polynomial with each monomial in descending order according to degree.

–4d4 – d

2 + 1

8. 2x5 – 12 + 3x

SOLUTION: Find the degree of each term.

2x5 → 5

–12 → 0 3x → 1

The greatest degree is 5, from the term 2x5, so the leading coefficient of 2x

5 – 12 + 3x is 2.

Rewrite the polynomial with each monomial in descending order according to degree.

2x5 + 3x – 12

9. 4z – 2z2 – 5z

4

SOLUTION: Find the degree of each term. 4z → 1

– 2z2 → 2

– 5z4→ 4

The greatest degree is 4, from the term – 5z4, so the leading coefficient of 4z – 2z

2 – 5z

4 is –5.

Rewrite the polynomial with each monomial in descending order according to degree.

–5z4 – 2z

2 + 4z

10. 2a + 4a3 – 5a

2 – 1

SOLUTION: Find the degree of each term. 2a → 1

4a3→ 3

– 5a2 → 2

– 1 → 0

The greatest degree is 3, from the term 4a3, so the leading coefficient of 2a + 4a

3 – 5a

2 – 1 is 4.

Rewrite the polynomial with each monomial in descending order according to degree.

4a3– 5a

2 + 2a – 1

Find each sum or difference.

11. (6x3 − 4) + (−2x

3 + 9)

SOLUTION:

12. (g3 − 2g2 + 5g + 6) − (g2 + 2g)

SOLUTION:

13. (4 + 2a2 − 2a) − (3a

2 − 8a + 7)

SOLUTION:

14. (8y − 4y2) + (3y − 9y

2)

SOLUTION:

15. (−4z3 − 2z + 8) − (4z

3 + 3z2 − 5)

SOLUTION:

16. (−3d2 − 8 + 2d) + (4d − 12 + d2

)

SOLUTION:

17. (y + 5) + (2y + 4y2 – 2)

SOLUTION:

18. (3n3 − 5n + n2

) − (−8n2 + 3n

3)

SOLUTION:

19. CCSS SENSE-MAKING The total number of students T who traveled for spring break consists of two groups: students who flew to their destinations F and students who drove to their destination D. The number (in thousands) of students who flew and the total number of students who flew or drove can be modeled by the following equations,where n is the number of years since 1995. T = 14n + 21

F = 8n + 7 a. Write an equation that models the number of students who drove to their destination for this time period. b. Predict the number of students who will drive to their destination in 2012. c. How many students will drive or fly to their destination in 2015?

SOLUTION: a.

D = 6n + 14 b. n = 2012 – 1995 = 17

The number of students who will drive to their destination in 2012 is 116,000 students. c. n = 2015 – 1995 = 20

The number of students who will drive or fly to their destination in 2015 is 301,000 students.

Determine whether each expression is a polynomial. If it is a polynomial, find the degree and determine whether it is a monomial, binomial, or trinomial.

20.

SOLUTION: A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer exponents. It has

only one term. is a division of two monomials, so it is not a monomial.

21.

SOLUTION: A polynomial is a monomial or the sum of monomials. 21 is a monomial, so it is also a polynomial. The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of 21 is 0. The polynomial has only one term, so it is a monomial.

22. c4 – 2c

2 + 1

SOLUTION:

A polynomial is a monomial or the sum of monomials. c4 – 2c

2 + 1

is the sum of 3 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 4, 2, and

0, so the degree of c4 – 2c2 + 1 is 4. The polynomial has three terms, so it is a trinomial.

23. d + 3dc

SOLUTION: A polynomial is a monomial or the sum of monomials. A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer

exponents. 3dc

has a variable in the exponent, so it is not a monomial.

24. a – a2

SOLUTION:

A polynomial is a monomial or the sum of monomials. a – a2 is the sum of 2 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 1 and 2,

so the degree of a – a2 is 2. The polynomial has two terms, so it is a binomial.

25. 5n3 + nq

3

SOLUTION:

A polynomial is a monomial or the sum of monomials. 5n3 + nq3 is the sum of 2 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 3 and 4,

so the degree of 5n3 + nq3 is 4. The polynomial has two terms, so it is a binomial.

Write each polynomial in standard form. Identify the leading coefficient.

26. 5x2 – 2 + 3x

SOLUTION: Find the degree of each term.

5x2 → 2

– 2 → 0

3x → 1

The greatest degree is 2, from the term 5x2, so the leading coefficient of 5x

2 – 2 + 3x is 5.

Rewrite the polynomial with each monomial in descending order according to degree.

5x2 + 3x – 2

27. 8y + 7y3

SOLUTION: Find the degree of each term.

7y3 → 3

8y → 1

The greatest degree is 3, from the term 7y3, so the leading coefficient of 8y + 7y

3 is 7.

Rewrite the polynomial with each monomial in descending order according to degree.

7y3 + 8y

28. 4 – 3c – 5c2

SOLUTION: Find the degree of each term. 4 → 0 3c → 1

– 5c2 → 2

The greatest degree is 2, from the term – 5c2, so the leading coefficient of 4 – 3c – 5c

2 is –5.

Rewrite the polynomial with each monomial in descending order according to degree.

–5c2 – 3c + 4

29. –y3 + 3y – 3y

2 + 2

SOLUTION: Find the degree of each term.

–y3 → 3

3y → 1

3y2 → 2

2 → 0

The greatest degree is 3, from the term –y3, so the leading coefficient of –y

3 + 3y – 3y

2 + 2 is –1.

Rewrite the polynomial with each monomial in descending order according to degree. –y 3 – 3y 2 + 3y + 2

30. 11t + 2t2 – 3 + t

5

SOLUTION: Find the degree of each term. 11t → 1

2t2 → 2

–3 → 0

t5 → 5

The greatest degree is 5, from the term t5, so the leading coefficient of 11t + 2t2 – 3 + t

5 is 1.

Rewrite the polynomial with each monomial in descending order according to degree.

t5 + 2t

2 + 11t – 3

31. 2 + r – r3

SOLUTION: Find the degree of each term. 2 → 0 r → 1

– r3→ 3

The greatest degree is 3, from the term – r3, so the leading coefficient of 2 + r – r3

is –1. Rewrite the polynomial with each monomial in descending order according to degree.

–r3 + r + 2

32.

SOLUTION: Find the degree of each term.

→ 0

–3x4 → 4

7 → 0

The greatest degree is 4, from the term –3x4, so the leading coefficient of is –3.

Rewrite the polynomial with each monomial in descending order according to degree.

33. –9b2 + 10b – b

6

SOLUTION: Find the degree of each term.

–9b2 → 2

10b → 1

–b6→ 6

The greatest degree is 6, from the term –b6, so the leading coefficient of –9b2 + 10b – b6 is –1. Rewrite the polynomial with each monomial in descending order according to degree.

–b6 – 9b

2 + 10b

Find each sum or difference.

34. (2c2 + 6c + 4) + (5c – 7)

SOLUTION:

35. (2x + 3x2) − (7 − 8x

2)

SOLUTION:

36. (3c3 − c + 11) − (c2 + 2c + 8)

SOLUTION:

37. (z2 + z) + (z

2 − 11)

SOLUTION:

38. (2x − 2y + 1) − (3y + 4x)

SOLUTION:

39. (4a − 5b2 + 3) + (6 − 2a + 3b

2)

SOLUTION:

40. (x2y − 3x

2 + y) + (3y − 2x2y)

SOLUTION:

41. (−8xy + 3x2 − 5y) + (4x

2 − 2y + 6xy)

SOLUTION:

42. (5n − 2p2 + 2np) − (4p

2 + 4n)

SOLUTION:

43. (4rxt − 8r2x + x2

) − (6rx2 + 5rxt − 2x

2)

SOLUTION:

44. PETS From 1999 through 2009, the number of dogs D and the number of cats C (in hundreds) adopted from animal

shelters in the United States are modeled by the equations D = 2n + 3 and C = n + 4, where n is the number of yearssince 1999. a. Write an equation that models the total number T of dogs and cats adopted in hundreds for this time period. b. If this trend continues, how many dogs and cats will be adopted in 2013?

SOLUTION: a.

So, an equation that models the total number of dogs and cats adopted is T = 3n + 7. b. Evaluate the equation for the total number of dogs and cats for n = 2013 – 1999 = 14.

The number of cats and dogs adopted in 2013 will be 49 × 100 or 4900 cats and dogs.

Classify each polynomial according to its degree and number of terms.

45. 4x – 3x2 + 5

SOLUTION:

Find the degree of each term of 4x – 3x2 + 5.

4x → 1

– 3x2 → 2

5 → 0

The greatest degree is 2 and there are 3 terms, so 4x – 3x2 + 5 is a quadratic trinomial.

46. 11z3

SOLUTION: Find the degree of each term. 11z3

→ 3

The greatest degree is 3 and there is one term, so 11z3 is a cubic monomial.

47. 9 + y4

SOLUTION:

Find the degree of each term of 9 + y 4.

9 → 1

y4→ 4

The greatest degree is 4 and there are 2 terms, so 9 + y 4 is a quartic binomial.

48. 3x3 – 7

SOLUTION:

Find the degree of each term of 3x3 – 7.

3x3 → 3

–7 → 0

The greatest degree is 3 and there are 2 terms, so 3x3 – 7 is a cubic binomial.

49. –2x5 – x

2 + 5x – 8

SOLUTION:

Find the degree of each term of –2x5 – x2 + 5x – 8.

–2x5 → 5

–x2 → 2

5x → 1 – 8 → 0

The greatest degree is 5 and there are 4 terms, so –2x5 – x2 + 5x – 8 is a quintic polynomial.

50. 10t – 4t2 + 6t

3

SOLUTION:

Find the degree of each term of 10t – 4t2 + 6t

3.

10t → 1

4t2 → 2

6t3 → 3

The greatest degree is 3 and there are 3 terms, so 10t – 4t2 + 6t

3 is a cubic trinomial.

51. ENROLLMENT In a rapidly growing school system, the numbers (in hundreds) of total students N and

K-5 students P enrolled from 2000 to 2009 are modeled by the equations N = 1.25t2 – t + 7.5 and P =

0.7t2 – 0.95t + 3.8, where t is the number of years since 2000.

a. Write an equation modeling the number of 6-12 students S enrolled for this time period. b. How many 6-12 students were enrolled in the school system in 2007?

SOLUTION: a. To write an equation that represents the number of 6-12 students enrolled, subtract the equations that represent the total number of students and the number of K-5 students.

b. Replace t with 7 in the equation for S to determine the number of students enrolled in 6-12 in 2007.

In 2007 there were 30.3 hundreds or 3030 students enrolled in 6-12.

52. CCSS REASONING The perimeter of the figure shown is represented by the expression 3x2 − 7x + 2. Write a

polynomial that represents the measure of the third side.

SOLUTION:

53. GEOMETRY Consider the rectangle.

a. What does (4x2 + 2x – 1)(2x

2 – x + 3) represent?

b. What does 2(4x2 + 2x – 1) + 2(2x

2 – x + 3) represent?

SOLUTION:

a. (4x2 + 2x – 1)(2x

2 – x + 3) is a multiplication of the length and the width of the rectangle, which is the formula for

the area of a rectangle.

b. 2(4x2 + 2x – 1) + 2(2x

2 – x + 3) is the sum of twice the length and twice the width of a rectangle, which is the

formula for the perimeter of the rectangle

Find each sum or difference.

54. (4x + 2y − 6z) + (5y − 2z + 7x) + (−9z − 2x − 3y)

SOLUTION:

55. (5a2 − 4) + (a

2 − 2a + 12) + (4a2 − 6a + 8)

SOLUTION:

56. (3c2 − 7) + (4c + 7) − (c

2 + 5c − 8)

SOLUTION:

57. (3n3 + 3n − 10) − (4n

2 − 5n) + (4n3 − 3n

2 − 9n + 4)

SOLUTION:

58. FOOTBALL The National Football League is divided into two conferences, the American A and the National N. From 2002 through 2009, the total attendance T (in thousands) for both conferences and for the American Conference games are modeled by the following equations, where x is the number of years since 2002.

T = –0.69x3 + 55.83x

2 + 643.31x + 10,538

A = –3.78x3 + 58.96x

2 + 265.96x + 5257

Determine how many people attended National Conference football games in 2009.

SOLUTION:

Let x = 7 represent 2009, then find how many people attended a National Conference football game in 2009.

In 2009 the number of people who attended National Conference football games was about 8829 thousand. Multiply by 1000 to find the attendance in standard form. 8829 ×1000 = 8,829,000. So, about 8,829,000 people attended National Conference football games in 2009.

59. CAR RENTAL The cost to rent a car for a day is $15 plus $0.15 for each mile driven. a. Write a polynomial that represents the cost of renting a car for m miles. b. If a car is driven 145 miles, how much would it cost to rent? c. If a car is driven 105 miles each day for four days, how much would it cost to rent a car? d. If a car is driven 220 miles each day for seven days, how much would it cost to rent a car?

SOLUTION: a. The cost to rent a car is the daily rate and the mileage cost or 15 + 0.15m. b. Substitute 145 for m to find the daily cost to drive 145 miles.

The cost to rent the car would be $36.75. c. The expression represents the cost per day. For 4 days, multiply the entire expression by 4 to find the cost for a 4 day trip. Substitute 105 for m.

The cost to rent the car would be $123. d. The expression represents the cost per day. For 7 days, multiply the entire expression by 7 to find the cost for a 7 day trip. Substitute 220 for m.

The cost to rent the car would be $336.

60. MULTIPLE REPRESENTATIONS In this problem, you will explore perimeter and area. a. Geometric Draw three rectangles that each have a perimeter of 400 feet. b. Tabular Record the width and length of each rectangle in a table like the one shown below. Find the area of eachrectangle.

c. Graphical On a coordinate system, graph the area of rectangle 4 in terms of the length, x. Use the graph to determine the largest area possible. d. Analytical Determine the length and width that produce the largest area.

SOLUTION: a. For the perimeters to be 400, create lengths and widths that sum to 200. Note that some lengths are already provided in the table in part b.

b. The area is length multiplied by width. Make sure the units are squared for area. The sum of the length and width must be 200, so if the length is x, the width must be 200 – x.

c. The length will be the x-values on the horizontal and the area will be the y-values on the vertical. The length cannot pass 200 since the sum of the length and width is 200. Set the intervals for the x-axis to 25 feet. Extend the table of values to find more points to plot on the graph. It appears that 10,000 is the greatest area, so set the intervals

for the y-axis to 1000 ft2. After graphing, it appears that the highest point on the graph is at an area of 10,000 ft

2.

d. The associated x-value with the maximum area is x = 100, so the length must be 100 and the width must be 200 – 100, or 100. The length and width of the rectangle must be 100 feet each to have the largest area.

61. CCSS CRITIQUE Cheyenne and Sebastian are finding (2x2 − x) − (3x + 3x

2 − 2). Is either of them correct? Explain your reasoning.

SOLUTION:

Neither is correct. Cheyenne, did not distribute the negative to the 2nd and 3rd terms when she found the additive inverse. Sebastian did not distribute the negate to the 3rd terms when he found the additive inverse. To find the additive inverse, all terms should be multiplied by −1.

62. REASONING Determine whether each of the following statements is true or false . Explain your reasoning. a. A binomial can have a degree of zero. b. The order in which polynomials are subtracted does not matter.

SOLUTION: a. If a binomial has two terms that are each a degree of 0, then those terms can be combined and the binomial becomes a monomial. For example, 18 + 7 = 25. If one of the terms of the binomial does not have a degree of 0, then the binomial cannot have a degree of 0, since the degree of a polynomial is the greatest degree of any term in the polynomial. b. Subtraction is not commutative. While 2 + 5 = 5 + 2, 2 – 5 ≠ 5 – 2. This is also true for polynomials. Sample answer: (2x – 3) – (4x – 3) = –2x, but (4x – 3) – (2x – 3) = 2x

63. CHALLENGE Write a polynomial that represents the sum of an odd integer 2n + 1 and the next two consecutive odd integers.

SOLUTION:

64. WRITING IN MATH Why would you add or subtract equations that represent real-world situations? Explain.

SOLUTION:

65. WRITING IN MATH Describe how to add and subtract polynomials using both the vertical and horizontal formats.

SOLUTION: To add polynomials in a horizontal format, you combine like terms. For the vertical format, you write the polynomials in standard form, align like terms in columns, and combine like terms.

To subtract polynomials in a horizontal format you find the additive inverse of the polynomial you are subtracting, andthen combine like terms. For the vertical format, you write the polynomials in standard form, align like terms in columns, and subtract by adding the additive inverse.

66. Three consecutive integers can be represented by x, x + 1, and x + 2. What is the sum of these three integers? A x(x + 1)(x + 2)

B x3 + 3

C 3x + 3 D x + 3

SOLUTION:

The correct choice is C.

67. SHORT RESPONSE What is the perimeter of a square with sides that measure 2x + 3 units?

SOLUTION:

The perimeter of the square is 8x + 12 units.

68. Jim cuts a board in the shape of a regular hexagon and pounds in a nail at each vertex, as shown. How many rubber bands will he need to stretch a rubber band across every possible pair of nails?

F 15 G 14 H 12 J 9

SOLUTION: The first nail would connect to 5 others, the second to 4 others, the third to 3 others, etc.5 + 4 + 3 + 2 + 1 = 15 The correct choice is F.

69. Which ordered pair is in the solution set of the system of inequalities shown in the graph?

A (−3, 0) B (0, −3) C (5, 0) D (0, 5)

SOLUTION: Choice A is outside the shaded area for both inequalities. Choices B and D are inside the shaded area for only one inequality. Choice C is the only point in the solution for both inequalities. So, the correct choice is C.

70. COMPUTERS A computer technician charges by the hour to fix and repair computer equipment. The total cost of the technician for one hour is $75, for two hours is $125, for three hours is $175, for four hours is $225, and so on. Write a recursive formula for the sequence.

SOLUTION: Write out the terms. $75, $125, $175, $225, ... The first term is 75, and 50 is added to form each following term. Therefore, we have a1 = 75, an = an – 1 + 50, n ≥ 2.

Determine whether each sequence is arithmetic, geometric, or neither. Explain.71. 8, –32, 128, –512, ...

SOLUTION: Check for a common difference. –32 – 8 = –40 128 – (–32) = 160 There is no common difference. Check for a common ratio. –32 ÷ 8 = –4 128 ÷ (–32) = –4 Geometric; the common ratio is –4.

72. 25, 8, –9, –26, ...

SOLUTION: Check for a common difference. 8 – 25 = –17 –9 – 8 = –17 Arithmetic; the common difference is –17.

73.

SOLUTION: Check for a common difference.

There is no common difference. Check for a common ratio.

There is no common ratio, so the sequence is not arithmetic or geometric.

74. 43, 52, 61, 70, ...

SOLUTION: Check for a common difference. 52 – 43 = 9 61 – 52 = 9 Arithmetic; the common difference is 9.

75. –27, –16, –5, 6, ...

SOLUTION: Check for a common difference. –16 – (–27) = 11 –5 – (–16) = 11 Arithmetic; the common difference is 11.

76. 200, 100, 50, 25, …

SOLUTION: Check for a common difference. 100 – 200 = –100 50 – 100 = –50 There is no common difference. Check for a common ratio. 100 ÷ 200 = 0.5 50 ÷ 100 = 0.5

Geometric; the common ratio is 0.5 or .

77. JOBS Kimi received an offer for a new job. She wants to compare the offer with her current job. What is total amount of sales that Kimi must get each month to make the same income at either job?

SOLUTION: Let s be Kimi’s monthly sales.

Kimi must sell $80,000 each month to make the same income at either job.

Determine whether each sequence is an arithmetic sequence. If it is, state the common difference.78. 24, 16, 8, 0, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is –8, because 16 – 24 = –8; 8 –16 = –8; etc.

79. , 13, 26, …

SOLUTION: Find the difference between the terms.

– =

13 – =

26 – 13 = 13 There is not a common different. The sequence is not an arithmetic sequence

80. 7, 6, 5, 4, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is –1, because 6 – 7 = –1; 5 – 6 = –1; 4 – 5 = –1;etc.

81. 10, 12, 15, 18, …

SOLUTION: Find the difference between the terms. 12 – 10 = 2 15 – 12 = 3 18 – 15 = 3 There is no common difference. The sequence is not an arithmetic sequence.

82. −15, −11, −7, −3, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is 4, because –11 – (–15) = 4; –7 – (–11) = 4; –3 – (–7) = 4;etc.

83. −0.3, 0.2, 0.7, 1.2, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is 0.5, because 0.2 – (–0.3) = 0.5; 0.7 – 0.2 = 0.5; 1.2 – 0.7 = 0.5; etc.

Simplify.

84. t(t5)(t

7)

SOLUTION:

85. n3(n

2)(−2n

3)

SOLUTION:

86. (5t5v

2)(10t

3v

4)

SOLUTION:

87. (−8u4z

5)(5uz

4)

SOLUTION:

88. [(3)2]3

SOLUTION:

89. [(2)3]2

SOLUTION:

90. (2m4k

3)2(−3mk

2)3

SOLUTION:

91. (6xy2)2(2x

2y

2z

2)3

SOLUTION:

eSolutions Manual - Powered by Cognero Page 15

8-1 Adding and Subtracting Polynomials

Determine whether each expression is a polynomial. If it is a polynomial, find the degree and determine whether it is a monomial, binomial, or trinomial.

1. 7ab + 6b2 – 2a

3

SOLUTION:

A polynomial is a monomial or the sum of monomials. 7ab + 6b2 – 2a

3 is the sum of 3 monomials, so it is a

polynomial. The degree of a polynomial is the greatest degree of any term in the polynomial.The degree of each term is 2, 2, and

3, so the degree of 7ab + 6b2 – 2a

3 is 3. The polynomial has three terms, so it is a trinomial.

2. 2y – 5 + 3y2

SOLUTION:

2y – 5 + 3y2

is the sum of monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 1, 0, and

2, so the degree of 2y – 5 + 3y2 is 2. The polynomial has three terms, so it is a trinomial.

3. 3x2

SOLUTION:

A polynomial is a monomial or the sum of monomials, so 3x2 is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of 3x2 is 2. The

polynomial has one term, so it is a monomial.

4.

SOLUTION: A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer exponents. It has

only one term. is a division of two monomials, so it is not a monomial.

5. 5m2p

3 + 6

SOLUTION:

A polynomial is a monomial or the sum of monomials. 5m2p

3 + 6

is the sum of 2 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 5 and 0,

so the degree of 5m2p

3 + 6 is 5. The polynomial has two terms, so it is a binomial.

6. 5q–4

+ 6q

SOLUTION: A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer exponents. It has

only one term. 5q-4 is equal to , which is a division of two monomials, so it is not a monomial.

Write each polynomial in standard form. Identify the leading coefficient.

7. –4d4 + 1 – d

2

SOLUTION: Find the degree of each term.

–4d4→ 4

1 → 0

–d2 → 2

The greatest degree is 4, from the term –4d4, so the leading coefficient of –4d

4 + 1 – d

2 is –4.

Rewrite the polynomial with each monomial in descending order according to degree.

–4d4 – d

2 + 1

8. 2x5 – 12 + 3x

SOLUTION: Find the degree of each term.

2x5 → 5

–12 → 0 3x → 1

The greatest degree is 5, from the term 2x5, so the leading coefficient of 2x

5 – 12 + 3x is 2.

Rewrite the polynomial with each monomial in descending order according to degree.

2x5 + 3x – 12

9. 4z – 2z2 – 5z

4

SOLUTION: Find the degree of each term. 4z → 1

– 2z2 → 2

– 5z4→ 4

The greatest degree is 4, from the term – 5z4, so the leading coefficient of 4z – 2z

2 – 5z

4 is –5.

Rewrite the polynomial with each monomial in descending order according to degree.

–5z4 – 2z

2 + 4z

10. 2a + 4a3 – 5a

2 – 1

SOLUTION: Find the degree of each term. 2a → 1

4a3→ 3

– 5a2 → 2

– 1 → 0

The greatest degree is 3, from the term 4a3, so the leading coefficient of 2a + 4a

3 – 5a

2 – 1 is 4.

Rewrite the polynomial with each monomial in descending order according to degree.

4a3– 5a

2 + 2a – 1

Find each sum or difference.

11. (6x3 − 4) + (−2x

3 + 9)

SOLUTION:

12. (g3 − 2g2 + 5g + 6) − (g2 + 2g)

SOLUTION:

13. (4 + 2a2 − 2a) − (3a

2 − 8a + 7)

SOLUTION:

14. (8y − 4y2) + (3y − 9y

2)

SOLUTION:

15. (−4z3 − 2z + 8) − (4z

3 + 3z2 − 5)

SOLUTION:

16. (−3d2 − 8 + 2d) + (4d − 12 + d2

)

SOLUTION:

17. (y + 5) + (2y + 4y2 – 2)

SOLUTION:

18. (3n3 − 5n + n2

) − (−8n2 + 3n

3)

SOLUTION:

19. CCSS SENSE-MAKING The total number of students T who traveled for spring break consists of two groups: students who flew to their destinations F and students who drove to their destination D. The number (in thousands) of students who flew and the total number of students who flew or drove can be modeled by the following equations,where n is the number of years since 1995. T = 14n + 21

F = 8n + 7 a. Write an equation that models the number of students who drove to their destination for this time period. b. Predict the number of students who will drive to their destination in 2012. c. How many students will drive or fly to their destination in 2015?

SOLUTION: a.

D = 6n + 14 b. n = 2012 – 1995 = 17

The number of students who will drive to their destination in 2012 is 116,000 students. c. n = 2015 – 1995 = 20

The number of students who will drive or fly to their destination in 2015 is 301,000 students.

Determine whether each expression is a polynomial. If it is a polynomial, find the degree and determine whether it is a monomial, binomial, or trinomial.

20.

SOLUTION: A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer exponents. It has

only one term. is a division of two monomials, so it is not a monomial.

21.

SOLUTION: A polynomial is a monomial or the sum of monomials. 21 is a monomial, so it is also a polynomial. The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of 21 is 0. The polynomial has only one term, so it is a monomial.

22. c4 – 2c

2 + 1

SOLUTION:

A polynomial is a monomial or the sum of monomials. c4 – 2c

2 + 1

is the sum of 3 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 4, 2, and

0, so the degree of c4 – 2c2 + 1 is 4. The polynomial has three terms, so it is a trinomial.

23. d + 3dc

SOLUTION: A polynomial is a monomial or the sum of monomials. A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer

exponents. 3dc

has a variable in the exponent, so it is not a monomial.

24. a – a2

SOLUTION:

A polynomial is a monomial or the sum of monomials. a – a2 is the sum of 2 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 1 and 2,

so the degree of a – a2 is 2. The polynomial has two terms, so it is a binomial.

25. 5n3 + nq

3

SOLUTION:

A polynomial is a monomial or the sum of monomials. 5n3 + nq3 is the sum of 2 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 3 and 4,

so the degree of 5n3 + nq3 is 4. The polynomial has two terms, so it is a binomial.

Write each polynomial in standard form. Identify the leading coefficient.

26. 5x2 – 2 + 3x

SOLUTION: Find the degree of each term.

5x2 → 2

– 2 → 0

3x → 1

The greatest degree is 2, from the term 5x2, so the leading coefficient of 5x

2 – 2 + 3x is 5.

Rewrite the polynomial with each monomial in descending order according to degree.

5x2 + 3x – 2

27. 8y + 7y3

SOLUTION: Find the degree of each term.

7y3 → 3

8y → 1

The greatest degree is 3, from the term 7y3, so the leading coefficient of 8y + 7y

3 is 7.

Rewrite the polynomial with each monomial in descending order according to degree.

7y3 + 8y

28. 4 – 3c – 5c2

SOLUTION: Find the degree of each term. 4 → 0 3c → 1

– 5c2 → 2

The greatest degree is 2, from the term – 5c2, so the leading coefficient of 4 – 3c – 5c

2 is –5.

Rewrite the polynomial with each monomial in descending order according to degree.

–5c2 – 3c + 4

29. –y3 + 3y – 3y

2 + 2

SOLUTION: Find the degree of each term.

–y3 → 3

3y → 1

3y2 → 2

2 → 0

The greatest degree is 3, from the term –y3, so the leading coefficient of –y

3 + 3y – 3y

2 + 2 is –1.

Rewrite the polynomial with each monomial in descending order according to degree. –y 3 – 3y 2 + 3y + 2

30. 11t + 2t2 – 3 + t

5

SOLUTION: Find the degree of each term. 11t → 1

2t2 → 2

–3 → 0

t5 → 5

The greatest degree is 5, from the term t5, so the leading coefficient of 11t + 2t2 – 3 + t

5 is 1.

Rewrite the polynomial with each monomial in descending order according to degree.

t5 + 2t

2 + 11t – 3

31. 2 + r – r3

SOLUTION: Find the degree of each term. 2 → 0 r → 1

– r3→ 3

The greatest degree is 3, from the term – r3, so the leading coefficient of 2 + r – r3

is –1. Rewrite the polynomial with each monomial in descending order according to degree.

–r3 + r + 2

32.

SOLUTION: Find the degree of each term.

→ 0

–3x4 → 4

7 → 0

The greatest degree is 4, from the term –3x4, so the leading coefficient of is –3.

Rewrite the polynomial with each monomial in descending order according to degree.

33. –9b2 + 10b – b

6

SOLUTION: Find the degree of each term.

–9b2 → 2

10b → 1

–b6→ 6

The greatest degree is 6, from the term –b6, so the leading coefficient of –9b2 + 10b – b6 is –1. Rewrite the polynomial with each monomial in descending order according to degree.

–b6 – 9b

2 + 10b

Find each sum or difference.

34. (2c2 + 6c + 4) + (5c – 7)

SOLUTION:

35. (2x + 3x2) − (7 − 8x

2)

SOLUTION:

36. (3c3 − c + 11) − (c2 + 2c + 8)

SOLUTION:

37. (z2 + z) + (z

2 − 11)

SOLUTION:

38. (2x − 2y + 1) − (3y + 4x)

SOLUTION:

39. (4a − 5b2 + 3) + (6 − 2a + 3b

2)

SOLUTION:

40. (x2y − 3x

2 + y) + (3y − 2x2y)

SOLUTION:

41. (−8xy + 3x2 − 5y) + (4x

2 − 2y + 6xy)

SOLUTION:

42. (5n − 2p2 + 2np) − (4p

2 + 4n)

SOLUTION:

43. (4rxt − 8r2x + x2

) − (6rx2 + 5rxt − 2x

2)

SOLUTION:

44. PETS From 1999 through 2009, the number of dogs D and the number of cats C (in hundreds) adopted from animal

shelters in the United States are modeled by the equations D = 2n + 3 and C = n + 4, where n is the number of yearssince 1999. a. Write an equation that models the total number T of dogs and cats adopted in hundreds for this time period. b. If this trend continues, how many dogs and cats will be adopted in 2013?

SOLUTION: a.

So, an equation that models the total number of dogs and cats adopted is T = 3n + 7. b. Evaluate the equation for the total number of dogs and cats for n = 2013 – 1999 = 14.

The number of cats and dogs adopted in 2013 will be 49 × 100 or 4900 cats and dogs.

Classify each polynomial according to its degree and number of terms.

45. 4x – 3x2 + 5

SOLUTION:

Find the degree of each term of 4x – 3x2 + 5.

4x → 1

– 3x2 → 2

5 → 0

The greatest degree is 2 and there are 3 terms, so 4x – 3x2 + 5 is a quadratic trinomial.

46. 11z3

SOLUTION: Find the degree of each term. 11z3

→ 3

The greatest degree is 3 and there is one term, so 11z3 is a cubic monomial.

47. 9 + y4

SOLUTION:

Find the degree of each term of 9 + y 4.

9 → 1

y4→ 4

The greatest degree is 4 and there are 2 terms, so 9 + y 4 is a quartic binomial.

48. 3x3 – 7

SOLUTION:

Find the degree of each term of 3x3 – 7.

3x3 → 3

–7 → 0

The greatest degree is 3 and there are 2 terms, so 3x3 – 7 is a cubic binomial.

49. –2x5 – x

2 + 5x – 8

SOLUTION:

Find the degree of each term of –2x5 – x2 + 5x – 8.

–2x5 → 5

–x2 → 2

5x → 1 – 8 → 0

The greatest degree is 5 and there are 4 terms, so –2x5 – x2 + 5x – 8 is a quintic polynomial.

50. 10t – 4t2 + 6t

3

SOLUTION:

Find the degree of each term of 10t – 4t2 + 6t

3.

10t → 1

4t2 → 2

6t3 → 3

The greatest degree is 3 and there are 3 terms, so 10t – 4t2 + 6t

3 is a cubic trinomial.

51. ENROLLMENT In a rapidly growing school system, the numbers (in hundreds) of total students N and

K-5 students P enrolled from 2000 to 2009 are modeled by the equations N = 1.25t2 – t + 7.5 and P =

0.7t2 – 0.95t + 3.8, where t is the number of years since 2000.

a. Write an equation modeling the number of 6-12 students S enrolled for this time period. b. How many 6-12 students were enrolled in the school system in 2007?

SOLUTION: a. To write an equation that represents the number of 6-12 students enrolled, subtract the equations that represent the total number of students and the number of K-5 students.

b. Replace t with 7 in the equation for S to determine the number of students enrolled in 6-12 in 2007.

In 2007 there were 30.3 hundreds or 3030 students enrolled in 6-12.

52. CCSS REASONING The perimeter of the figure shown is represented by the expression 3x2 − 7x + 2. Write a

polynomial that represents the measure of the third side.

SOLUTION:

53. GEOMETRY Consider the rectangle.

a. What does (4x2 + 2x – 1)(2x

2 – x + 3) represent?

b. What does 2(4x2 + 2x – 1) + 2(2x

2 – x + 3) represent?

SOLUTION:

a. (4x2 + 2x – 1)(2x

2 – x + 3) is a multiplication of the length and the width of the rectangle, which is the formula for

the area of a rectangle.

b. 2(4x2 + 2x – 1) + 2(2x

2 – x + 3) is the sum of twice the length and twice the width of a rectangle, which is the

formula for the perimeter of the rectangle

Find each sum or difference.

54. (4x + 2y − 6z) + (5y − 2z + 7x) + (−9z − 2x − 3y)

SOLUTION:

55. (5a2 − 4) + (a

2 − 2a + 12) + (4a2 − 6a + 8)

SOLUTION:

56. (3c2 − 7) + (4c + 7) − (c

2 + 5c − 8)

SOLUTION:

57. (3n3 + 3n − 10) − (4n

2 − 5n) + (4n3 − 3n

2 − 9n + 4)

SOLUTION:

58. FOOTBALL The National Football League is divided into two conferences, the American A and the National N. From 2002 through 2009, the total attendance T (in thousands) for both conferences and for the American Conference games are modeled by the following equations, where x is the number of years since 2002.

T = –0.69x3 + 55.83x

2 + 643.31x + 10,538

A = –3.78x3 + 58.96x

2 + 265.96x + 5257

Determine how many people attended National Conference football games in 2009.

SOLUTION:

Let x = 7 represent 2009, then find how many people attended a National Conference football game in 2009.

In 2009 the number of people who attended National Conference football games was about 8829 thousand. Multiply by 1000 to find the attendance in standard form. 8829 ×1000 = 8,829,000. So, about 8,829,000 people attended National Conference football games in 2009.

59. CAR RENTAL The cost to rent a car for a day is $15 plus $0.15 for each mile driven. a. Write a polynomial that represents the cost of renting a car for m miles. b. If a car is driven 145 miles, how much would it cost to rent? c. If a car is driven 105 miles each day for four days, how much would it cost to rent a car? d. If a car is driven 220 miles each day for seven days, how much would it cost to rent a car?

SOLUTION: a. The cost to rent a car is the daily rate and the mileage cost or 15 + 0.15m. b. Substitute 145 for m to find the daily cost to drive 145 miles.

The cost to rent the car would be $36.75. c. The expression represents the cost per day. For 4 days, multiply the entire expression by 4 to find the cost for a 4 day trip. Substitute 105 for m.

The cost to rent the car would be $123. d. The expression represents the cost per day. For 7 days, multiply the entire expression by 7 to find the cost for a 7 day trip. Substitute 220 for m.

The cost to rent the car would be $336.

60. MULTIPLE REPRESENTATIONS In this problem, you will explore perimeter and area. a. Geometric Draw three rectangles that each have a perimeter of 400 feet. b. Tabular Record the width and length of each rectangle in a table like the one shown below. Find the area of eachrectangle.

c. Graphical On a coordinate system, graph the area of rectangle 4 in terms of the length, x. Use the graph to determine the largest area possible. d. Analytical Determine the length and width that produce the largest area.

SOLUTION: a. For the perimeters to be 400, create lengths and widths that sum to 200. Note that some lengths are already provided in the table in part b.

b. The area is length multiplied by width. Make sure the units are squared for area. The sum of the length and width must be 200, so if the length is x, the width must be 200 – x.

c. The length will be the x-values on the horizontal and the area will be the y-values on the vertical. The length cannot pass 200 since the sum of the length and width is 200. Set the intervals for the x-axis to 25 feet. Extend the table of values to find more points to plot on the graph. It appears that 10,000 is the greatest area, so set the intervals

for the y-axis to 1000 ft2. After graphing, it appears that the highest point on the graph is at an area of 10,000 ft

2.

d. The associated x-value with the maximum area is x = 100, so the length must be 100 and the width must be 200 – 100, or 100. The length and width of the rectangle must be 100 feet each to have the largest area.

61. CCSS CRITIQUE Cheyenne and Sebastian are finding (2x2 − x) − (3x + 3x

2 − 2). Is either of them correct? Explain your reasoning.

SOLUTION:

Neither is correct. Cheyenne, did not distribute the negative to the 2nd and 3rd terms when she found the additive inverse. Sebastian did not distribute the negate to the 3rd terms when he found the additive inverse. To find the additive inverse, all terms should be multiplied by −1.

62. REASONING Determine whether each of the following statements is true or false . Explain your reasoning. a. A binomial can have a degree of zero. b. The order in which polynomials are subtracted does not matter.

SOLUTION: a. If a binomial has two terms that are each a degree of 0, then those terms can be combined and the binomial becomes a monomial. For example, 18 + 7 = 25. If one of the terms of the binomial does not have a degree of 0, then the binomial cannot have a degree of 0, since the degree of a polynomial is the greatest degree of any term in the polynomial. b. Subtraction is not commutative. While 2 + 5 = 5 + 2, 2 – 5 ≠ 5 – 2. This is also true for polynomials. Sample answer: (2x – 3) – (4x – 3) = –2x, but (4x – 3) – (2x – 3) = 2x

63. CHALLENGE Write a polynomial that represents the sum of an odd integer 2n + 1 and the next two consecutive odd integers.

SOLUTION:

64. WRITING IN MATH Why would you add or subtract equations that represent real-world situations? Explain.

SOLUTION:

65. WRITING IN MATH Describe how to add and subtract polynomials using both the vertical and horizontal formats.

SOLUTION: To add polynomials in a horizontal format, you combine like terms. For the vertical format, you write the polynomials in standard form, align like terms in columns, and combine like terms.

To subtract polynomials in a horizontal format you find the additive inverse of the polynomial you are subtracting, andthen combine like terms. For the vertical format, you write the polynomials in standard form, align like terms in columns, and subtract by adding the additive inverse.

66. Three consecutive integers can be represented by x, x + 1, and x + 2. What is the sum of these three integers? A x(x + 1)(x + 2)

B x3 + 3

C 3x + 3 D x + 3

SOLUTION:

The correct choice is C.

67. SHORT RESPONSE What is the perimeter of a square with sides that measure 2x + 3 units?

SOLUTION:

The perimeter of the square is 8x + 12 units.

68. Jim cuts a board in the shape of a regular hexagon and pounds in a nail at each vertex, as shown. How many rubber bands will he need to stretch a rubber band across every possible pair of nails?

F 15 G 14 H 12 J 9

SOLUTION: The first nail would connect to 5 others, the second to 4 others, the third to 3 others, etc.5 + 4 + 3 + 2 + 1 = 15 The correct choice is F.

69. Which ordered pair is in the solution set of the system of inequalities shown in the graph?

A (−3, 0) B (0, −3) C (5, 0) D (0, 5)

SOLUTION: Choice A is outside the shaded area for both inequalities. Choices B and D are inside the shaded area for only one inequality. Choice C is the only point in the solution for both inequalities. So, the correct choice is C.

70. COMPUTERS A computer technician charges by the hour to fix and repair computer equipment. The total cost of the technician for one hour is $75, for two hours is $125, for three hours is $175, for four hours is $225, and so on. Write a recursive formula for the sequence.

SOLUTION: Write out the terms. $75, $125, $175, $225, ... The first term is 75, and 50 is added to form each following term. Therefore, we have a1 = 75, an = an – 1 + 50, n ≥ 2.

Determine whether each sequence is arithmetic, geometric, or neither. Explain.71. 8, –32, 128, –512, ...

SOLUTION: Check for a common difference. –32 – 8 = –40 128 – (–32) = 160 There is no common difference. Check for a common ratio. –32 ÷ 8 = –4 128 ÷ (–32) = –4 Geometric; the common ratio is –4.

72. 25, 8, –9, –26, ...

SOLUTION: Check for a common difference. 8 – 25 = –17 –9 – 8 = –17 Arithmetic; the common difference is –17.

73.

SOLUTION: Check for a common difference.

There is no common difference. Check for a common ratio.

There is no common ratio, so the sequence is not arithmetic or geometric.

74. 43, 52, 61, 70, ...

SOLUTION: Check for a common difference. 52 – 43 = 9 61 – 52 = 9 Arithmetic; the common difference is 9.

75. –27, –16, –5, 6, ...

SOLUTION: Check for a common difference. –16 – (–27) = 11 –5 – (–16) = 11 Arithmetic; the common difference is 11.

76. 200, 100, 50, 25, …

SOLUTION: Check for a common difference. 100 – 200 = –100 50 – 100 = –50 There is no common difference. Check for a common ratio. 100 ÷ 200 = 0.5 50 ÷ 100 = 0.5

Geometric; the common ratio is 0.5 or .

77. JOBS Kimi received an offer for a new job. She wants to compare the offer with her current job. What is total amount of sales that Kimi must get each month to make the same income at either job?

SOLUTION: Let s be Kimi’s monthly sales.

Kimi must sell $80,000 each month to make the same income at either job.

Determine whether each sequence is an arithmetic sequence. If it is, state the common difference.78. 24, 16, 8, 0, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is –8, because 16 – 24 = –8; 8 –16 = –8; etc.

79. , 13, 26, …

SOLUTION: Find the difference between the terms.

– =

13 – =

26 – 13 = 13 There is not a common different. The sequence is not an arithmetic sequence

80. 7, 6, 5, 4, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is –1, because 6 – 7 = –1; 5 – 6 = –1; 4 – 5 = –1;etc.

81. 10, 12, 15, 18, …

SOLUTION: Find the difference between the terms. 12 – 10 = 2 15 – 12 = 3 18 – 15 = 3 There is no common difference. The sequence is not an arithmetic sequence.

82. −15, −11, −7, −3, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is 4, because –11 – (–15) = 4; –7 – (–11) = 4; –3 – (–7) = 4;etc.

83. −0.3, 0.2, 0.7, 1.2, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is 0.5, because 0.2 – (–0.3) = 0.5; 0.7 – 0.2 = 0.5; 1.2 – 0.7 = 0.5; etc.

Simplify.

84. t(t5)(t

7)

SOLUTION:

85. n3(n

2)(−2n

3)

SOLUTION:

86. (5t5v

2)(10t

3v

4)

SOLUTION:

87. (−8u4z

5)(5uz

4)

SOLUTION:

88. [(3)2]3

SOLUTION:

89. [(2)3]2

SOLUTION:

90. (2m4k

3)2(−3mk

2)3

SOLUTION:

91. (6xy2)2(2x

2y

2z

2)3

SOLUTION:

eSolutions Manual - Powered by Cognero Page 16

8-1 Adding and Subtracting Polynomials

Determine whether each expression is a polynomial. If it is a polynomial, find the degree and determine whether it is a monomial, binomial, or trinomial.

1. 7ab + 6b2 – 2a

3

SOLUTION:

A polynomial is a monomial or the sum of monomials. 7ab + 6b2 – 2a

3 is the sum of 3 monomials, so it is a

polynomial. The degree of a polynomial is the greatest degree of any term in the polynomial.The degree of each term is 2, 2, and

3, so the degree of 7ab + 6b2 – 2a

3 is 3. The polynomial has three terms, so it is a trinomial.

2. 2y – 5 + 3y2

SOLUTION:

2y – 5 + 3y2

is the sum of monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 1, 0, and

2, so the degree of 2y – 5 + 3y2 is 2. The polynomial has three terms, so it is a trinomial.

3. 3x2

SOLUTION:

A polynomial is a monomial or the sum of monomials, so 3x2 is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of 3x2 is 2. The

polynomial has one term, so it is a monomial.

4.

SOLUTION: A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer exponents. It has

only one term. is a division of two monomials, so it is not a monomial.

5. 5m2p

3 + 6

SOLUTION:

A polynomial is a monomial or the sum of monomials. 5m2p

3 + 6

is the sum of 2 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 5 and 0,

so the degree of 5m2p

3 + 6 is 5. The polynomial has two terms, so it is a binomial.

6. 5q–4

+ 6q

SOLUTION: A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer exponents. It has

only one term. 5q-4 is equal to , which is a division of two monomials, so it is not a monomial.

Write each polynomial in standard form. Identify the leading coefficient.

7. –4d4 + 1 – d

2

SOLUTION: Find the degree of each term.

–4d4→ 4

1 → 0

–d2 → 2

The greatest degree is 4, from the term –4d4, so the leading coefficient of –4d

4 + 1 – d

2 is –4.

Rewrite the polynomial with each monomial in descending order according to degree.

–4d4 – d

2 + 1

8. 2x5 – 12 + 3x

SOLUTION: Find the degree of each term.

2x5 → 5

–12 → 0 3x → 1

The greatest degree is 5, from the term 2x5, so the leading coefficient of 2x

5 – 12 + 3x is 2.

Rewrite the polynomial with each monomial in descending order according to degree.

2x5 + 3x – 12

9. 4z – 2z2 – 5z

4

SOLUTION: Find the degree of each term. 4z → 1

– 2z2 → 2

– 5z4→ 4

The greatest degree is 4, from the term – 5z4, so the leading coefficient of 4z – 2z

2 – 5z

4 is –5.

Rewrite the polynomial with each monomial in descending order according to degree.

–5z4 – 2z

2 + 4z

10. 2a + 4a3 – 5a

2 – 1

SOLUTION: Find the degree of each term. 2a → 1

4a3→ 3

– 5a2 → 2

– 1 → 0

The greatest degree is 3, from the term 4a3, so the leading coefficient of 2a + 4a

3 – 5a

2 – 1 is 4.

Rewrite the polynomial with each monomial in descending order according to degree.

4a3– 5a

2 + 2a – 1

Find each sum or difference.

11. (6x3 − 4) + (−2x

3 + 9)

SOLUTION:

12. (g3 − 2g2 + 5g + 6) − (g2 + 2g)

SOLUTION:

13. (4 + 2a2 − 2a) − (3a

2 − 8a + 7)

SOLUTION:

14. (8y − 4y2) + (3y − 9y

2)

SOLUTION:

15. (−4z3 − 2z + 8) − (4z

3 + 3z2 − 5)

SOLUTION:

16. (−3d2 − 8 + 2d) + (4d − 12 + d2

)

SOLUTION:

17. (y + 5) + (2y + 4y2 – 2)

SOLUTION:

18. (3n3 − 5n + n2

) − (−8n2 + 3n

3)

SOLUTION:

19. CCSS SENSE-MAKING The total number of students T who traveled for spring break consists of two groups: students who flew to their destinations F and students who drove to their destination D. The number (in thousands) of students who flew and the total number of students who flew or drove can be modeled by the following equations,where n is the number of years since 1995. T = 14n + 21

F = 8n + 7 a. Write an equation that models the number of students who drove to their destination for this time period. b. Predict the number of students who will drive to their destination in 2012. c. How many students will drive or fly to their destination in 2015?

SOLUTION: a.

D = 6n + 14 b. n = 2012 – 1995 = 17

The number of students who will drive to their destination in 2012 is 116,000 students. c. n = 2015 – 1995 = 20

The number of students who will drive or fly to their destination in 2015 is 301,000 students.

Determine whether each expression is a polynomial. If it is a polynomial, find the degree and determine whether it is a monomial, binomial, or trinomial.

20.

SOLUTION: A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer exponents. It has

only one term. is a division of two monomials, so it is not a monomial.

21.

SOLUTION: A polynomial is a monomial or the sum of monomials. 21 is a monomial, so it is also a polynomial. The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of 21 is 0. The polynomial has only one term, so it is a monomial.

22. c4 – 2c

2 + 1

SOLUTION:

A polynomial is a monomial or the sum of monomials. c4 – 2c

2 + 1

is the sum of 3 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 4, 2, and

0, so the degree of c4 – 2c2 + 1 is 4. The polynomial has three terms, so it is a trinomial.

23. d + 3dc

SOLUTION: A polynomial is a monomial or the sum of monomials. A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer

exponents. 3dc

has a variable in the exponent, so it is not a monomial.

24. a – a2

SOLUTION:

A polynomial is a monomial or the sum of monomials. a – a2 is the sum of 2 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 1 and 2,

so the degree of a – a2 is 2. The polynomial has two terms, so it is a binomial.

25. 5n3 + nq

3

SOLUTION:

A polynomial is a monomial or the sum of monomials. 5n3 + nq3 is the sum of 2 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 3 and 4,

so the degree of 5n3 + nq3 is 4. The polynomial has two terms, so it is a binomial.

Write each polynomial in standard form. Identify the leading coefficient.

26. 5x2 – 2 + 3x

SOLUTION: Find the degree of each term.

5x2 → 2

– 2 → 0

3x → 1

The greatest degree is 2, from the term 5x2, so the leading coefficient of 5x

2 – 2 + 3x is 5.

Rewrite the polynomial with each monomial in descending order according to degree.

5x2 + 3x – 2

27. 8y + 7y3

SOLUTION: Find the degree of each term.

7y3 → 3

8y → 1

The greatest degree is 3, from the term 7y3, so the leading coefficient of 8y + 7y

3 is 7.

Rewrite the polynomial with each monomial in descending order according to degree.

7y3 + 8y

28. 4 – 3c – 5c2

SOLUTION: Find the degree of each term. 4 → 0 3c → 1

– 5c2 → 2

The greatest degree is 2, from the term – 5c2, so the leading coefficient of 4 – 3c – 5c

2 is –5.

Rewrite the polynomial with each monomial in descending order according to degree.

–5c2 – 3c + 4

29. –y3 + 3y – 3y

2 + 2

SOLUTION: Find the degree of each term.

–y3 → 3

3y → 1

3y2 → 2

2 → 0

The greatest degree is 3, from the term –y3, so the leading coefficient of –y

3 + 3y – 3y

2 + 2 is –1.

Rewrite the polynomial with each monomial in descending order according to degree. –y 3 – 3y 2 + 3y + 2

30. 11t + 2t2 – 3 + t

5

SOLUTION: Find the degree of each term. 11t → 1

2t2 → 2

–3 → 0

t5 → 5

The greatest degree is 5, from the term t5, so the leading coefficient of 11t + 2t2 – 3 + t

5 is 1.

Rewrite the polynomial with each monomial in descending order according to degree.

t5 + 2t

2 + 11t – 3

31. 2 + r – r3

SOLUTION: Find the degree of each term. 2 → 0 r → 1

– r3→ 3

The greatest degree is 3, from the term – r3, so the leading coefficient of 2 + r – r3

is –1. Rewrite the polynomial with each monomial in descending order according to degree.

–r3 + r + 2

32.

SOLUTION: Find the degree of each term.

→ 0

–3x4 → 4

7 → 0

The greatest degree is 4, from the term –3x4, so the leading coefficient of is –3.

Rewrite the polynomial with each monomial in descending order according to degree.

33. –9b2 + 10b – b

6

SOLUTION: Find the degree of each term.

–9b2 → 2

10b → 1

–b6→ 6

The greatest degree is 6, from the term –b6, so the leading coefficient of –9b2 + 10b – b6 is –1. Rewrite the polynomial with each monomial in descending order according to degree.

–b6 – 9b

2 + 10b

Find each sum or difference.

34. (2c2 + 6c + 4) + (5c – 7)

SOLUTION:

35. (2x + 3x2) − (7 − 8x

2)

SOLUTION:

36. (3c3 − c + 11) − (c2 + 2c + 8)

SOLUTION:

37. (z2 + z) + (z

2 − 11)

SOLUTION:

38. (2x − 2y + 1) − (3y + 4x)

SOLUTION:

39. (4a − 5b2 + 3) + (6 − 2a + 3b

2)

SOLUTION:

40. (x2y − 3x

2 + y) + (3y − 2x2y)

SOLUTION:

41. (−8xy + 3x2 − 5y) + (4x

2 − 2y + 6xy)

SOLUTION:

42. (5n − 2p2 + 2np) − (4p

2 + 4n)

SOLUTION:

43. (4rxt − 8r2x + x2

) − (6rx2 + 5rxt − 2x

2)

SOLUTION:

44. PETS From 1999 through 2009, the number of dogs D and the number of cats C (in hundreds) adopted from animal

shelters in the United States are modeled by the equations D = 2n + 3 and C = n + 4, where n is the number of yearssince 1999. a. Write an equation that models the total number T of dogs and cats adopted in hundreds for this time period. b. If this trend continues, how many dogs and cats will be adopted in 2013?

SOLUTION: a.

So, an equation that models the total number of dogs and cats adopted is T = 3n + 7. b. Evaluate the equation for the total number of dogs and cats for n = 2013 – 1999 = 14.

The number of cats and dogs adopted in 2013 will be 49 × 100 or 4900 cats and dogs.

Classify each polynomial according to its degree and number of terms.

45. 4x – 3x2 + 5

SOLUTION:

Find the degree of each term of 4x – 3x2 + 5.

4x → 1

– 3x2 → 2

5 → 0

The greatest degree is 2 and there are 3 terms, so 4x – 3x2 + 5 is a quadratic trinomial.

46. 11z3

SOLUTION: Find the degree of each term. 11z3

→ 3

The greatest degree is 3 and there is one term, so 11z3 is a cubic monomial.

47. 9 + y4

SOLUTION:

Find the degree of each term of 9 + y 4.

9 → 1

y4→ 4

The greatest degree is 4 and there are 2 terms, so 9 + y 4 is a quartic binomial.

48. 3x3 – 7

SOLUTION:

Find the degree of each term of 3x3 – 7.

3x3 → 3

–7 → 0

The greatest degree is 3 and there are 2 terms, so 3x3 – 7 is a cubic binomial.

49. –2x5 – x

2 + 5x – 8

SOLUTION:

Find the degree of each term of –2x5 – x2 + 5x – 8.

–2x5 → 5

–x2 → 2

5x → 1 – 8 → 0

The greatest degree is 5 and there are 4 terms, so –2x5 – x2 + 5x – 8 is a quintic polynomial.

50. 10t – 4t2 + 6t

3

SOLUTION:

Find the degree of each term of 10t – 4t2 + 6t

3.

10t → 1

4t2 → 2

6t3 → 3

The greatest degree is 3 and there are 3 terms, so 10t – 4t2 + 6t

3 is a cubic trinomial.

51. ENROLLMENT In a rapidly growing school system, the numbers (in hundreds) of total students N and

K-5 students P enrolled from 2000 to 2009 are modeled by the equations N = 1.25t2 – t + 7.5 and P =

0.7t2 – 0.95t + 3.8, where t is the number of years since 2000.

a. Write an equation modeling the number of 6-12 students S enrolled for this time period. b. How many 6-12 students were enrolled in the school system in 2007?

SOLUTION: a. To write an equation that represents the number of 6-12 students enrolled, subtract the equations that represent the total number of students and the number of K-5 students.

b. Replace t with 7 in the equation for S to determine the number of students enrolled in 6-12 in 2007.

In 2007 there were 30.3 hundreds or 3030 students enrolled in 6-12.

52. CCSS REASONING The perimeter of the figure shown is represented by the expression 3x2 − 7x + 2. Write a

polynomial that represents the measure of the third side.

SOLUTION:

53. GEOMETRY Consider the rectangle.

a. What does (4x2 + 2x – 1)(2x

2 – x + 3) represent?

b. What does 2(4x2 + 2x – 1) + 2(2x

2 – x + 3) represent?

SOLUTION:

a. (4x2 + 2x – 1)(2x

2 – x + 3) is a multiplication of the length and the width of the rectangle, which is the formula for

the area of a rectangle.

b. 2(4x2 + 2x – 1) + 2(2x

2 – x + 3) is the sum of twice the length and twice the width of a rectangle, which is the

formula for the perimeter of the rectangle

Find each sum or difference.

54. (4x + 2y − 6z) + (5y − 2z + 7x) + (−9z − 2x − 3y)

SOLUTION:

55. (5a2 − 4) + (a

2 − 2a + 12) + (4a2 − 6a + 8)

SOLUTION:

56. (3c2 − 7) + (4c + 7) − (c

2 + 5c − 8)

SOLUTION:

57. (3n3 + 3n − 10) − (4n

2 − 5n) + (4n3 − 3n

2 − 9n + 4)

SOLUTION:

58. FOOTBALL The National Football League is divided into two conferences, the American A and the National N. From 2002 through 2009, the total attendance T (in thousands) for both conferences and for the American Conference games are modeled by the following equations, where x is the number of years since 2002.

T = –0.69x3 + 55.83x

2 + 643.31x + 10,538

A = –3.78x3 + 58.96x

2 + 265.96x + 5257

Determine how many people attended National Conference football games in 2009.

SOLUTION:

Let x = 7 represent 2009, then find how many people attended a National Conference football game in 2009.

In 2009 the number of people who attended National Conference football games was about 8829 thousand. Multiply by 1000 to find the attendance in standard form. 8829 ×1000 = 8,829,000. So, about 8,829,000 people attended National Conference football games in 2009.

59. CAR RENTAL The cost to rent a car for a day is $15 plus $0.15 for each mile driven. a. Write a polynomial that represents the cost of renting a car for m miles. b. If a car is driven 145 miles, how much would it cost to rent? c. If a car is driven 105 miles each day for four days, how much would it cost to rent a car? d. If a car is driven 220 miles each day for seven days, how much would it cost to rent a car?

SOLUTION: a. The cost to rent a car is the daily rate and the mileage cost or 15 + 0.15m. b. Substitute 145 for m to find the daily cost to drive 145 miles.

The cost to rent the car would be $36.75. c. The expression represents the cost per day. For 4 days, multiply the entire expression by 4 to find the cost for a 4 day trip. Substitute 105 for m.

The cost to rent the car would be $123. d. The expression represents the cost per day. For 7 days, multiply the entire expression by 7 to find the cost for a 7 day trip. Substitute 220 for m.

The cost to rent the car would be $336.

60. MULTIPLE REPRESENTATIONS In this problem, you will explore perimeter and area. a. Geometric Draw three rectangles that each have a perimeter of 400 feet. b. Tabular Record the width and length of each rectangle in a table like the one shown below. Find the area of eachrectangle.

c. Graphical On a coordinate system, graph the area of rectangle 4 in terms of the length, x. Use the graph to determine the largest area possible. d. Analytical Determine the length and width that produce the largest area.

SOLUTION: a. For the perimeters to be 400, create lengths and widths that sum to 200. Note that some lengths are already provided in the table in part b.

b. The area is length multiplied by width. Make sure the units are squared for area. The sum of the length and width must be 200, so if the length is x, the width must be 200 – x.

c. The length will be the x-values on the horizontal and the area will be the y-values on the vertical. The length cannot pass 200 since the sum of the length and width is 200. Set the intervals for the x-axis to 25 feet. Extend the table of values to find more points to plot on the graph. It appears that 10,000 is the greatest area, so set the intervals

for the y-axis to 1000 ft2. After graphing, it appears that the highest point on the graph is at an area of 10,000 ft

2.

d. The associated x-value with the maximum area is x = 100, so the length must be 100 and the width must be 200 – 100, or 100. The length and width of the rectangle must be 100 feet each to have the largest area.

61. CCSS CRITIQUE Cheyenne and Sebastian are finding (2x2 − x) − (3x + 3x

2 − 2). Is either of them correct? Explain your reasoning.

SOLUTION:

Neither is correct. Cheyenne, did not distribute the negative to the 2nd and 3rd terms when she found the additive inverse. Sebastian did not distribute the negate to the 3rd terms when he found the additive inverse. To find the additive inverse, all terms should be multiplied by −1.

62. REASONING Determine whether each of the following statements is true or false . Explain your reasoning. a. A binomial can have a degree of zero. b. The order in which polynomials are subtracted does not matter.

SOLUTION: a. If a binomial has two terms that are each a degree of 0, then those terms can be combined and the binomial becomes a monomial. For example, 18 + 7 = 25. If one of the terms of the binomial does not have a degree of 0, then the binomial cannot have a degree of 0, since the degree of a polynomial is the greatest degree of any term in the polynomial. b. Subtraction is not commutative. While 2 + 5 = 5 + 2, 2 – 5 ≠ 5 – 2. This is also true for polynomials. Sample answer: (2x – 3) – (4x – 3) = –2x, but (4x – 3) – (2x – 3) = 2x

63. CHALLENGE Write a polynomial that represents the sum of an odd integer 2n + 1 and the next two consecutive odd integers.

SOLUTION:

64. WRITING IN MATH Why would you add or subtract equations that represent real-world situations? Explain.

SOLUTION:

65. WRITING IN MATH Describe how to add and subtract polynomials using both the vertical and horizontal formats.

SOLUTION: To add polynomials in a horizontal format, you combine like terms. For the vertical format, you write the polynomials in standard form, align like terms in columns, and combine like terms.

To subtract polynomials in a horizontal format you find the additive inverse of the polynomial you are subtracting, andthen combine like terms. For the vertical format, you write the polynomials in standard form, align like terms in columns, and subtract by adding the additive inverse.

66. Three consecutive integers can be represented by x, x + 1, and x + 2. What is the sum of these three integers? A x(x + 1)(x + 2)

B x3 + 3

C 3x + 3 D x + 3

SOLUTION:

The correct choice is C.

67. SHORT RESPONSE What is the perimeter of a square with sides that measure 2x + 3 units?

SOLUTION:

The perimeter of the square is 8x + 12 units.

68. Jim cuts a board in the shape of a regular hexagon and pounds in a nail at each vertex, as shown. How many rubber bands will he need to stretch a rubber band across every possible pair of nails?

F 15 G 14 H 12 J 9

SOLUTION: The first nail would connect to 5 others, the second to 4 others, the third to 3 others, etc.5 + 4 + 3 + 2 + 1 = 15 The correct choice is F.

69. Which ordered pair is in the solution set of the system of inequalities shown in the graph?

A (−3, 0) B (0, −3) C (5, 0) D (0, 5)

SOLUTION: Choice A is outside the shaded area for both inequalities. Choices B and D are inside the shaded area for only one inequality. Choice C is the only point in the solution for both inequalities. So, the correct choice is C.

70. COMPUTERS A computer technician charges by the hour to fix and repair computer equipment. The total cost of the technician for one hour is $75, for two hours is $125, for three hours is $175, for four hours is $225, and so on. Write a recursive formula for the sequence.

SOLUTION: Write out the terms. $75, $125, $175, $225, ... The first term is 75, and 50 is added to form each following term. Therefore, we have a1 = 75, an = an – 1 + 50, n ≥ 2.

Determine whether each sequence is arithmetic, geometric, or neither. Explain.71. 8, –32, 128, –512, ...

SOLUTION: Check for a common difference. –32 – 8 = –40 128 – (–32) = 160 There is no common difference. Check for a common ratio. –32 ÷ 8 = –4 128 ÷ (–32) = –4 Geometric; the common ratio is –4.

72. 25, 8, –9, –26, ...

SOLUTION: Check for a common difference. 8 – 25 = –17 –9 – 8 = –17 Arithmetic; the common difference is –17.

73.

SOLUTION: Check for a common difference.

There is no common difference. Check for a common ratio.

There is no common ratio, so the sequence is not arithmetic or geometric.

74. 43, 52, 61, 70, ...

SOLUTION: Check for a common difference. 52 – 43 = 9 61 – 52 = 9 Arithmetic; the common difference is 9.

75. –27, –16, –5, 6, ...

SOLUTION: Check for a common difference. –16 – (–27) = 11 –5 – (–16) = 11 Arithmetic; the common difference is 11.

76. 200, 100, 50, 25, …

SOLUTION: Check for a common difference. 100 – 200 = –100 50 – 100 = –50 There is no common difference. Check for a common ratio. 100 ÷ 200 = 0.5 50 ÷ 100 = 0.5

Geometric; the common ratio is 0.5 or .

77. JOBS Kimi received an offer for a new job. She wants to compare the offer with her current job. What is total amount of sales that Kimi must get each month to make the same income at either job?

SOLUTION: Let s be Kimi’s monthly sales.

Kimi must sell $80,000 each month to make the same income at either job.

Determine whether each sequence is an arithmetic sequence. If it is, state the common difference.78. 24, 16, 8, 0, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is –8, because 16 – 24 = –8; 8 –16 = –8; etc.

79. , 13, 26, …

SOLUTION: Find the difference between the terms.

– =

13 – =

26 – 13 = 13 There is not a common different. The sequence is not an arithmetic sequence

80. 7, 6, 5, 4, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is –1, because 6 – 7 = –1; 5 – 6 = –1; 4 – 5 = –1;etc.

81. 10, 12, 15, 18, …

SOLUTION: Find the difference between the terms. 12 – 10 = 2 15 – 12 = 3 18 – 15 = 3 There is no common difference. The sequence is not an arithmetic sequence.

82. −15, −11, −7, −3, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is 4, because –11 – (–15) = 4; –7 – (–11) = 4; –3 – (–7) = 4;etc.

83. −0.3, 0.2, 0.7, 1.2, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is 0.5, because 0.2 – (–0.3) = 0.5; 0.7 – 0.2 = 0.5; 1.2 – 0.7 = 0.5; etc.

Simplify.

84. t(t5)(t

7)

SOLUTION:

85. n3(n

2)(−2n

3)

SOLUTION:

86. (5t5v

2)(10t

3v

4)

SOLUTION:

87. (−8u4z

5)(5uz

4)

SOLUTION:

88. [(3)2]3

SOLUTION:

89. [(2)3]2

SOLUTION:

90. (2m4k

3)2(−3mk

2)3

SOLUTION:

91. (6xy2)2(2x

2y

2z

2)3

SOLUTION:

eSolutions Manual - Powered by Cognero Page 17

8-1 Adding and Subtracting Polynomials

Determine whether each expression is a polynomial. If it is a polynomial, find the degree and determine whether it is a monomial, binomial, or trinomial.

1. 7ab + 6b2 – 2a

3

SOLUTION:

A polynomial is a monomial or the sum of monomials. 7ab + 6b2 – 2a

3 is the sum of 3 monomials, so it is a

polynomial. The degree of a polynomial is the greatest degree of any term in the polynomial.The degree of each term is 2, 2, and

3, so the degree of 7ab + 6b2 – 2a

3 is 3. The polynomial has three terms, so it is a trinomial.

2. 2y – 5 + 3y2

SOLUTION:

2y – 5 + 3y2

is the sum of monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 1, 0, and

2, so the degree of 2y – 5 + 3y2 is 2. The polynomial has three terms, so it is a trinomial.

3. 3x2

SOLUTION:

A polynomial is a monomial or the sum of monomials, so 3x2 is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of 3x2 is 2. The

polynomial has one term, so it is a monomial.

4.

SOLUTION: A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer exponents. It has

only one term. is a division of two monomials, so it is not a monomial.

5. 5m2p

3 + 6

SOLUTION:

A polynomial is a monomial or the sum of monomials. 5m2p

3 + 6

is the sum of 2 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 5 and 0,

so the degree of 5m2p

3 + 6 is 5. The polynomial has two terms, so it is a binomial.

6. 5q–4

+ 6q

SOLUTION: A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer exponents. It has

only one term. 5q-4 is equal to , which is a division of two monomials, so it is not a monomial.

Write each polynomial in standard form. Identify the leading coefficient.

7. –4d4 + 1 – d

2

SOLUTION: Find the degree of each term.

–4d4→ 4

1 → 0

–d2 → 2

The greatest degree is 4, from the term –4d4, so the leading coefficient of –4d

4 + 1 – d

2 is –4.

Rewrite the polynomial with each monomial in descending order according to degree.

–4d4 – d

2 + 1

8. 2x5 – 12 + 3x

SOLUTION: Find the degree of each term.

2x5 → 5

–12 → 0 3x → 1

The greatest degree is 5, from the term 2x5, so the leading coefficient of 2x

5 – 12 + 3x is 2.

Rewrite the polynomial with each monomial in descending order according to degree.

2x5 + 3x – 12

9. 4z – 2z2 – 5z

4

SOLUTION: Find the degree of each term. 4z → 1

– 2z2 → 2

– 5z4→ 4

The greatest degree is 4, from the term – 5z4, so the leading coefficient of 4z – 2z

2 – 5z

4 is –5.

Rewrite the polynomial with each monomial in descending order according to degree.

–5z4 – 2z

2 + 4z

10. 2a + 4a3 – 5a

2 – 1

SOLUTION: Find the degree of each term. 2a → 1

4a3→ 3

– 5a2 → 2

– 1 → 0

The greatest degree is 3, from the term 4a3, so the leading coefficient of 2a + 4a

3 – 5a

2 – 1 is 4.

Rewrite the polynomial with each monomial in descending order according to degree.

4a3– 5a

2 + 2a – 1

Find each sum or difference.

11. (6x3 − 4) + (−2x

3 + 9)

SOLUTION:

12. (g3 − 2g2 + 5g + 6) − (g2 + 2g)

SOLUTION:

13. (4 + 2a2 − 2a) − (3a

2 − 8a + 7)

SOLUTION:

14. (8y − 4y2) + (3y − 9y

2)

SOLUTION:

15. (−4z3 − 2z + 8) − (4z

3 + 3z2 − 5)

SOLUTION:

16. (−3d2 − 8 + 2d) + (4d − 12 + d2

)

SOLUTION:

17. (y + 5) + (2y + 4y2 – 2)

SOLUTION:

18. (3n3 − 5n + n2

) − (−8n2 + 3n

3)

SOLUTION:

19. CCSS SENSE-MAKING The total number of students T who traveled for spring break consists of two groups: students who flew to their destinations F and students who drove to their destination D. The number (in thousands) of students who flew and the total number of students who flew or drove can be modeled by the following equations,where n is the number of years since 1995. T = 14n + 21

F = 8n + 7 a. Write an equation that models the number of students who drove to their destination for this time period. b. Predict the number of students who will drive to their destination in 2012. c. How many students will drive or fly to their destination in 2015?

SOLUTION: a.

D = 6n + 14 b. n = 2012 – 1995 = 17

The number of students who will drive to their destination in 2012 is 116,000 students. c. n = 2015 – 1995 = 20

The number of students who will drive or fly to their destination in 2015 is 301,000 students.

Determine whether each expression is a polynomial. If it is a polynomial, find the degree and determine whether it is a monomial, binomial, or trinomial.

20.

SOLUTION: A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer exponents. It has

only one term. is a division of two monomials, so it is not a monomial.

21.

SOLUTION: A polynomial is a monomial or the sum of monomials. 21 is a monomial, so it is also a polynomial. The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of 21 is 0. The polynomial has only one term, so it is a monomial.

22. c4 – 2c

2 + 1

SOLUTION:

A polynomial is a monomial or the sum of monomials. c4 – 2c

2 + 1

is the sum of 3 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 4, 2, and

0, so the degree of c4 – 2c2 + 1 is 4. The polynomial has three terms, so it is a trinomial.

23. d + 3dc

SOLUTION: A polynomial is a monomial or the sum of monomials. A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer

exponents. 3dc

has a variable in the exponent, so it is not a monomial.

24. a – a2

SOLUTION:

A polynomial is a monomial or the sum of monomials. a – a2 is the sum of 2 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 1 and 2,

so the degree of a – a2 is 2. The polynomial has two terms, so it is a binomial.

25. 5n3 + nq

3

SOLUTION:

A polynomial is a monomial or the sum of monomials. 5n3 + nq3 is the sum of 2 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 3 and 4,

so the degree of 5n3 + nq3 is 4. The polynomial has two terms, so it is a binomial.

Write each polynomial in standard form. Identify the leading coefficient.

26. 5x2 – 2 + 3x

SOLUTION: Find the degree of each term.

5x2 → 2

– 2 → 0

3x → 1

The greatest degree is 2, from the term 5x2, so the leading coefficient of 5x

2 – 2 + 3x is 5.

Rewrite the polynomial with each monomial in descending order according to degree.

5x2 + 3x – 2

27. 8y + 7y3

SOLUTION: Find the degree of each term.

7y3 → 3

8y → 1

The greatest degree is 3, from the term 7y3, so the leading coefficient of 8y + 7y

3 is 7.

Rewrite the polynomial with each monomial in descending order according to degree.

7y3 + 8y

28. 4 – 3c – 5c2

SOLUTION: Find the degree of each term. 4 → 0 3c → 1

– 5c2 → 2

The greatest degree is 2, from the term – 5c2, so the leading coefficient of 4 – 3c – 5c

2 is –5.

Rewrite the polynomial with each monomial in descending order according to degree.

–5c2 – 3c + 4

29. –y3 + 3y – 3y

2 + 2

SOLUTION: Find the degree of each term.

–y3 → 3

3y → 1

3y2 → 2

2 → 0

The greatest degree is 3, from the term –y3, so the leading coefficient of –y

3 + 3y – 3y

2 + 2 is –1.

Rewrite the polynomial with each monomial in descending order according to degree. –y 3 – 3y 2 + 3y + 2

30. 11t + 2t2 – 3 + t

5

SOLUTION: Find the degree of each term. 11t → 1

2t2 → 2

–3 → 0

t5 → 5

The greatest degree is 5, from the term t5, so the leading coefficient of 11t + 2t2 – 3 + t

5 is 1.

Rewrite the polynomial with each monomial in descending order according to degree.

t5 + 2t

2 + 11t – 3

31. 2 + r – r3

SOLUTION: Find the degree of each term. 2 → 0 r → 1

– r3→ 3

The greatest degree is 3, from the term – r3, so the leading coefficient of 2 + r – r3

is –1. Rewrite the polynomial with each monomial in descending order according to degree.

–r3 + r + 2

32.

SOLUTION: Find the degree of each term.

→ 0

–3x4 → 4

7 → 0

The greatest degree is 4, from the term –3x4, so the leading coefficient of is –3.

Rewrite the polynomial with each monomial in descending order according to degree.

33. –9b2 + 10b – b

6

SOLUTION: Find the degree of each term.

–9b2 → 2

10b → 1

–b6→ 6

The greatest degree is 6, from the term –b6, so the leading coefficient of –9b2 + 10b – b6 is –1. Rewrite the polynomial with each monomial in descending order according to degree.

–b6 – 9b

2 + 10b

Find each sum or difference.

34. (2c2 + 6c + 4) + (5c – 7)

SOLUTION:

35. (2x + 3x2) − (7 − 8x

2)

SOLUTION:

36. (3c3 − c + 11) − (c2 + 2c + 8)

SOLUTION:

37. (z2 + z) + (z

2 − 11)

SOLUTION:

38. (2x − 2y + 1) − (3y + 4x)

SOLUTION:

39. (4a − 5b2 + 3) + (6 − 2a + 3b

2)

SOLUTION:

40. (x2y − 3x

2 + y) + (3y − 2x2y)

SOLUTION:

41. (−8xy + 3x2 − 5y) + (4x

2 − 2y + 6xy)

SOLUTION:

42. (5n − 2p2 + 2np) − (4p

2 + 4n)

SOLUTION:

43. (4rxt − 8r2x + x2

) − (6rx2 + 5rxt − 2x

2)

SOLUTION:

44. PETS From 1999 through 2009, the number of dogs D and the number of cats C (in hundreds) adopted from animal

shelters in the United States are modeled by the equations D = 2n + 3 and C = n + 4, where n is the number of yearssince 1999. a. Write an equation that models the total number T of dogs and cats adopted in hundreds for this time period. b. If this trend continues, how many dogs and cats will be adopted in 2013?

SOLUTION: a.

So, an equation that models the total number of dogs and cats adopted is T = 3n + 7. b. Evaluate the equation for the total number of dogs and cats for n = 2013 – 1999 = 14.

The number of cats and dogs adopted in 2013 will be 49 × 100 or 4900 cats and dogs.

Classify each polynomial according to its degree and number of terms.

45. 4x – 3x2 + 5

SOLUTION:

Find the degree of each term of 4x – 3x2 + 5.

4x → 1

– 3x2 → 2

5 → 0

The greatest degree is 2 and there are 3 terms, so 4x – 3x2 + 5 is a quadratic trinomial.

46. 11z3

SOLUTION: Find the degree of each term. 11z3

→ 3

The greatest degree is 3 and there is one term, so 11z3 is a cubic monomial.

47. 9 + y4

SOLUTION:

Find the degree of each term of 9 + y 4.

9 → 1

y4→ 4

The greatest degree is 4 and there are 2 terms, so 9 + y 4 is a quartic binomial.

48. 3x3 – 7

SOLUTION:

Find the degree of each term of 3x3 – 7.

3x3 → 3

–7 → 0

The greatest degree is 3 and there are 2 terms, so 3x3 – 7 is a cubic binomial.

49. –2x5 – x

2 + 5x – 8

SOLUTION:

Find the degree of each term of –2x5 – x2 + 5x – 8.

–2x5 → 5

–x2 → 2

5x → 1 – 8 → 0

The greatest degree is 5 and there are 4 terms, so –2x5 – x2 + 5x – 8 is a quintic polynomial.

50. 10t – 4t2 + 6t

3

SOLUTION:

Find the degree of each term of 10t – 4t2 + 6t

3.

10t → 1

4t2 → 2

6t3 → 3

The greatest degree is 3 and there are 3 terms, so 10t – 4t2 + 6t

3 is a cubic trinomial.

51. ENROLLMENT In a rapidly growing school system, the numbers (in hundreds) of total students N and

K-5 students P enrolled from 2000 to 2009 are modeled by the equations N = 1.25t2 – t + 7.5 and P =

0.7t2 – 0.95t + 3.8, where t is the number of years since 2000.

a. Write an equation modeling the number of 6-12 students S enrolled for this time period. b. How many 6-12 students were enrolled in the school system in 2007?

SOLUTION: a. To write an equation that represents the number of 6-12 students enrolled, subtract the equations that represent the total number of students and the number of K-5 students.

b. Replace t with 7 in the equation for S to determine the number of students enrolled in 6-12 in 2007.

In 2007 there were 30.3 hundreds or 3030 students enrolled in 6-12.

52. CCSS REASONING The perimeter of the figure shown is represented by the expression 3x2 − 7x + 2. Write a

polynomial that represents the measure of the third side.

SOLUTION:

53. GEOMETRY Consider the rectangle.

a. What does (4x2 + 2x – 1)(2x

2 – x + 3) represent?

b. What does 2(4x2 + 2x – 1) + 2(2x

2 – x + 3) represent?

SOLUTION:

a. (4x2 + 2x – 1)(2x

2 – x + 3) is a multiplication of the length and the width of the rectangle, which is the formula for

the area of a rectangle.

b. 2(4x2 + 2x – 1) + 2(2x

2 – x + 3) is the sum of twice the length and twice the width of a rectangle, which is the

formula for the perimeter of the rectangle

Find each sum or difference.

54. (4x + 2y − 6z) + (5y − 2z + 7x) + (−9z − 2x − 3y)

SOLUTION:

55. (5a2 − 4) + (a

2 − 2a + 12) + (4a2 − 6a + 8)

SOLUTION:

56. (3c2 − 7) + (4c + 7) − (c

2 + 5c − 8)

SOLUTION:

57. (3n3 + 3n − 10) − (4n

2 − 5n) + (4n3 − 3n

2 − 9n + 4)

SOLUTION:

58. FOOTBALL The National Football League is divided into two conferences, the American A and the National N. From 2002 through 2009, the total attendance T (in thousands) for both conferences and for the American Conference games are modeled by the following equations, where x is the number of years since 2002.

T = –0.69x3 + 55.83x

2 + 643.31x + 10,538

A = –3.78x3 + 58.96x

2 + 265.96x + 5257

Determine how many people attended National Conference football games in 2009.

SOLUTION:

Let x = 7 represent 2009, then find how many people attended a National Conference football game in 2009.

In 2009 the number of people who attended National Conference football games was about 8829 thousand. Multiply by 1000 to find the attendance in standard form. 8829 ×1000 = 8,829,000. So, about 8,829,000 people attended National Conference football games in 2009.

59. CAR RENTAL The cost to rent a car for a day is $15 plus $0.15 for each mile driven. a. Write a polynomial that represents the cost of renting a car for m miles. b. If a car is driven 145 miles, how much would it cost to rent? c. If a car is driven 105 miles each day for four days, how much would it cost to rent a car? d. If a car is driven 220 miles each day for seven days, how much would it cost to rent a car?

SOLUTION: a. The cost to rent a car is the daily rate and the mileage cost or 15 + 0.15m. b. Substitute 145 for m to find the daily cost to drive 145 miles.

The cost to rent the car would be $36.75. c. The expression represents the cost per day. For 4 days, multiply the entire expression by 4 to find the cost for a 4 day trip. Substitute 105 for m.

The cost to rent the car would be $123. d. The expression represents the cost per day. For 7 days, multiply the entire expression by 7 to find the cost for a 7 day trip. Substitute 220 for m.

The cost to rent the car would be $336.

60. MULTIPLE REPRESENTATIONS In this problem, you will explore perimeter and area. a. Geometric Draw three rectangles that each have a perimeter of 400 feet. b. Tabular Record the width and length of each rectangle in a table like the one shown below. Find the area of eachrectangle.

c. Graphical On a coordinate system, graph the area of rectangle 4 in terms of the length, x. Use the graph to determine the largest area possible. d. Analytical Determine the length and width that produce the largest area.

SOLUTION: a. For the perimeters to be 400, create lengths and widths that sum to 200. Note that some lengths are already provided in the table in part b.

b. The area is length multiplied by width. Make sure the units are squared for area. The sum of the length and width must be 200, so if the length is x, the width must be 200 – x.

c. The length will be the x-values on the horizontal and the area will be the y-values on the vertical. The length cannot pass 200 since the sum of the length and width is 200. Set the intervals for the x-axis to 25 feet. Extend the table of values to find more points to plot on the graph. It appears that 10,000 is the greatest area, so set the intervals

for the y-axis to 1000 ft2. After graphing, it appears that the highest point on the graph is at an area of 10,000 ft

2.

d. The associated x-value with the maximum area is x = 100, so the length must be 100 and the width must be 200 – 100, or 100. The length and width of the rectangle must be 100 feet each to have the largest area.

61. CCSS CRITIQUE Cheyenne and Sebastian are finding (2x2 − x) − (3x + 3x

2 − 2). Is either of them correct? Explain your reasoning.

SOLUTION:

Neither is correct. Cheyenne, did not distribute the negative to the 2nd and 3rd terms when she found the additive inverse. Sebastian did not distribute the negate to the 3rd terms when he found the additive inverse. To find the additive inverse, all terms should be multiplied by −1.

62. REASONING Determine whether each of the following statements is true or false . Explain your reasoning. a. A binomial can have a degree of zero. b. The order in which polynomials are subtracted does not matter.

SOLUTION: a. If a binomial has two terms that are each a degree of 0, then those terms can be combined and the binomial becomes a monomial. For example, 18 + 7 = 25. If one of the terms of the binomial does not have a degree of 0, then the binomial cannot have a degree of 0, since the degree of a polynomial is the greatest degree of any term in the polynomial. b. Subtraction is not commutative. While 2 + 5 = 5 + 2, 2 – 5 ≠ 5 – 2. This is also true for polynomials. Sample answer: (2x – 3) – (4x – 3) = –2x, but (4x – 3) – (2x – 3) = 2x

63. CHALLENGE Write a polynomial that represents the sum of an odd integer 2n + 1 and the next two consecutive odd integers.

SOLUTION:

64. WRITING IN MATH Why would you add or subtract equations that represent real-world situations? Explain.

SOLUTION:

65. WRITING IN MATH Describe how to add and subtract polynomials using both the vertical and horizontal formats.

SOLUTION: To add polynomials in a horizontal format, you combine like terms. For the vertical format, you write the polynomials in standard form, align like terms in columns, and combine like terms.

To subtract polynomials in a horizontal format you find the additive inverse of the polynomial you are subtracting, andthen combine like terms. For the vertical format, you write the polynomials in standard form, align like terms in columns, and subtract by adding the additive inverse.

66. Three consecutive integers can be represented by x, x + 1, and x + 2. What is the sum of these three integers? A x(x + 1)(x + 2)

B x3 + 3

C 3x + 3 D x + 3

SOLUTION:

The correct choice is C.

67. SHORT RESPONSE What is the perimeter of a square with sides that measure 2x + 3 units?

SOLUTION:

The perimeter of the square is 8x + 12 units.

68. Jim cuts a board in the shape of a regular hexagon and pounds in a nail at each vertex, as shown. How many rubber bands will he need to stretch a rubber band across every possible pair of nails?

F 15 G 14 H 12 J 9

SOLUTION: The first nail would connect to 5 others, the second to 4 others, the third to 3 others, etc.5 + 4 + 3 + 2 + 1 = 15 The correct choice is F.

69. Which ordered pair is in the solution set of the system of inequalities shown in the graph?

A (−3, 0) B (0, −3) C (5, 0) D (0, 5)

SOLUTION: Choice A is outside the shaded area for both inequalities. Choices B and D are inside the shaded area for only one inequality. Choice C is the only point in the solution for both inequalities. So, the correct choice is C.

70. COMPUTERS A computer technician charges by the hour to fix and repair computer equipment. The total cost of the technician for one hour is $75, for two hours is $125, for three hours is $175, for four hours is $225, and so on. Write a recursive formula for the sequence.

SOLUTION: Write out the terms. $75, $125, $175, $225, ... The first term is 75, and 50 is added to form each following term. Therefore, we have a1 = 75, an = an – 1 + 50, n ≥ 2.

Determine whether each sequence is arithmetic, geometric, or neither. Explain.71. 8, –32, 128, –512, ...

SOLUTION: Check for a common difference. –32 – 8 = –40 128 – (–32) = 160 There is no common difference. Check for a common ratio. –32 ÷ 8 = –4 128 ÷ (–32) = –4 Geometric; the common ratio is –4.

72. 25, 8, –9, –26, ...

SOLUTION: Check for a common difference. 8 – 25 = –17 –9 – 8 = –17 Arithmetic; the common difference is –17.

73.

SOLUTION: Check for a common difference.

There is no common difference. Check for a common ratio.

There is no common ratio, so the sequence is not arithmetic or geometric.

74. 43, 52, 61, 70, ...

SOLUTION: Check for a common difference. 52 – 43 = 9 61 – 52 = 9 Arithmetic; the common difference is 9.

75. –27, –16, –5, 6, ...

SOLUTION: Check for a common difference. –16 – (–27) = 11 –5 – (–16) = 11 Arithmetic; the common difference is 11.

76. 200, 100, 50, 25, …

SOLUTION: Check for a common difference. 100 – 200 = –100 50 – 100 = –50 There is no common difference. Check for a common ratio. 100 ÷ 200 = 0.5 50 ÷ 100 = 0.5

Geometric; the common ratio is 0.5 or .

77. JOBS Kimi received an offer for a new job. She wants to compare the offer with her current job. What is total amount of sales that Kimi must get each month to make the same income at either job?

SOLUTION: Let s be Kimi’s monthly sales.

Kimi must sell $80,000 each month to make the same income at either job.

Determine whether each sequence is an arithmetic sequence. If it is, state the common difference.78. 24, 16, 8, 0, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is –8, because 16 – 24 = –8; 8 –16 = –8; etc.

79. , 13, 26, …

SOLUTION: Find the difference between the terms.

– =

13 – =

26 – 13 = 13 There is not a common different. The sequence is not an arithmetic sequence

80. 7, 6, 5, 4, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is –1, because 6 – 7 = –1; 5 – 6 = –1; 4 – 5 = –1;etc.

81. 10, 12, 15, 18, …

SOLUTION: Find the difference between the terms. 12 – 10 = 2 15 – 12 = 3 18 – 15 = 3 There is no common difference. The sequence is not an arithmetic sequence.

82. −15, −11, −7, −3, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is 4, because –11 – (–15) = 4; –7 – (–11) = 4; –3 – (–7) = 4;etc.

83. −0.3, 0.2, 0.7, 1.2, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is 0.5, because 0.2 – (–0.3) = 0.5; 0.7 – 0.2 = 0.5; 1.2 – 0.7 = 0.5; etc.

Simplify.

84. t(t5)(t

7)

SOLUTION:

85. n3(n

2)(−2n

3)

SOLUTION:

86. (5t5v

2)(10t

3v

4)

SOLUTION:

87. (−8u4z

5)(5uz

4)

SOLUTION:

88. [(3)2]3

SOLUTION:

89. [(2)3]2

SOLUTION:

90. (2m4k

3)2(−3mk

2)3

SOLUTION:

91. (6xy2)2(2x

2y

2z

2)3

SOLUTION:

eSolutions Manual - Powered by Cognero Page 18

8-1 Adding and Subtracting Polynomials

Determine whether each expression is a polynomial. If it is a polynomial, find the degree and determine whether it is a monomial, binomial, or trinomial.

1. 7ab + 6b2 – 2a

3

SOLUTION:

A polynomial is a monomial or the sum of monomials. 7ab + 6b2 – 2a

3 is the sum of 3 monomials, so it is a

polynomial. The degree of a polynomial is the greatest degree of any term in the polynomial.The degree of each term is 2, 2, and

3, so the degree of 7ab + 6b2 – 2a

3 is 3. The polynomial has three terms, so it is a trinomial.

2. 2y – 5 + 3y2

SOLUTION:

2y – 5 + 3y2

is the sum of monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 1, 0, and

2, so the degree of 2y – 5 + 3y2 is 2. The polynomial has three terms, so it is a trinomial.

3. 3x2

SOLUTION:

A polynomial is a monomial or the sum of monomials, so 3x2 is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of 3x2 is 2. The

polynomial has one term, so it is a monomial.

4.

SOLUTION: A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer exponents. It has

only one term. is a division of two monomials, so it is not a monomial.

5. 5m2p

3 + 6

SOLUTION:

A polynomial is a monomial or the sum of monomials. 5m2p

3 + 6

is the sum of 2 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 5 and 0,

so the degree of 5m2p

3 + 6 is 5. The polynomial has two terms, so it is a binomial.

6. 5q–4

+ 6q

SOLUTION: A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer exponents. It has

only one term. 5q-4 is equal to , which is a division of two monomials, so it is not a monomial.

Write each polynomial in standard form. Identify the leading coefficient.

7. –4d4 + 1 – d

2

SOLUTION: Find the degree of each term.

–4d4→ 4

1 → 0

–d2 → 2

The greatest degree is 4, from the term –4d4, so the leading coefficient of –4d

4 + 1 – d

2 is –4.

Rewrite the polynomial with each monomial in descending order according to degree.

–4d4 – d

2 + 1

8. 2x5 – 12 + 3x

SOLUTION: Find the degree of each term.

2x5 → 5

–12 → 0 3x → 1

The greatest degree is 5, from the term 2x5, so the leading coefficient of 2x

5 – 12 + 3x is 2.

Rewrite the polynomial with each monomial in descending order according to degree.

2x5 + 3x – 12

9. 4z – 2z2 – 5z

4

SOLUTION: Find the degree of each term. 4z → 1

– 2z2 → 2

– 5z4→ 4

The greatest degree is 4, from the term – 5z4, so the leading coefficient of 4z – 2z

2 – 5z

4 is –5.

Rewrite the polynomial with each monomial in descending order according to degree.

–5z4 – 2z

2 + 4z

10. 2a + 4a3 – 5a

2 – 1

SOLUTION: Find the degree of each term. 2a → 1

4a3→ 3

– 5a2 → 2

– 1 → 0

The greatest degree is 3, from the term 4a3, so the leading coefficient of 2a + 4a

3 – 5a

2 – 1 is 4.

Rewrite the polynomial with each monomial in descending order according to degree.

4a3– 5a

2 + 2a – 1

Find each sum or difference.

11. (6x3 − 4) + (−2x

3 + 9)

SOLUTION:

12. (g3 − 2g2 + 5g + 6) − (g2 + 2g)

SOLUTION:

13. (4 + 2a2 − 2a) − (3a

2 − 8a + 7)

SOLUTION:

14. (8y − 4y2) + (3y − 9y

2)

SOLUTION:

15. (−4z3 − 2z + 8) − (4z

3 + 3z2 − 5)

SOLUTION:

16. (−3d2 − 8 + 2d) + (4d − 12 + d2

)

SOLUTION:

17. (y + 5) + (2y + 4y2 – 2)

SOLUTION:

18. (3n3 − 5n + n2

) − (−8n2 + 3n

3)

SOLUTION:

19. CCSS SENSE-MAKING The total number of students T who traveled for spring break consists of two groups: students who flew to their destinations F and students who drove to their destination D. The number (in thousands) of students who flew and the total number of students who flew or drove can be modeled by the following equations,where n is the number of years since 1995. T = 14n + 21

F = 8n + 7 a. Write an equation that models the number of students who drove to their destination for this time period. b. Predict the number of students who will drive to their destination in 2012. c. How many students will drive or fly to their destination in 2015?

SOLUTION: a.

D = 6n + 14 b. n = 2012 – 1995 = 17

The number of students who will drive to their destination in 2012 is 116,000 students. c. n = 2015 – 1995 = 20

The number of students who will drive or fly to their destination in 2015 is 301,000 students.

Determine whether each expression is a polynomial. If it is a polynomial, find the degree and determine whether it is a monomial, binomial, or trinomial.

20.

SOLUTION: A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer exponents. It has

only one term. is a division of two monomials, so it is not a monomial.

21.

SOLUTION: A polynomial is a monomial or the sum of monomials. 21 is a monomial, so it is also a polynomial. The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of 21 is 0. The polynomial has only one term, so it is a monomial.

22. c4 – 2c

2 + 1

SOLUTION:

A polynomial is a monomial or the sum of monomials. c4 – 2c

2 + 1

is the sum of 3 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 4, 2, and

0, so the degree of c4 – 2c2 + 1 is 4. The polynomial has three terms, so it is a trinomial.

23. d + 3dc

SOLUTION: A polynomial is a monomial or the sum of monomials. A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer

exponents. 3dc

has a variable in the exponent, so it is not a monomial.

24. a – a2

SOLUTION:

A polynomial is a monomial or the sum of monomials. a – a2 is the sum of 2 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 1 and 2,

so the degree of a – a2 is 2. The polynomial has two terms, so it is a binomial.

25. 5n3 + nq

3

SOLUTION:

A polynomial is a monomial or the sum of monomials. 5n3 + nq3 is the sum of 2 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 3 and 4,

so the degree of 5n3 + nq3 is 4. The polynomial has two terms, so it is a binomial.

Write each polynomial in standard form. Identify the leading coefficient.

26. 5x2 – 2 + 3x

SOLUTION: Find the degree of each term.

5x2 → 2

– 2 → 0

3x → 1

The greatest degree is 2, from the term 5x2, so the leading coefficient of 5x

2 – 2 + 3x is 5.

Rewrite the polynomial with each monomial in descending order according to degree.

5x2 + 3x – 2

27. 8y + 7y3

SOLUTION: Find the degree of each term.

7y3 → 3

8y → 1

The greatest degree is 3, from the term 7y3, so the leading coefficient of 8y + 7y

3 is 7.

Rewrite the polynomial with each monomial in descending order according to degree.

7y3 + 8y

28. 4 – 3c – 5c2

SOLUTION: Find the degree of each term. 4 → 0 3c → 1

– 5c2 → 2

The greatest degree is 2, from the term – 5c2, so the leading coefficient of 4 – 3c – 5c

2 is –5.

Rewrite the polynomial with each monomial in descending order according to degree.

–5c2 – 3c + 4

29. –y3 + 3y – 3y

2 + 2

SOLUTION: Find the degree of each term.

–y3 → 3

3y → 1

3y2 → 2

2 → 0

The greatest degree is 3, from the term –y3, so the leading coefficient of –y

3 + 3y – 3y

2 + 2 is –1.

Rewrite the polynomial with each monomial in descending order according to degree. –y 3 – 3y 2 + 3y + 2

30. 11t + 2t2 – 3 + t

5

SOLUTION: Find the degree of each term. 11t → 1

2t2 → 2

–3 → 0

t5 → 5

The greatest degree is 5, from the term t5, so the leading coefficient of 11t + 2t2 – 3 + t

5 is 1.

Rewrite the polynomial with each monomial in descending order according to degree.

t5 + 2t

2 + 11t – 3

31. 2 + r – r3

SOLUTION: Find the degree of each term. 2 → 0 r → 1

– r3→ 3

The greatest degree is 3, from the term – r3, so the leading coefficient of 2 + r – r3

is –1. Rewrite the polynomial with each monomial in descending order according to degree.

–r3 + r + 2

32.

SOLUTION: Find the degree of each term.

→ 0

–3x4 → 4

7 → 0

The greatest degree is 4, from the term –3x4, so the leading coefficient of is –3.

Rewrite the polynomial with each monomial in descending order according to degree.

33. –9b2 + 10b – b

6

SOLUTION: Find the degree of each term.

–9b2 → 2

10b → 1

–b6→ 6

The greatest degree is 6, from the term –b6, so the leading coefficient of –9b2 + 10b – b6 is –1. Rewrite the polynomial with each monomial in descending order according to degree.

–b6 – 9b

2 + 10b

Find each sum or difference.

34. (2c2 + 6c + 4) + (5c – 7)

SOLUTION:

35. (2x + 3x2) − (7 − 8x

2)

SOLUTION:

36. (3c3 − c + 11) − (c2 + 2c + 8)

SOLUTION:

37. (z2 + z) + (z

2 − 11)

SOLUTION:

38. (2x − 2y + 1) − (3y + 4x)

SOLUTION:

39. (4a − 5b2 + 3) + (6 − 2a + 3b

2)

SOLUTION:

40. (x2y − 3x

2 + y) + (3y − 2x2y)

SOLUTION:

41. (−8xy + 3x2 − 5y) + (4x

2 − 2y + 6xy)

SOLUTION:

42. (5n − 2p2 + 2np) − (4p

2 + 4n)

SOLUTION:

43. (4rxt − 8r2x + x2

) − (6rx2 + 5rxt − 2x

2)

SOLUTION:

44. PETS From 1999 through 2009, the number of dogs D and the number of cats C (in hundreds) adopted from animal

shelters in the United States are modeled by the equations D = 2n + 3 and C = n + 4, where n is the number of yearssince 1999. a. Write an equation that models the total number T of dogs and cats adopted in hundreds for this time period. b. If this trend continues, how many dogs and cats will be adopted in 2013?

SOLUTION: a.

So, an equation that models the total number of dogs and cats adopted is T = 3n + 7. b. Evaluate the equation for the total number of dogs and cats for n = 2013 – 1999 = 14.

The number of cats and dogs adopted in 2013 will be 49 × 100 or 4900 cats and dogs.

Classify each polynomial according to its degree and number of terms.

45. 4x – 3x2 + 5

SOLUTION:

Find the degree of each term of 4x – 3x2 + 5.

4x → 1

– 3x2 → 2

5 → 0

The greatest degree is 2 and there are 3 terms, so 4x – 3x2 + 5 is a quadratic trinomial.

46. 11z3

SOLUTION: Find the degree of each term. 11z3

→ 3

The greatest degree is 3 and there is one term, so 11z3 is a cubic monomial.

47. 9 + y4

SOLUTION:

Find the degree of each term of 9 + y 4.

9 → 1

y4→ 4

The greatest degree is 4 and there are 2 terms, so 9 + y 4 is a quartic binomial.

48. 3x3 – 7

SOLUTION:

Find the degree of each term of 3x3 – 7.

3x3 → 3

–7 → 0

The greatest degree is 3 and there are 2 terms, so 3x3 – 7 is a cubic binomial.

49. –2x5 – x

2 + 5x – 8

SOLUTION:

Find the degree of each term of –2x5 – x2 + 5x – 8.

–2x5 → 5

–x2 → 2

5x → 1 – 8 → 0

The greatest degree is 5 and there are 4 terms, so –2x5 – x2 + 5x – 8 is a quintic polynomial.

50. 10t – 4t2 + 6t

3

SOLUTION:

Find the degree of each term of 10t – 4t2 + 6t

3.

10t → 1

4t2 → 2

6t3 → 3

The greatest degree is 3 and there are 3 terms, so 10t – 4t2 + 6t

3 is a cubic trinomial.

51. ENROLLMENT In a rapidly growing school system, the numbers (in hundreds) of total students N and

K-5 students P enrolled from 2000 to 2009 are modeled by the equations N = 1.25t2 – t + 7.5 and P =

0.7t2 – 0.95t + 3.8, where t is the number of years since 2000.

a. Write an equation modeling the number of 6-12 students S enrolled for this time period. b. How many 6-12 students were enrolled in the school system in 2007?

SOLUTION: a. To write an equation that represents the number of 6-12 students enrolled, subtract the equations that represent the total number of students and the number of K-5 students.

b. Replace t with 7 in the equation for S to determine the number of students enrolled in 6-12 in 2007.

In 2007 there were 30.3 hundreds or 3030 students enrolled in 6-12.

52. CCSS REASONING The perimeter of the figure shown is represented by the expression 3x2 − 7x + 2. Write a

polynomial that represents the measure of the third side.

SOLUTION:

53. GEOMETRY Consider the rectangle.

a. What does (4x2 + 2x – 1)(2x

2 – x + 3) represent?

b. What does 2(4x2 + 2x – 1) + 2(2x

2 – x + 3) represent?

SOLUTION:

a. (4x2 + 2x – 1)(2x

2 – x + 3) is a multiplication of the length and the width of the rectangle, which is the formula for

the area of a rectangle.

b. 2(4x2 + 2x – 1) + 2(2x

2 – x + 3) is the sum of twice the length and twice the width of a rectangle, which is the

formula for the perimeter of the rectangle

Find each sum or difference.

54. (4x + 2y − 6z) + (5y − 2z + 7x) + (−9z − 2x − 3y)

SOLUTION:

55. (5a2 − 4) + (a

2 − 2a + 12) + (4a2 − 6a + 8)

SOLUTION:

56. (3c2 − 7) + (4c + 7) − (c

2 + 5c − 8)

SOLUTION:

57. (3n3 + 3n − 10) − (4n

2 − 5n) + (4n3 − 3n

2 − 9n + 4)

SOLUTION:

58. FOOTBALL The National Football League is divided into two conferences, the American A and the National N. From 2002 through 2009, the total attendance T (in thousands) for both conferences and for the American Conference games are modeled by the following equations, where x is the number of years since 2002.

T = –0.69x3 + 55.83x

2 + 643.31x + 10,538

A = –3.78x3 + 58.96x

2 + 265.96x + 5257

Determine how many people attended National Conference football games in 2009.

SOLUTION:

Let x = 7 represent 2009, then find how many people attended a National Conference football game in 2009.

In 2009 the number of people who attended National Conference football games was about 8829 thousand. Multiply by 1000 to find the attendance in standard form. 8829 ×1000 = 8,829,000. So, about 8,829,000 people attended National Conference football games in 2009.

59. CAR RENTAL The cost to rent a car for a day is $15 plus $0.15 for each mile driven. a. Write a polynomial that represents the cost of renting a car for m miles. b. If a car is driven 145 miles, how much would it cost to rent? c. If a car is driven 105 miles each day for four days, how much would it cost to rent a car? d. If a car is driven 220 miles each day for seven days, how much would it cost to rent a car?

SOLUTION: a. The cost to rent a car is the daily rate and the mileage cost or 15 + 0.15m. b. Substitute 145 for m to find the daily cost to drive 145 miles.

The cost to rent the car would be $36.75. c. The expression represents the cost per day. For 4 days, multiply the entire expression by 4 to find the cost for a 4 day trip. Substitute 105 for m.

The cost to rent the car would be $123. d. The expression represents the cost per day. For 7 days, multiply the entire expression by 7 to find the cost for a 7 day trip. Substitute 220 for m.

The cost to rent the car would be $336.

60. MULTIPLE REPRESENTATIONS In this problem, you will explore perimeter and area. a. Geometric Draw three rectangles that each have a perimeter of 400 feet. b. Tabular Record the width and length of each rectangle in a table like the one shown below. Find the area of eachrectangle.

c. Graphical On a coordinate system, graph the area of rectangle 4 in terms of the length, x. Use the graph to determine the largest area possible. d. Analytical Determine the length and width that produce the largest area.

SOLUTION: a. For the perimeters to be 400, create lengths and widths that sum to 200. Note that some lengths are already provided in the table in part b.

b. The area is length multiplied by width. Make sure the units are squared for area. The sum of the length and width must be 200, so if the length is x, the width must be 200 – x.

c. The length will be the x-values on the horizontal and the area will be the y-values on the vertical. The length cannot pass 200 since the sum of the length and width is 200. Set the intervals for the x-axis to 25 feet. Extend the table of values to find more points to plot on the graph. It appears that 10,000 is the greatest area, so set the intervals

for the y-axis to 1000 ft2. After graphing, it appears that the highest point on the graph is at an area of 10,000 ft

2.

d. The associated x-value with the maximum area is x = 100, so the length must be 100 and the width must be 200 – 100, or 100. The length and width of the rectangle must be 100 feet each to have the largest area.

61. CCSS CRITIQUE Cheyenne and Sebastian are finding (2x2 − x) − (3x + 3x

2 − 2). Is either of them correct? Explain your reasoning.

SOLUTION:

Neither is correct. Cheyenne, did not distribute the negative to the 2nd and 3rd terms when she found the additive inverse. Sebastian did not distribute the negate to the 3rd terms when he found the additive inverse. To find the additive inverse, all terms should be multiplied by −1.

62. REASONING Determine whether each of the following statements is true or false . Explain your reasoning. a. A binomial can have a degree of zero. b. The order in which polynomials are subtracted does not matter.

SOLUTION: a. If a binomial has two terms that are each a degree of 0, then those terms can be combined and the binomial becomes a monomial. For example, 18 + 7 = 25. If one of the terms of the binomial does not have a degree of 0, then the binomial cannot have a degree of 0, since the degree of a polynomial is the greatest degree of any term in the polynomial. b. Subtraction is not commutative. While 2 + 5 = 5 + 2, 2 – 5 ≠ 5 – 2. This is also true for polynomials. Sample answer: (2x – 3) – (4x – 3) = –2x, but (4x – 3) – (2x – 3) = 2x

63. CHALLENGE Write a polynomial that represents the sum of an odd integer 2n + 1 and the next two consecutive odd integers.

SOLUTION:

64. WRITING IN MATH Why would you add or subtract equations that represent real-world situations? Explain.

SOLUTION:

65. WRITING IN MATH Describe how to add and subtract polynomials using both the vertical and horizontal formats.

SOLUTION: To add polynomials in a horizontal format, you combine like terms. For the vertical format, you write the polynomials in standard form, align like terms in columns, and combine like terms.

To subtract polynomials in a horizontal format you find the additive inverse of the polynomial you are subtracting, andthen combine like terms. For the vertical format, you write the polynomials in standard form, align like terms in columns, and subtract by adding the additive inverse.

66. Three consecutive integers can be represented by x, x + 1, and x + 2. What is the sum of these three integers? A x(x + 1)(x + 2)

B x3 + 3

C 3x + 3 D x + 3

SOLUTION:

The correct choice is C.

67. SHORT RESPONSE What is the perimeter of a square with sides that measure 2x + 3 units?

SOLUTION:

The perimeter of the square is 8x + 12 units.

68. Jim cuts a board in the shape of a regular hexagon and pounds in a nail at each vertex, as shown. How many rubber bands will he need to stretch a rubber band across every possible pair of nails?

F 15 G 14 H 12 J 9

SOLUTION: The first nail would connect to 5 others, the second to 4 others, the third to 3 others, etc.5 + 4 + 3 + 2 + 1 = 15 The correct choice is F.

69. Which ordered pair is in the solution set of the system of inequalities shown in the graph?

A (−3, 0) B (0, −3) C (5, 0) D (0, 5)

SOLUTION: Choice A is outside the shaded area for both inequalities. Choices B and D are inside the shaded area for only one inequality. Choice C is the only point in the solution for both inequalities. So, the correct choice is C.

70. COMPUTERS A computer technician charges by the hour to fix and repair computer equipment. The total cost of the technician for one hour is $75, for two hours is $125, for three hours is $175, for four hours is $225, and so on. Write a recursive formula for the sequence.

SOLUTION: Write out the terms. $75, $125, $175, $225, ... The first term is 75, and 50 is added to form each following term. Therefore, we have a1 = 75, an = an – 1 + 50, n ≥ 2.

Determine whether each sequence is arithmetic, geometric, or neither. Explain.71. 8, –32, 128, –512, ...

SOLUTION: Check for a common difference. –32 – 8 = –40 128 – (–32) = 160 There is no common difference. Check for a common ratio. –32 ÷ 8 = –4 128 ÷ (–32) = –4 Geometric; the common ratio is –4.

72. 25, 8, –9, –26, ...

SOLUTION: Check for a common difference. 8 – 25 = –17 –9 – 8 = –17 Arithmetic; the common difference is –17.

73.

SOLUTION: Check for a common difference.

There is no common difference. Check for a common ratio.

There is no common ratio, so the sequence is not arithmetic or geometric.

74. 43, 52, 61, 70, ...

SOLUTION: Check for a common difference. 52 – 43 = 9 61 – 52 = 9 Arithmetic; the common difference is 9.

75. –27, –16, –5, 6, ...

SOLUTION: Check for a common difference. –16 – (–27) = 11 –5 – (–16) = 11 Arithmetic; the common difference is 11.

76. 200, 100, 50, 25, …

SOLUTION: Check for a common difference. 100 – 200 = –100 50 – 100 = –50 There is no common difference. Check for a common ratio. 100 ÷ 200 = 0.5 50 ÷ 100 = 0.5

Geometric; the common ratio is 0.5 or .

77. JOBS Kimi received an offer for a new job. She wants to compare the offer with her current job. What is total amount of sales that Kimi must get each month to make the same income at either job?

SOLUTION: Let s be Kimi’s monthly sales.

Kimi must sell $80,000 each month to make the same income at either job.

Determine whether each sequence is an arithmetic sequence. If it is, state the common difference.78. 24, 16, 8, 0, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is –8, because 16 – 24 = –8; 8 –16 = –8; etc.

79. , 13, 26, …

SOLUTION: Find the difference between the terms.

– =

13 – =

26 – 13 = 13 There is not a common different. The sequence is not an arithmetic sequence

80. 7, 6, 5, 4, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is –1, because 6 – 7 = –1; 5 – 6 = –1; 4 – 5 = –1;etc.

81. 10, 12, 15, 18, …

SOLUTION: Find the difference between the terms. 12 – 10 = 2 15 – 12 = 3 18 – 15 = 3 There is no common difference. The sequence is not an arithmetic sequence.

82. −15, −11, −7, −3, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is 4, because –11 – (–15) = 4; –7 – (–11) = 4; –3 – (–7) = 4;etc.

83. −0.3, 0.2, 0.7, 1.2, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is 0.5, because 0.2 – (–0.3) = 0.5; 0.7 – 0.2 = 0.5; 1.2 – 0.7 = 0.5; etc.

Simplify.

84. t(t5)(t

7)

SOLUTION:

85. n3(n

2)(−2n

3)

SOLUTION:

86. (5t5v

2)(10t

3v

4)

SOLUTION:

87. (−8u4z

5)(5uz

4)

SOLUTION:

88. [(3)2]3

SOLUTION:

89. [(2)3]2

SOLUTION:

90. (2m4k

3)2(−3mk

2)3

SOLUTION:

91. (6xy2)2(2x

2y

2z

2)3

SOLUTION:

eSolutions Manual - Powered by Cognero Page 19

8-1 Adding and Subtracting Polynomials

Determine whether each expression is a polynomial. If it is a polynomial, find the degree and determine whether it is a monomial, binomial, or trinomial.

1. 7ab + 6b2 – 2a

3

SOLUTION:

A polynomial is a monomial or the sum of monomials. 7ab + 6b2 – 2a

3 is the sum of 3 monomials, so it is a

polynomial. The degree of a polynomial is the greatest degree of any term in the polynomial.The degree of each term is 2, 2, and

3, so the degree of 7ab + 6b2 – 2a

3 is 3. The polynomial has three terms, so it is a trinomial.

2. 2y – 5 + 3y2

SOLUTION:

2y – 5 + 3y2

is the sum of monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 1, 0, and

2, so the degree of 2y – 5 + 3y2 is 2. The polynomial has three terms, so it is a trinomial.

3. 3x2

SOLUTION:

A polynomial is a monomial or the sum of monomials, so 3x2 is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of 3x2 is 2. The

polynomial has one term, so it is a monomial.

4.

SOLUTION: A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer exponents. It has

only one term. is a division of two monomials, so it is not a monomial.

5. 5m2p

3 + 6

SOLUTION:

A polynomial is a monomial or the sum of monomials. 5m2p

3 + 6

is the sum of 2 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 5 and 0,

so the degree of 5m2p

3 + 6 is 5. The polynomial has two terms, so it is a binomial.

6. 5q–4

+ 6q

SOLUTION: A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer exponents. It has

only one term. 5q-4 is equal to , which is a division of two monomials, so it is not a monomial.

Write each polynomial in standard form. Identify the leading coefficient.

7. –4d4 + 1 – d

2

SOLUTION: Find the degree of each term.

–4d4→ 4

1 → 0

–d2 → 2

The greatest degree is 4, from the term –4d4, so the leading coefficient of –4d

4 + 1 – d

2 is –4.

Rewrite the polynomial with each monomial in descending order according to degree.

–4d4 – d

2 + 1

8. 2x5 – 12 + 3x

SOLUTION: Find the degree of each term.

2x5 → 5

–12 → 0 3x → 1

The greatest degree is 5, from the term 2x5, so the leading coefficient of 2x

5 – 12 + 3x is 2.

Rewrite the polynomial with each monomial in descending order according to degree.

2x5 + 3x – 12

9. 4z – 2z2 – 5z

4

SOLUTION: Find the degree of each term. 4z → 1

– 2z2 → 2

– 5z4→ 4

The greatest degree is 4, from the term – 5z4, so the leading coefficient of 4z – 2z

2 – 5z

4 is –5.

Rewrite the polynomial with each monomial in descending order according to degree.

–5z4 – 2z

2 + 4z

10. 2a + 4a3 – 5a

2 – 1

SOLUTION: Find the degree of each term. 2a → 1

4a3→ 3

– 5a2 → 2

– 1 → 0

The greatest degree is 3, from the term 4a3, so the leading coefficient of 2a + 4a

3 – 5a

2 – 1 is 4.

Rewrite the polynomial with each monomial in descending order according to degree.

4a3– 5a

2 + 2a – 1

Find each sum or difference.

11. (6x3 − 4) + (−2x

3 + 9)

SOLUTION:

12. (g3 − 2g2 + 5g + 6) − (g2 + 2g)

SOLUTION:

13. (4 + 2a2 − 2a) − (3a

2 − 8a + 7)

SOLUTION:

14. (8y − 4y2) + (3y − 9y

2)

SOLUTION:

15. (−4z3 − 2z + 8) − (4z

3 + 3z2 − 5)

SOLUTION:

16. (−3d2 − 8 + 2d) + (4d − 12 + d2

)

SOLUTION:

17. (y + 5) + (2y + 4y2 – 2)

SOLUTION:

18. (3n3 − 5n + n2

) − (−8n2 + 3n

3)

SOLUTION:

19. CCSS SENSE-MAKING The total number of students T who traveled for spring break consists of two groups: students who flew to their destinations F and students who drove to their destination D. The number (in thousands) of students who flew and the total number of students who flew or drove can be modeled by the following equations,where n is the number of years since 1995. T = 14n + 21

F = 8n + 7 a. Write an equation that models the number of students who drove to their destination for this time period. b. Predict the number of students who will drive to their destination in 2012. c. How many students will drive or fly to their destination in 2015?

SOLUTION: a.

D = 6n + 14 b. n = 2012 – 1995 = 17

The number of students who will drive to their destination in 2012 is 116,000 students. c. n = 2015 – 1995 = 20

The number of students who will drive or fly to their destination in 2015 is 301,000 students.

Determine whether each expression is a polynomial. If it is a polynomial, find the degree and determine whether it is a monomial, binomial, or trinomial.

20.

SOLUTION: A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer exponents. It has

only one term. is a division of two monomials, so it is not a monomial.

21.

SOLUTION: A polynomial is a monomial or the sum of monomials. 21 is a monomial, so it is also a polynomial. The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of 21 is 0. The polynomial has only one term, so it is a monomial.

22. c4 – 2c

2 + 1

SOLUTION:

A polynomial is a monomial or the sum of monomials. c4 – 2c

2 + 1

is the sum of 3 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 4, 2, and

0, so the degree of c4 – 2c2 + 1 is 4. The polynomial has three terms, so it is a trinomial.

23. d + 3dc

SOLUTION: A polynomial is a monomial or the sum of monomials. A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer

exponents. 3dc

has a variable in the exponent, so it is not a monomial.

24. a – a2

SOLUTION:

A polynomial is a monomial or the sum of monomials. a – a2 is the sum of 2 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 1 and 2,

so the degree of a – a2 is 2. The polynomial has two terms, so it is a binomial.

25. 5n3 + nq

3

SOLUTION:

A polynomial is a monomial or the sum of monomials. 5n3 + nq3 is the sum of 2 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 3 and 4,

so the degree of 5n3 + nq3 is 4. The polynomial has two terms, so it is a binomial.

Write each polynomial in standard form. Identify the leading coefficient.

26. 5x2 – 2 + 3x

SOLUTION: Find the degree of each term.

5x2 → 2

– 2 → 0

3x → 1

The greatest degree is 2, from the term 5x2, so the leading coefficient of 5x

2 – 2 + 3x is 5.

Rewrite the polynomial with each monomial in descending order according to degree.

5x2 + 3x – 2

27. 8y + 7y3

SOLUTION: Find the degree of each term.

7y3 → 3

8y → 1

The greatest degree is 3, from the term 7y3, so the leading coefficient of 8y + 7y

3 is 7.

Rewrite the polynomial with each monomial in descending order according to degree.

7y3 + 8y

28. 4 – 3c – 5c2

SOLUTION: Find the degree of each term. 4 → 0 3c → 1

– 5c2 → 2

The greatest degree is 2, from the term – 5c2, so the leading coefficient of 4 – 3c – 5c

2 is –5.

Rewrite the polynomial with each monomial in descending order according to degree.

–5c2 – 3c + 4

29. –y3 + 3y – 3y

2 + 2

SOLUTION: Find the degree of each term.

–y3 → 3

3y → 1

3y2 → 2

2 → 0

The greatest degree is 3, from the term –y3, so the leading coefficient of –y

3 + 3y – 3y

2 + 2 is –1.

Rewrite the polynomial with each monomial in descending order according to degree. –y 3 – 3y 2 + 3y + 2

30. 11t + 2t2 – 3 + t

5

SOLUTION: Find the degree of each term. 11t → 1

2t2 → 2

–3 → 0

t5 → 5

The greatest degree is 5, from the term t5, so the leading coefficient of 11t + 2t2 – 3 + t

5 is 1.

Rewrite the polynomial with each monomial in descending order according to degree.

t5 + 2t

2 + 11t – 3

31. 2 + r – r3

SOLUTION: Find the degree of each term. 2 → 0 r → 1

– r3→ 3

The greatest degree is 3, from the term – r3, so the leading coefficient of 2 + r – r3

is –1. Rewrite the polynomial with each monomial in descending order according to degree.

–r3 + r + 2

32.

SOLUTION: Find the degree of each term.

→ 0

–3x4 → 4

7 → 0

The greatest degree is 4, from the term –3x4, so the leading coefficient of is –3.

Rewrite the polynomial with each monomial in descending order according to degree.

33. –9b2 + 10b – b

6

SOLUTION: Find the degree of each term.

–9b2 → 2

10b → 1

–b6→ 6

The greatest degree is 6, from the term –b6, so the leading coefficient of –9b2 + 10b – b6 is –1. Rewrite the polynomial with each monomial in descending order according to degree.

–b6 – 9b

2 + 10b

Find each sum or difference.

34. (2c2 + 6c + 4) + (5c – 7)

SOLUTION:

35. (2x + 3x2) − (7 − 8x

2)

SOLUTION:

36. (3c3 − c + 11) − (c2 + 2c + 8)

SOLUTION:

37. (z2 + z) + (z

2 − 11)

SOLUTION:

38. (2x − 2y + 1) − (3y + 4x)

SOLUTION:

39. (4a − 5b2 + 3) + (6 − 2a + 3b

2)

SOLUTION:

40. (x2y − 3x

2 + y) + (3y − 2x2y)

SOLUTION:

41. (−8xy + 3x2 − 5y) + (4x

2 − 2y + 6xy)

SOLUTION:

42. (5n − 2p2 + 2np) − (4p

2 + 4n)

SOLUTION:

43. (4rxt − 8r2x + x2

) − (6rx2 + 5rxt − 2x

2)

SOLUTION:

44. PETS From 1999 through 2009, the number of dogs D and the number of cats C (in hundreds) adopted from animal

shelters in the United States are modeled by the equations D = 2n + 3 and C = n + 4, where n is the number of yearssince 1999. a. Write an equation that models the total number T of dogs and cats adopted in hundreds for this time period. b. If this trend continues, how many dogs and cats will be adopted in 2013?

SOLUTION: a.

So, an equation that models the total number of dogs and cats adopted is T = 3n + 7. b. Evaluate the equation for the total number of dogs and cats for n = 2013 – 1999 = 14.

The number of cats and dogs adopted in 2013 will be 49 × 100 or 4900 cats and dogs.

Classify each polynomial according to its degree and number of terms.

45. 4x – 3x2 + 5

SOLUTION:

Find the degree of each term of 4x – 3x2 + 5.

4x → 1

– 3x2 → 2

5 → 0

The greatest degree is 2 and there are 3 terms, so 4x – 3x2 + 5 is a quadratic trinomial.

46. 11z3

SOLUTION: Find the degree of each term. 11z3

→ 3

The greatest degree is 3 and there is one term, so 11z3 is a cubic monomial.

47. 9 + y4

SOLUTION:

Find the degree of each term of 9 + y 4.

9 → 1

y4→ 4

The greatest degree is 4 and there are 2 terms, so 9 + y 4 is a quartic binomial.

48. 3x3 – 7

SOLUTION:

Find the degree of each term of 3x3 – 7.

3x3 → 3

–7 → 0

The greatest degree is 3 and there are 2 terms, so 3x3 – 7 is a cubic binomial.

49. –2x5 – x

2 + 5x – 8

SOLUTION:

Find the degree of each term of –2x5 – x2 + 5x – 8.

–2x5 → 5

–x2 → 2

5x → 1 – 8 → 0

The greatest degree is 5 and there are 4 terms, so –2x5 – x2 + 5x – 8 is a quintic polynomial.

50. 10t – 4t2 + 6t

3

SOLUTION:

Find the degree of each term of 10t – 4t2 + 6t

3.

10t → 1

4t2 → 2

6t3 → 3

The greatest degree is 3 and there are 3 terms, so 10t – 4t2 + 6t

3 is a cubic trinomial.

51. ENROLLMENT In a rapidly growing school system, the numbers (in hundreds) of total students N and

K-5 students P enrolled from 2000 to 2009 are modeled by the equations N = 1.25t2 – t + 7.5 and P =

0.7t2 – 0.95t + 3.8, where t is the number of years since 2000.

a. Write an equation modeling the number of 6-12 students S enrolled for this time period. b. How many 6-12 students were enrolled in the school system in 2007?

SOLUTION: a. To write an equation that represents the number of 6-12 students enrolled, subtract the equations that represent the total number of students and the number of K-5 students.

b. Replace t with 7 in the equation for S to determine the number of students enrolled in 6-12 in 2007.

In 2007 there were 30.3 hundreds or 3030 students enrolled in 6-12.

52. CCSS REASONING The perimeter of the figure shown is represented by the expression 3x2 − 7x + 2. Write a

polynomial that represents the measure of the third side.

SOLUTION:

53. GEOMETRY Consider the rectangle.

a. What does (4x2 + 2x – 1)(2x

2 – x + 3) represent?

b. What does 2(4x2 + 2x – 1) + 2(2x

2 – x + 3) represent?

SOLUTION:

a. (4x2 + 2x – 1)(2x

2 – x + 3) is a multiplication of the length and the width of the rectangle, which is the formula for

the area of a rectangle.

b. 2(4x2 + 2x – 1) + 2(2x

2 – x + 3) is the sum of twice the length and twice the width of a rectangle, which is the

formula for the perimeter of the rectangle

Find each sum or difference.

54. (4x + 2y − 6z) + (5y − 2z + 7x) + (−9z − 2x − 3y)

SOLUTION:

55. (5a2 − 4) + (a

2 − 2a + 12) + (4a2 − 6a + 8)

SOLUTION:

56. (3c2 − 7) + (4c + 7) − (c

2 + 5c − 8)

SOLUTION:

57. (3n3 + 3n − 10) − (4n

2 − 5n) + (4n3 − 3n

2 − 9n + 4)

SOLUTION:

58. FOOTBALL The National Football League is divided into two conferences, the American A and the National N. From 2002 through 2009, the total attendance T (in thousands) for both conferences and for the American Conference games are modeled by the following equations, where x is the number of years since 2002.

T = –0.69x3 + 55.83x

2 + 643.31x + 10,538

A = –3.78x3 + 58.96x

2 + 265.96x + 5257

Determine how many people attended National Conference football games in 2009.

SOLUTION:

Let x = 7 represent 2009, then find how many people attended a National Conference football game in 2009.

In 2009 the number of people who attended National Conference football games was about 8829 thousand. Multiply by 1000 to find the attendance in standard form. 8829 ×1000 = 8,829,000. So, about 8,829,000 people attended National Conference football games in 2009.

59. CAR RENTAL The cost to rent a car for a day is $15 plus $0.15 for each mile driven. a. Write a polynomial that represents the cost of renting a car for m miles. b. If a car is driven 145 miles, how much would it cost to rent? c. If a car is driven 105 miles each day for four days, how much would it cost to rent a car? d. If a car is driven 220 miles each day for seven days, how much would it cost to rent a car?

SOLUTION: a. The cost to rent a car is the daily rate and the mileage cost or 15 + 0.15m. b. Substitute 145 for m to find the daily cost to drive 145 miles.

The cost to rent the car would be $36.75. c. The expression represents the cost per day. For 4 days, multiply the entire expression by 4 to find the cost for a 4 day trip. Substitute 105 for m.

The cost to rent the car would be $123. d. The expression represents the cost per day. For 7 days, multiply the entire expression by 7 to find the cost for a 7 day trip. Substitute 220 for m.

The cost to rent the car would be $336.

60. MULTIPLE REPRESENTATIONS In this problem, you will explore perimeter and area. a. Geometric Draw three rectangles that each have a perimeter of 400 feet. b. Tabular Record the width and length of each rectangle in a table like the one shown below. Find the area of eachrectangle.

c. Graphical On a coordinate system, graph the area of rectangle 4 in terms of the length, x. Use the graph to determine the largest area possible. d. Analytical Determine the length and width that produce the largest area.

SOLUTION: a. For the perimeters to be 400, create lengths and widths that sum to 200. Note that some lengths are already provided in the table in part b.

b. The area is length multiplied by width. Make sure the units are squared for area. The sum of the length and width must be 200, so if the length is x, the width must be 200 – x.

c. The length will be the x-values on the horizontal and the area will be the y-values on the vertical. The length cannot pass 200 since the sum of the length and width is 200. Set the intervals for the x-axis to 25 feet. Extend the table of values to find more points to plot on the graph. It appears that 10,000 is the greatest area, so set the intervals

for the y-axis to 1000 ft2. After graphing, it appears that the highest point on the graph is at an area of 10,000 ft

2.

d. The associated x-value with the maximum area is x = 100, so the length must be 100 and the width must be 200 – 100, or 100. The length and width of the rectangle must be 100 feet each to have the largest area.

61. CCSS CRITIQUE Cheyenne and Sebastian are finding (2x2 − x) − (3x + 3x

2 − 2). Is either of them correct? Explain your reasoning.

SOLUTION:

Neither is correct. Cheyenne, did not distribute the negative to the 2nd and 3rd terms when she found the additive inverse. Sebastian did not distribute the negate to the 3rd terms when he found the additive inverse. To find the additive inverse, all terms should be multiplied by −1.

62. REASONING Determine whether each of the following statements is true or false . Explain your reasoning. a. A binomial can have a degree of zero. b. The order in which polynomials are subtracted does not matter.

SOLUTION: a. If a binomial has two terms that are each a degree of 0, then those terms can be combined and the binomial becomes a monomial. For example, 18 + 7 = 25. If one of the terms of the binomial does not have a degree of 0, then the binomial cannot have a degree of 0, since the degree of a polynomial is the greatest degree of any term in the polynomial. b. Subtraction is not commutative. While 2 + 5 = 5 + 2, 2 – 5 ≠ 5 – 2. This is also true for polynomials. Sample answer: (2x – 3) – (4x – 3) = –2x, but (4x – 3) – (2x – 3) = 2x

63. CHALLENGE Write a polynomial that represents the sum of an odd integer 2n + 1 and the next two consecutive odd integers.

SOLUTION:

64. WRITING IN MATH Why would you add or subtract equations that represent real-world situations? Explain.

SOLUTION:

65. WRITING IN MATH Describe how to add and subtract polynomials using both the vertical and horizontal formats.

SOLUTION: To add polynomials in a horizontal format, you combine like terms. For the vertical format, you write the polynomials in standard form, align like terms in columns, and combine like terms.

To subtract polynomials in a horizontal format you find the additive inverse of the polynomial you are subtracting, andthen combine like terms. For the vertical format, you write the polynomials in standard form, align like terms in columns, and subtract by adding the additive inverse.

66. Three consecutive integers can be represented by x, x + 1, and x + 2. What is the sum of these three integers? A x(x + 1)(x + 2)

B x3 + 3

C 3x + 3 D x + 3

SOLUTION:

The correct choice is C.

67. SHORT RESPONSE What is the perimeter of a square with sides that measure 2x + 3 units?

SOLUTION:

The perimeter of the square is 8x + 12 units.

68. Jim cuts a board in the shape of a regular hexagon and pounds in a nail at each vertex, as shown. How many rubber bands will he need to stretch a rubber band across every possible pair of nails?

F 15 G 14 H 12 J 9

SOLUTION: The first nail would connect to 5 others, the second to 4 others, the third to 3 others, etc.5 + 4 + 3 + 2 + 1 = 15 The correct choice is F.

69. Which ordered pair is in the solution set of the system of inequalities shown in the graph?

A (−3, 0) B (0, −3) C (5, 0) D (0, 5)

SOLUTION: Choice A is outside the shaded area for both inequalities. Choices B and D are inside the shaded area for only one inequality. Choice C is the only point in the solution for both inequalities. So, the correct choice is C.

70. COMPUTERS A computer technician charges by the hour to fix and repair computer equipment. The total cost of the technician for one hour is $75, for two hours is $125, for three hours is $175, for four hours is $225, and so on. Write a recursive formula for the sequence.

SOLUTION: Write out the terms. $75, $125, $175, $225, ... The first term is 75, and 50 is added to form each following term. Therefore, we have a1 = 75, an = an – 1 + 50, n ≥ 2.

Determine whether each sequence is arithmetic, geometric, or neither. Explain.71. 8, –32, 128, –512, ...

SOLUTION: Check for a common difference. –32 – 8 = –40 128 – (–32) = 160 There is no common difference. Check for a common ratio. –32 ÷ 8 = –4 128 ÷ (–32) = –4 Geometric; the common ratio is –4.

72. 25, 8, –9, –26, ...

SOLUTION: Check for a common difference. 8 – 25 = –17 –9 – 8 = –17 Arithmetic; the common difference is –17.

73.

SOLUTION: Check for a common difference.

There is no common difference. Check for a common ratio.

There is no common ratio, so the sequence is not arithmetic or geometric.

74. 43, 52, 61, 70, ...

SOLUTION: Check for a common difference. 52 – 43 = 9 61 – 52 = 9 Arithmetic; the common difference is 9.

75. –27, –16, –5, 6, ...

SOLUTION: Check for a common difference. –16 – (–27) = 11 –5 – (–16) = 11 Arithmetic; the common difference is 11.

76. 200, 100, 50, 25, …

SOLUTION: Check for a common difference. 100 – 200 = –100 50 – 100 = –50 There is no common difference. Check for a common ratio. 100 ÷ 200 = 0.5 50 ÷ 100 = 0.5

Geometric; the common ratio is 0.5 or .

77. JOBS Kimi received an offer for a new job. She wants to compare the offer with her current job. What is total amount of sales that Kimi must get each month to make the same income at either job?

SOLUTION: Let s be Kimi’s monthly sales.

Kimi must sell $80,000 each month to make the same income at either job.

Determine whether each sequence is an arithmetic sequence. If it is, state the common difference.78. 24, 16, 8, 0, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is –8, because 16 – 24 = –8; 8 –16 = –8; etc.

79. , 13, 26, …

SOLUTION: Find the difference between the terms.

– =

13 – =

26 – 13 = 13 There is not a common different. The sequence is not an arithmetic sequence

80. 7, 6, 5, 4, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is –1, because 6 – 7 = –1; 5 – 6 = –1; 4 – 5 = –1;etc.

81. 10, 12, 15, 18, …

SOLUTION: Find the difference between the terms. 12 – 10 = 2 15 – 12 = 3 18 – 15 = 3 There is no common difference. The sequence is not an arithmetic sequence.

82. −15, −11, −7, −3, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is 4, because –11 – (–15) = 4; –7 – (–11) = 4; –3 – (–7) = 4;etc.

83. −0.3, 0.2, 0.7, 1.2, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is 0.5, because 0.2 – (–0.3) = 0.5; 0.7 – 0.2 = 0.5; 1.2 – 0.7 = 0.5; etc.

Simplify.

84. t(t5)(t

7)

SOLUTION:

85. n3(n

2)(−2n

3)

SOLUTION:

86. (5t5v

2)(10t

3v

4)

SOLUTION:

87. (−8u4z

5)(5uz

4)

SOLUTION:

88. [(3)2]3

SOLUTION:

89. [(2)3]2

SOLUTION:

90. (2m4k

3)2(−3mk

2)3

SOLUTION:

91. (6xy2)2(2x

2y

2z

2)3

SOLUTION:

eSolutions Manual - Powered by Cognero Page 20

8-1 Adding and Subtracting Polynomials

Determine whether each expression is a polynomial. If it is a polynomial, find the degree and determine whether it is a monomial, binomial, or trinomial.

1. 7ab + 6b2 – 2a

3

SOLUTION:

A polynomial is a monomial or the sum of monomials. 7ab + 6b2 – 2a

3 is the sum of 3 monomials, so it is a

polynomial. The degree of a polynomial is the greatest degree of any term in the polynomial.The degree of each term is 2, 2, and

3, so the degree of 7ab + 6b2 – 2a

3 is 3. The polynomial has three terms, so it is a trinomial.

2. 2y – 5 + 3y2

SOLUTION:

2y – 5 + 3y2

is the sum of monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 1, 0, and

2, so the degree of 2y – 5 + 3y2 is 2. The polynomial has three terms, so it is a trinomial.

3. 3x2

SOLUTION:

A polynomial is a monomial or the sum of monomials, so 3x2 is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of 3x2 is 2. The

polynomial has one term, so it is a monomial.

4.

SOLUTION: A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer exponents. It has

only one term. is a division of two monomials, so it is not a monomial.

5. 5m2p

3 + 6

SOLUTION:

A polynomial is a monomial or the sum of monomials. 5m2p

3 + 6

is the sum of 2 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 5 and 0,

so the degree of 5m2p

3 + 6 is 5. The polynomial has two terms, so it is a binomial.

6. 5q–4

+ 6q

SOLUTION: A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer exponents. It has

only one term. 5q-4 is equal to , which is a division of two monomials, so it is not a monomial.

Write each polynomial in standard form. Identify the leading coefficient.

7. –4d4 + 1 – d

2

SOLUTION: Find the degree of each term.

–4d4→ 4

1 → 0

–d2 → 2

The greatest degree is 4, from the term –4d4, so the leading coefficient of –4d

4 + 1 – d

2 is –4.

Rewrite the polynomial with each monomial in descending order according to degree.

–4d4 – d

2 + 1

8. 2x5 – 12 + 3x

SOLUTION: Find the degree of each term.

2x5 → 5

–12 → 0 3x → 1

The greatest degree is 5, from the term 2x5, so the leading coefficient of 2x

5 – 12 + 3x is 2.

Rewrite the polynomial with each monomial in descending order according to degree.

2x5 + 3x – 12

9. 4z – 2z2 – 5z

4

SOLUTION: Find the degree of each term. 4z → 1

– 2z2 → 2

– 5z4→ 4

The greatest degree is 4, from the term – 5z4, so the leading coefficient of 4z – 2z

2 – 5z

4 is –5.

Rewrite the polynomial with each monomial in descending order according to degree.

–5z4 – 2z

2 + 4z

10. 2a + 4a3 – 5a

2 – 1

SOLUTION: Find the degree of each term. 2a → 1

4a3→ 3

– 5a2 → 2

– 1 → 0

The greatest degree is 3, from the term 4a3, so the leading coefficient of 2a + 4a

3 – 5a

2 – 1 is 4.

Rewrite the polynomial with each monomial in descending order according to degree.

4a3– 5a

2 + 2a – 1

Find each sum or difference.

11. (6x3 − 4) + (−2x

3 + 9)

SOLUTION:

12. (g3 − 2g2 + 5g + 6) − (g2 + 2g)

SOLUTION:

13. (4 + 2a2 − 2a) − (3a

2 − 8a + 7)

SOLUTION:

14. (8y − 4y2) + (3y − 9y

2)

SOLUTION:

15. (−4z3 − 2z + 8) − (4z

3 + 3z2 − 5)

SOLUTION:

16. (−3d2 − 8 + 2d) + (4d − 12 + d2

)

SOLUTION:

17. (y + 5) + (2y + 4y2 – 2)

SOLUTION:

18. (3n3 − 5n + n2

) − (−8n2 + 3n

3)

SOLUTION:

19. CCSS SENSE-MAKING The total number of students T who traveled for spring break consists of two groups: students who flew to their destinations F and students who drove to their destination D. The number (in thousands) of students who flew and the total number of students who flew or drove can be modeled by the following equations,where n is the number of years since 1995. T = 14n + 21

F = 8n + 7 a. Write an equation that models the number of students who drove to their destination for this time period. b. Predict the number of students who will drive to their destination in 2012. c. How many students will drive or fly to their destination in 2015?

SOLUTION: a.

D = 6n + 14 b. n = 2012 – 1995 = 17

The number of students who will drive to their destination in 2012 is 116,000 students. c. n = 2015 – 1995 = 20

The number of students who will drive or fly to their destination in 2015 is 301,000 students.

Determine whether each expression is a polynomial. If it is a polynomial, find the degree and determine whether it is a monomial, binomial, or trinomial.

20.

SOLUTION: A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer exponents. It has

only one term. is a division of two monomials, so it is not a monomial.

21.

SOLUTION: A polynomial is a monomial or the sum of monomials. 21 is a monomial, so it is also a polynomial. The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of 21 is 0. The polynomial has only one term, so it is a monomial.

22. c4 – 2c

2 + 1

SOLUTION:

A polynomial is a monomial or the sum of monomials. c4 – 2c

2 + 1

is the sum of 3 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 4, 2, and

0, so the degree of c4 – 2c2 + 1 is 4. The polynomial has three terms, so it is a trinomial.

23. d + 3dc

SOLUTION: A polynomial is a monomial or the sum of monomials. A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer

exponents. 3dc

has a variable in the exponent, so it is not a monomial.

24. a – a2

SOLUTION:

A polynomial is a monomial or the sum of monomials. a – a2 is the sum of 2 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 1 and 2,

so the degree of a – a2 is 2. The polynomial has two terms, so it is a binomial.

25. 5n3 + nq

3

SOLUTION:

A polynomial is a monomial or the sum of monomials. 5n3 + nq3 is the sum of 2 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 3 and 4,

so the degree of 5n3 + nq3 is 4. The polynomial has two terms, so it is a binomial.

Write each polynomial in standard form. Identify the leading coefficient.

26. 5x2 – 2 + 3x

SOLUTION: Find the degree of each term.

5x2 → 2

– 2 → 0

3x → 1

The greatest degree is 2, from the term 5x2, so the leading coefficient of 5x

2 – 2 + 3x is 5.

Rewrite the polynomial with each monomial in descending order according to degree.

5x2 + 3x – 2

27. 8y + 7y3

SOLUTION: Find the degree of each term.

7y3 → 3

8y → 1

The greatest degree is 3, from the term 7y3, so the leading coefficient of 8y + 7y

3 is 7.

Rewrite the polynomial with each monomial in descending order according to degree.

7y3 + 8y

28. 4 – 3c – 5c2

SOLUTION: Find the degree of each term. 4 → 0 3c → 1

– 5c2 → 2

The greatest degree is 2, from the term – 5c2, so the leading coefficient of 4 – 3c – 5c

2 is –5.

Rewrite the polynomial with each monomial in descending order according to degree.

–5c2 – 3c + 4

29. –y3 + 3y – 3y

2 + 2

SOLUTION: Find the degree of each term.

–y3 → 3

3y → 1

3y2 → 2

2 → 0

The greatest degree is 3, from the term –y3, so the leading coefficient of –y

3 + 3y – 3y

2 + 2 is –1.

Rewrite the polynomial with each monomial in descending order according to degree. –y 3 – 3y 2 + 3y + 2

30. 11t + 2t2 – 3 + t

5

SOLUTION: Find the degree of each term. 11t → 1

2t2 → 2

–3 → 0

t5 → 5

The greatest degree is 5, from the term t5, so the leading coefficient of 11t + 2t2 – 3 + t

5 is 1.

Rewrite the polynomial with each monomial in descending order according to degree.

t5 + 2t

2 + 11t – 3

31. 2 + r – r3

SOLUTION: Find the degree of each term. 2 → 0 r → 1

– r3→ 3

The greatest degree is 3, from the term – r3, so the leading coefficient of 2 + r – r3

is –1. Rewrite the polynomial with each monomial in descending order according to degree.

–r3 + r + 2

32.

SOLUTION: Find the degree of each term.

→ 0

–3x4 → 4

7 → 0

The greatest degree is 4, from the term –3x4, so the leading coefficient of is –3.

Rewrite the polynomial with each monomial in descending order according to degree.

33. –9b2 + 10b – b

6

SOLUTION: Find the degree of each term.

–9b2 → 2

10b → 1

–b6→ 6

The greatest degree is 6, from the term –b6, so the leading coefficient of –9b2 + 10b – b6 is –1. Rewrite the polynomial with each monomial in descending order according to degree.

–b6 – 9b

2 + 10b

Find each sum or difference.

34. (2c2 + 6c + 4) + (5c – 7)

SOLUTION:

35. (2x + 3x2) − (7 − 8x

2)

SOLUTION:

36. (3c3 − c + 11) − (c2 + 2c + 8)

SOLUTION:

37. (z2 + z) + (z

2 − 11)

SOLUTION:

38. (2x − 2y + 1) − (3y + 4x)

SOLUTION:

39. (4a − 5b2 + 3) + (6 − 2a + 3b

2)

SOLUTION:

40. (x2y − 3x

2 + y) + (3y − 2x2y)

SOLUTION:

41. (−8xy + 3x2 − 5y) + (4x

2 − 2y + 6xy)

SOLUTION:

42. (5n − 2p2 + 2np) − (4p

2 + 4n)

SOLUTION:

43. (4rxt − 8r2x + x2

) − (6rx2 + 5rxt − 2x

2)

SOLUTION:

44. PETS From 1999 through 2009, the number of dogs D and the number of cats C (in hundreds) adopted from animal

shelters in the United States are modeled by the equations D = 2n + 3 and C = n + 4, where n is the number of yearssince 1999. a. Write an equation that models the total number T of dogs and cats adopted in hundreds for this time period. b. If this trend continues, how many dogs and cats will be adopted in 2013?

SOLUTION: a.

So, an equation that models the total number of dogs and cats adopted is T = 3n + 7. b. Evaluate the equation for the total number of dogs and cats for n = 2013 – 1999 = 14.

The number of cats and dogs adopted in 2013 will be 49 × 100 or 4900 cats and dogs.

Classify each polynomial according to its degree and number of terms.

45. 4x – 3x2 + 5

SOLUTION:

Find the degree of each term of 4x – 3x2 + 5.

4x → 1

– 3x2 → 2

5 → 0

The greatest degree is 2 and there are 3 terms, so 4x – 3x2 + 5 is a quadratic trinomial.

46. 11z3

SOLUTION: Find the degree of each term. 11z3

→ 3

The greatest degree is 3 and there is one term, so 11z3 is a cubic monomial.

47. 9 + y4

SOLUTION:

Find the degree of each term of 9 + y 4.

9 → 1

y4→ 4

The greatest degree is 4 and there are 2 terms, so 9 + y 4 is a quartic binomial.

48. 3x3 – 7

SOLUTION:

Find the degree of each term of 3x3 – 7.

3x3 → 3

–7 → 0

The greatest degree is 3 and there are 2 terms, so 3x3 – 7 is a cubic binomial.

49. –2x5 – x

2 + 5x – 8

SOLUTION:

Find the degree of each term of –2x5 – x2 + 5x – 8.

–2x5 → 5

–x2 → 2

5x → 1 – 8 → 0

The greatest degree is 5 and there are 4 terms, so –2x5 – x2 + 5x – 8 is a quintic polynomial.

50. 10t – 4t2 + 6t

3

SOLUTION:

Find the degree of each term of 10t – 4t2 + 6t

3.

10t → 1

4t2 → 2

6t3 → 3

The greatest degree is 3 and there are 3 terms, so 10t – 4t2 + 6t

3 is a cubic trinomial.

51. ENROLLMENT In a rapidly growing school system, the numbers (in hundreds) of total students N and

K-5 students P enrolled from 2000 to 2009 are modeled by the equations N = 1.25t2 – t + 7.5 and P =

0.7t2 – 0.95t + 3.8, where t is the number of years since 2000.

a. Write an equation modeling the number of 6-12 students S enrolled for this time period. b. How many 6-12 students were enrolled in the school system in 2007?

SOLUTION: a. To write an equation that represents the number of 6-12 students enrolled, subtract the equations that represent the total number of students and the number of K-5 students.

b. Replace t with 7 in the equation for S to determine the number of students enrolled in 6-12 in 2007.

In 2007 there were 30.3 hundreds or 3030 students enrolled in 6-12.

52. CCSS REASONING The perimeter of the figure shown is represented by the expression 3x2 − 7x + 2. Write a

polynomial that represents the measure of the third side.

SOLUTION:

53. GEOMETRY Consider the rectangle.

a. What does (4x2 + 2x – 1)(2x

2 – x + 3) represent?

b. What does 2(4x2 + 2x – 1) + 2(2x

2 – x + 3) represent?

SOLUTION:

a. (4x2 + 2x – 1)(2x

2 – x + 3) is a multiplication of the length and the width of the rectangle, which is the formula for

the area of a rectangle.

b. 2(4x2 + 2x – 1) + 2(2x

2 – x + 3) is the sum of twice the length and twice the width of a rectangle, which is the

formula for the perimeter of the rectangle

Find each sum or difference.

54. (4x + 2y − 6z) + (5y − 2z + 7x) + (−9z − 2x − 3y)

SOLUTION:

55. (5a2 − 4) + (a

2 − 2a + 12) + (4a2 − 6a + 8)

SOLUTION:

56. (3c2 − 7) + (4c + 7) − (c

2 + 5c − 8)

SOLUTION:

57. (3n3 + 3n − 10) − (4n

2 − 5n) + (4n3 − 3n

2 − 9n + 4)

SOLUTION:

58. FOOTBALL The National Football League is divided into two conferences, the American A and the National N. From 2002 through 2009, the total attendance T (in thousands) for both conferences and for the American Conference games are modeled by the following equations, where x is the number of years since 2002.

T = –0.69x3 + 55.83x

2 + 643.31x + 10,538

A = –3.78x3 + 58.96x

2 + 265.96x + 5257

Determine how many people attended National Conference football games in 2009.

SOLUTION:

Let x = 7 represent 2009, then find how many people attended a National Conference football game in 2009.

In 2009 the number of people who attended National Conference football games was about 8829 thousand. Multiply by 1000 to find the attendance in standard form. 8829 ×1000 = 8,829,000. So, about 8,829,000 people attended National Conference football games in 2009.

59. CAR RENTAL The cost to rent a car for a day is $15 plus $0.15 for each mile driven. a. Write a polynomial that represents the cost of renting a car for m miles. b. If a car is driven 145 miles, how much would it cost to rent? c. If a car is driven 105 miles each day for four days, how much would it cost to rent a car? d. If a car is driven 220 miles each day for seven days, how much would it cost to rent a car?

SOLUTION: a. The cost to rent a car is the daily rate and the mileage cost or 15 + 0.15m. b. Substitute 145 for m to find the daily cost to drive 145 miles.

The cost to rent the car would be $36.75. c. The expression represents the cost per day. For 4 days, multiply the entire expression by 4 to find the cost for a 4 day trip. Substitute 105 for m.

The cost to rent the car would be $123. d. The expression represents the cost per day. For 7 days, multiply the entire expression by 7 to find the cost for a 7 day trip. Substitute 220 for m.

The cost to rent the car would be $336.

60. MULTIPLE REPRESENTATIONS In this problem, you will explore perimeter and area. a. Geometric Draw three rectangles that each have a perimeter of 400 feet. b. Tabular Record the width and length of each rectangle in a table like the one shown below. Find the area of eachrectangle.

c. Graphical On a coordinate system, graph the area of rectangle 4 in terms of the length, x. Use the graph to determine the largest area possible. d. Analytical Determine the length and width that produce the largest area.

SOLUTION: a. For the perimeters to be 400, create lengths and widths that sum to 200. Note that some lengths are already provided in the table in part b.

b. The area is length multiplied by width. Make sure the units are squared for area. The sum of the length and width must be 200, so if the length is x, the width must be 200 – x.

c. The length will be the x-values on the horizontal and the area will be the y-values on the vertical. The length cannot pass 200 since the sum of the length and width is 200. Set the intervals for the x-axis to 25 feet. Extend the table of values to find more points to plot on the graph. It appears that 10,000 is the greatest area, so set the intervals

for the y-axis to 1000 ft2. After graphing, it appears that the highest point on the graph is at an area of 10,000 ft

2.

d. The associated x-value with the maximum area is x = 100, so the length must be 100 and the width must be 200 – 100, or 100. The length and width of the rectangle must be 100 feet each to have the largest area.

61. CCSS CRITIQUE Cheyenne and Sebastian are finding (2x2 − x) − (3x + 3x

2 − 2). Is either of them correct? Explain your reasoning.

SOLUTION:

Neither is correct. Cheyenne, did not distribute the negative to the 2nd and 3rd terms when she found the additive inverse. Sebastian did not distribute the negate to the 3rd terms when he found the additive inverse. To find the additive inverse, all terms should be multiplied by −1.

62. REASONING Determine whether each of the following statements is true or false . Explain your reasoning. a. A binomial can have a degree of zero. b. The order in which polynomials are subtracted does not matter.

SOLUTION: a. If a binomial has two terms that are each a degree of 0, then those terms can be combined and the binomial becomes a monomial. For example, 18 + 7 = 25. If one of the terms of the binomial does not have a degree of 0, then the binomial cannot have a degree of 0, since the degree of a polynomial is the greatest degree of any term in the polynomial. b. Subtraction is not commutative. While 2 + 5 = 5 + 2, 2 – 5 ≠ 5 – 2. This is also true for polynomials. Sample answer: (2x – 3) – (4x – 3) = –2x, but (4x – 3) – (2x – 3) = 2x

63. CHALLENGE Write a polynomial that represents the sum of an odd integer 2n + 1 and the next two consecutive odd integers.

SOLUTION:

64. WRITING IN MATH Why would you add or subtract equations that represent real-world situations? Explain.

SOLUTION:

65. WRITING IN MATH Describe how to add and subtract polynomials using both the vertical and horizontal formats.

SOLUTION: To add polynomials in a horizontal format, you combine like terms. For the vertical format, you write the polynomials in standard form, align like terms in columns, and combine like terms.

To subtract polynomials in a horizontal format you find the additive inverse of the polynomial you are subtracting, andthen combine like terms. For the vertical format, you write the polynomials in standard form, align like terms in columns, and subtract by adding the additive inverse.

66. Three consecutive integers can be represented by x, x + 1, and x + 2. What is the sum of these three integers? A x(x + 1)(x + 2)

B x3 + 3

C 3x + 3 D x + 3

SOLUTION:

The correct choice is C.

67. SHORT RESPONSE What is the perimeter of a square with sides that measure 2x + 3 units?

SOLUTION:

The perimeter of the square is 8x + 12 units.

68. Jim cuts a board in the shape of a regular hexagon and pounds in a nail at each vertex, as shown. How many rubber bands will he need to stretch a rubber band across every possible pair of nails?

F 15 G 14 H 12 J 9

SOLUTION: The first nail would connect to 5 others, the second to 4 others, the third to 3 others, etc.5 + 4 + 3 + 2 + 1 = 15 The correct choice is F.

69. Which ordered pair is in the solution set of the system of inequalities shown in the graph?

A (−3, 0) B (0, −3) C (5, 0) D (0, 5)

SOLUTION: Choice A is outside the shaded area for both inequalities. Choices B and D are inside the shaded area for only one inequality. Choice C is the only point in the solution for both inequalities. So, the correct choice is C.

70. COMPUTERS A computer technician charges by the hour to fix and repair computer equipment. The total cost of the technician for one hour is $75, for two hours is $125, for three hours is $175, for four hours is $225, and so on. Write a recursive formula for the sequence.

SOLUTION: Write out the terms. $75, $125, $175, $225, ... The first term is 75, and 50 is added to form each following term. Therefore, we have a1 = 75, an = an – 1 + 50, n ≥ 2.

Determine whether each sequence is arithmetic, geometric, or neither. Explain.71. 8, –32, 128, –512, ...

SOLUTION: Check for a common difference. –32 – 8 = –40 128 – (–32) = 160 There is no common difference. Check for a common ratio. –32 ÷ 8 = –4 128 ÷ (–32) = –4 Geometric; the common ratio is –4.

72. 25, 8, –9, –26, ...

SOLUTION: Check for a common difference. 8 – 25 = –17 –9 – 8 = –17 Arithmetic; the common difference is –17.

73.

SOLUTION: Check for a common difference.

There is no common difference. Check for a common ratio.

There is no common ratio, so the sequence is not arithmetic or geometric.

74. 43, 52, 61, 70, ...

SOLUTION: Check for a common difference. 52 – 43 = 9 61 – 52 = 9 Arithmetic; the common difference is 9.

75. –27, –16, –5, 6, ...

SOLUTION: Check for a common difference. –16 – (–27) = 11 –5 – (–16) = 11 Arithmetic; the common difference is 11.

76. 200, 100, 50, 25, …

SOLUTION: Check for a common difference. 100 – 200 = –100 50 – 100 = –50 There is no common difference. Check for a common ratio. 100 ÷ 200 = 0.5 50 ÷ 100 = 0.5

Geometric; the common ratio is 0.5 or .

77. JOBS Kimi received an offer for a new job. She wants to compare the offer with her current job. What is total amount of sales that Kimi must get each month to make the same income at either job?

SOLUTION: Let s be Kimi’s monthly sales.

Kimi must sell $80,000 each month to make the same income at either job.

Determine whether each sequence is an arithmetic sequence. If it is, state the common difference.78. 24, 16, 8, 0, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is –8, because 16 – 24 = –8; 8 –16 = –8; etc.

79. , 13, 26, …

SOLUTION: Find the difference between the terms.

– =

13 – =

26 – 13 = 13 There is not a common different. The sequence is not an arithmetic sequence

80. 7, 6, 5, 4, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is –1, because 6 – 7 = –1; 5 – 6 = –1; 4 – 5 = –1;etc.

81. 10, 12, 15, 18, …

SOLUTION: Find the difference between the terms. 12 – 10 = 2 15 – 12 = 3 18 – 15 = 3 There is no common difference. The sequence is not an arithmetic sequence.

82. −15, −11, −7, −3, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is 4, because –11 – (–15) = 4; –7 – (–11) = 4; –3 – (–7) = 4;etc.

83. −0.3, 0.2, 0.7, 1.2, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is 0.5, because 0.2 – (–0.3) = 0.5; 0.7 – 0.2 = 0.5; 1.2 – 0.7 = 0.5; etc.

Simplify.

84. t(t5)(t

7)

SOLUTION:

85. n3(n

2)(−2n

3)

SOLUTION:

86. (5t5v

2)(10t

3v

4)

SOLUTION:

87. (−8u4z

5)(5uz

4)

SOLUTION:

88. [(3)2]3

SOLUTION:

89. [(2)3]2

SOLUTION:

90. (2m4k

3)2(−3mk

2)3

SOLUTION:

91. (6xy2)2(2x

2y

2z

2)3

SOLUTION:

eSolutions Manual - Powered by Cognero Page 21

8-1 Adding and Subtracting Polynomials

Determine whether each expression is a polynomial. If it is a polynomial, find the degree and determine whether it is a monomial, binomial, or trinomial.

1. 7ab + 6b2 – 2a

3

SOLUTION:

A polynomial is a monomial or the sum of monomials. 7ab + 6b2 – 2a

3 is the sum of 3 monomials, so it is a

polynomial. The degree of a polynomial is the greatest degree of any term in the polynomial.The degree of each term is 2, 2, and

3, so the degree of 7ab + 6b2 – 2a

3 is 3. The polynomial has three terms, so it is a trinomial.

2. 2y – 5 + 3y2

SOLUTION:

2y – 5 + 3y2

is the sum of monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 1, 0, and

2, so the degree of 2y – 5 + 3y2 is 2. The polynomial has three terms, so it is a trinomial.

3. 3x2

SOLUTION:

A polynomial is a monomial or the sum of monomials, so 3x2 is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of 3x2 is 2. The

polynomial has one term, so it is a monomial.

4.

SOLUTION: A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer exponents. It has

only one term. is a division of two monomials, so it is not a monomial.

5. 5m2p

3 + 6

SOLUTION:

A polynomial is a monomial or the sum of monomials. 5m2p

3 + 6

is the sum of 2 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 5 and 0,

so the degree of 5m2p

3 + 6 is 5. The polynomial has two terms, so it is a binomial.

6. 5q–4

+ 6q

SOLUTION: A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer exponents. It has

only one term. 5q-4 is equal to , which is a division of two monomials, so it is not a monomial.

Write each polynomial in standard form. Identify the leading coefficient.

7. –4d4 + 1 – d

2

SOLUTION: Find the degree of each term.

–4d4→ 4

1 → 0

–d2 → 2

The greatest degree is 4, from the term –4d4, so the leading coefficient of –4d

4 + 1 – d

2 is –4.

Rewrite the polynomial with each monomial in descending order according to degree.

–4d4 – d

2 + 1

8. 2x5 – 12 + 3x

SOLUTION: Find the degree of each term.

2x5 → 5

–12 → 0 3x → 1

The greatest degree is 5, from the term 2x5, so the leading coefficient of 2x

5 – 12 + 3x is 2.

Rewrite the polynomial with each monomial in descending order according to degree.

2x5 + 3x – 12

9. 4z – 2z2 – 5z

4

SOLUTION: Find the degree of each term. 4z → 1

– 2z2 → 2

– 5z4→ 4

The greatest degree is 4, from the term – 5z4, so the leading coefficient of 4z – 2z

2 – 5z

4 is –5.

Rewrite the polynomial with each monomial in descending order according to degree.

–5z4 – 2z

2 + 4z

10. 2a + 4a3 – 5a

2 – 1

SOLUTION: Find the degree of each term. 2a → 1

4a3→ 3

– 5a2 → 2

– 1 → 0

The greatest degree is 3, from the term 4a3, so the leading coefficient of 2a + 4a

3 – 5a

2 – 1 is 4.

Rewrite the polynomial with each monomial in descending order according to degree.

4a3– 5a

2 + 2a – 1

Find each sum or difference.

11. (6x3 − 4) + (−2x

3 + 9)

SOLUTION:

12. (g3 − 2g2 + 5g + 6) − (g2 + 2g)

SOLUTION:

13. (4 + 2a2 − 2a) − (3a

2 − 8a + 7)

SOLUTION:

14. (8y − 4y2) + (3y − 9y

2)

SOLUTION:

15. (−4z3 − 2z + 8) − (4z

3 + 3z2 − 5)

SOLUTION:

16. (−3d2 − 8 + 2d) + (4d − 12 + d2

)

SOLUTION:

17. (y + 5) + (2y + 4y2 – 2)

SOLUTION:

18. (3n3 − 5n + n2

) − (−8n2 + 3n

3)

SOLUTION:

19. CCSS SENSE-MAKING The total number of students T who traveled for spring break consists of two groups: students who flew to their destinations F and students who drove to their destination D. The number (in thousands) of students who flew and the total number of students who flew or drove can be modeled by the following equations,where n is the number of years since 1995. T = 14n + 21

F = 8n + 7 a. Write an equation that models the number of students who drove to their destination for this time period. b. Predict the number of students who will drive to their destination in 2012. c. How many students will drive or fly to their destination in 2015?

SOLUTION: a.

D = 6n + 14 b. n = 2012 – 1995 = 17

The number of students who will drive to their destination in 2012 is 116,000 students. c. n = 2015 – 1995 = 20

The number of students who will drive or fly to their destination in 2015 is 301,000 students.

Determine whether each expression is a polynomial. If it is a polynomial, find the degree and determine whether it is a monomial, binomial, or trinomial.

20.

SOLUTION: A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer exponents. It has

only one term. is a division of two monomials, so it is not a monomial.

21.

SOLUTION: A polynomial is a monomial or the sum of monomials. 21 is a monomial, so it is also a polynomial. The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of 21 is 0. The polynomial has only one term, so it is a monomial.

22. c4 – 2c

2 + 1

SOLUTION:

A polynomial is a monomial or the sum of monomials. c4 – 2c

2 + 1

is the sum of 3 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 4, 2, and

0, so the degree of c4 – 2c2 + 1 is 4. The polynomial has three terms, so it is a trinomial.

23. d + 3dc

SOLUTION: A polynomial is a monomial or the sum of monomials. A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer

exponents. 3dc

has a variable in the exponent, so it is not a monomial.

24. a – a2

SOLUTION:

A polynomial is a monomial or the sum of monomials. a – a2 is the sum of 2 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 1 and 2,

so the degree of a – a2 is 2. The polynomial has two terms, so it is a binomial.

25. 5n3 + nq

3

SOLUTION:

A polynomial is a monomial or the sum of monomials. 5n3 + nq3 is the sum of 2 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 3 and 4,

so the degree of 5n3 + nq3 is 4. The polynomial has two terms, so it is a binomial.

Write each polynomial in standard form. Identify the leading coefficient.

26. 5x2 – 2 + 3x

SOLUTION: Find the degree of each term.

5x2 → 2

– 2 → 0

3x → 1

The greatest degree is 2, from the term 5x2, so the leading coefficient of 5x

2 – 2 + 3x is 5.

Rewrite the polynomial with each monomial in descending order according to degree.

5x2 + 3x – 2

27. 8y + 7y3

SOLUTION: Find the degree of each term.

7y3 → 3

8y → 1

The greatest degree is 3, from the term 7y3, so the leading coefficient of 8y + 7y

3 is 7.

Rewrite the polynomial with each monomial in descending order according to degree.

7y3 + 8y

28. 4 – 3c – 5c2

SOLUTION: Find the degree of each term. 4 → 0 3c → 1

– 5c2 → 2

The greatest degree is 2, from the term – 5c2, so the leading coefficient of 4 – 3c – 5c

2 is –5.

Rewrite the polynomial with each monomial in descending order according to degree.

–5c2 – 3c + 4

29. –y3 + 3y – 3y

2 + 2

SOLUTION: Find the degree of each term.

–y3 → 3

3y → 1

3y2 → 2

2 → 0

The greatest degree is 3, from the term –y3, so the leading coefficient of –y

3 + 3y – 3y

2 + 2 is –1.

Rewrite the polynomial with each monomial in descending order according to degree. –y 3 – 3y 2 + 3y + 2

30. 11t + 2t2 – 3 + t

5

SOLUTION: Find the degree of each term. 11t → 1

2t2 → 2

–3 → 0

t5 → 5

The greatest degree is 5, from the term t5, so the leading coefficient of 11t + 2t2 – 3 + t

5 is 1.

Rewrite the polynomial with each monomial in descending order according to degree.

t5 + 2t

2 + 11t – 3

31. 2 + r – r3

SOLUTION: Find the degree of each term. 2 → 0 r → 1

– r3→ 3

The greatest degree is 3, from the term – r3, so the leading coefficient of 2 + r – r3

is –1. Rewrite the polynomial with each monomial in descending order according to degree.

–r3 + r + 2

32.

SOLUTION: Find the degree of each term.

→ 0

–3x4 → 4

7 → 0

The greatest degree is 4, from the term –3x4, so the leading coefficient of is –3.

Rewrite the polynomial with each monomial in descending order according to degree.

33. –9b2 + 10b – b

6

SOLUTION: Find the degree of each term.

–9b2 → 2

10b → 1

–b6→ 6

The greatest degree is 6, from the term –b6, so the leading coefficient of –9b2 + 10b – b6 is –1. Rewrite the polynomial with each monomial in descending order according to degree.

–b6 – 9b

2 + 10b

Find each sum or difference.

34. (2c2 + 6c + 4) + (5c – 7)

SOLUTION:

35. (2x + 3x2) − (7 − 8x

2)

SOLUTION:

36. (3c3 − c + 11) − (c2 + 2c + 8)

SOLUTION:

37. (z2 + z) + (z

2 − 11)

SOLUTION:

38. (2x − 2y + 1) − (3y + 4x)

SOLUTION:

39. (4a − 5b2 + 3) + (6 − 2a + 3b

2)

SOLUTION:

40. (x2y − 3x

2 + y) + (3y − 2x2y)

SOLUTION:

41. (−8xy + 3x2 − 5y) + (4x

2 − 2y + 6xy)

SOLUTION:

42. (5n − 2p2 + 2np) − (4p

2 + 4n)

SOLUTION:

43. (4rxt − 8r2x + x2

) − (6rx2 + 5rxt − 2x

2)

SOLUTION:

44. PETS From 1999 through 2009, the number of dogs D and the number of cats C (in hundreds) adopted from animal

shelters in the United States are modeled by the equations D = 2n + 3 and C = n + 4, where n is the number of yearssince 1999. a. Write an equation that models the total number T of dogs and cats adopted in hundreds for this time period. b. If this trend continues, how many dogs and cats will be adopted in 2013?

SOLUTION: a.

So, an equation that models the total number of dogs and cats adopted is T = 3n + 7. b. Evaluate the equation for the total number of dogs and cats for n = 2013 – 1999 = 14.

The number of cats and dogs adopted in 2013 will be 49 × 100 or 4900 cats and dogs.

Classify each polynomial according to its degree and number of terms.

45. 4x – 3x2 + 5

SOLUTION:

Find the degree of each term of 4x – 3x2 + 5.

4x → 1

– 3x2 → 2

5 → 0

The greatest degree is 2 and there are 3 terms, so 4x – 3x2 + 5 is a quadratic trinomial.

46. 11z3

SOLUTION: Find the degree of each term. 11z3

→ 3

The greatest degree is 3 and there is one term, so 11z3 is a cubic monomial.

47. 9 + y4

SOLUTION:

Find the degree of each term of 9 + y 4.

9 → 1

y4→ 4

The greatest degree is 4 and there are 2 terms, so 9 + y 4 is a quartic binomial.

48. 3x3 – 7

SOLUTION:

Find the degree of each term of 3x3 – 7.

3x3 → 3

–7 → 0

The greatest degree is 3 and there are 2 terms, so 3x3 – 7 is a cubic binomial.

49. –2x5 – x

2 + 5x – 8

SOLUTION:

Find the degree of each term of –2x5 – x2 + 5x – 8.

–2x5 → 5

–x2 → 2

5x → 1 – 8 → 0

The greatest degree is 5 and there are 4 terms, so –2x5 – x2 + 5x – 8 is a quintic polynomial.

50. 10t – 4t2 + 6t

3

SOLUTION:

Find the degree of each term of 10t – 4t2 + 6t

3.

10t → 1

4t2 → 2

6t3 → 3

The greatest degree is 3 and there are 3 terms, so 10t – 4t2 + 6t

3 is a cubic trinomial.

51. ENROLLMENT In a rapidly growing school system, the numbers (in hundreds) of total students N and

K-5 students P enrolled from 2000 to 2009 are modeled by the equations N = 1.25t2 – t + 7.5 and P =

0.7t2 – 0.95t + 3.8, where t is the number of years since 2000.

a. Write an equation modeling the number of 6-12 students S enrolled for this time period. b. How many 6-12 students were enrolled in the school system in 2007?

SOLUTION: a. To write an equation that represents the number of 6-12 students enrolled, subtract the equations that represent the total number of students and the number of K-5 students.

b. Replace t with 7 in the equation for S to determine the number of students enrolled in 6-12 in 2007.

In 2007 there were 30.3 hundreds or 3030 students enrolled in 6-12.

52. CCSS REASONING The perimeter of the figure shown is represented by the expression 3x2 − 7x + 2. Write a

polynomial that represents the measure of the third side.

SOLUTION:

53. GEOMETRY Consider the rectangle.

a. What does (4x2 + 2x – 1)(2x

2 – x + 3) represent?

b. What does 2(4x2 + 2x – 1) + 2(2x

2 – x + 3) represent?

SOLUTION:

a. (4x2 + 2x – 1)(2x

2 – x + 3) is a multiplication of the length and the width of the rectangle, which is the formula for

the area of a rectangle.

b. 2(4x2 + 2x – 1) + 2(2x

2 – x + 3) is the sum of twice the length and twice the width of a rectangle, which is the

formula for the perimeter of the rectangle

Find each sum or difference.

54. (4x + 2y − 6z) + (5y − 2z + 7x) + (−9z − 2x − 3y)

SOLUTION:

55. (5a2 − 4) + (a

2 − 2a + 12) + (4a2 − 6a + 8)

SOLUTION:

56. (3c2 − 7) + (4c + 7) − (c

2 + 5c − 8)

SOLUTION:

57. (3n3 + 3n − 10) − (4n

2 − 5n) + (4n3 − 3n

2 − 9n + 4)

SOLUTION:

58. FOOTBALL The National Football League is divided into two conferences, the American A and the National N. From 2002 through 2009, the total attendance T (in thousands) for both conferences and for the American Conference games are modeled by the following equations, where x is the number of years since 2002.

T = –0.69x3 + 55.83x

2 + 643.31x + 10,538

A = –3.78x3 + 58.96x

2 + 265.96x + 5257

Determine how many people attended National Conference football games in 2009.

SOLUTION:

Let x = 7 represent 2009, then find how many people attended a National Conference football game in 2009.

In 2009 the number of people who attended National Conference football games was about 8829 thousand. Multiply by 1000 to find the attendance in standard form. 8829 ×1000 = 8,829,000. So, about 8,829,000 people attended National Conference football games in 2009.

59. CAR RENTAL The cost to rent a car for a day is $15 plus $0.15 for each mile driven. a. Write a polynomial that represents the cost of renting a car for m miles. b. If a car is driven 145 miles, how much would it cost to rent? c. If a car is driven 105 miles each day for four days, how much would it cost to rent a car? d. If a car is driven 220 miles each day for seven days, how much would it cost to rent a car?

SOLUTION: a. The cost to rent a car is the daily rate and the mileage cost or 15 + 0.15m. b. Substitute 145 for m to find the daily cost to drive 145 miles.

The cost to rent the car would be $36.75. c. The expression represents the cost per day. For 4 days, multiply the entire expression by 4 to find the cost for a 4 day trip. Substitute 105 for m.

The cost to rent the car would be $123. d. The expression represents the cost per day. For 7 days, multiply the entire expression by 7 to find the cost for a 7 day trip. Substitute 220 for m.

The cost to rent the car would be $336.

60. MULTIPLE REPRESENTATIONS In this problem, you will explore perimeter and area. a. Geometric Draw three rectangles that each have a perimeter of 400 feet. b. Tabular Record the width and length of each rectangle in a table like the one shown below. Find the area of eachrectangle.

c. Graphical On a coordinate system, graph the area of rectangle 4 in terms of the length, x. Use the graph to determine the largest area possible. d. Analytical Determine the length and width that produce the largest area.

SOLUTION: a. For the perimeters to be 400, create lengths and widths that sum to 200. Note that some lengths are already provided in the table in part b.

b. The area is length multiplied by width. Make sure the units are squared for area. The sum of the length and width must be 200, so if the length is x, the width must be 200 – x.

c. The length will be the x-values on the horizontal and the area will be the y-values on the vertical. The length cannot pass 200 since the sum of the length and width is 200. Set the intervals for the x-axis to 25 feet. Extend the table of values to find more points to plot on the graph. It appears that 10,000 is the greatest area, so set the intervals

for the y-axis to 1000 ft2. After graphing, it appears that the highest point on the graph is at an area of 10,000 ft

2.

d. The associated x-value with the maximum area is x = 100, so the length must be 100 and the width must be 200 – 100, or 100. The length and width of the rectangle must be 100 feet each to have the largest area.

61. CCSS CRITIQUE Cheyenne and Sebastian are finding (2x2 − x) − (3x + 3x

2 − 2). Is either of them correct? Explain your reasoning.

SOLUTION:

Neither is correct. Cheyenne, did not distribute the negative to the 2nd and 3rd terms when she found the additive inverse. Sebastian did not distribute the negate to the 3rd terms when he found the additive inverse. To find the additive inverse, all terms should be multiplied by −1.

62. REASONING Determine whether each of the following statements is true or false . Explain your reasoning. a. A binomial can have a degree of zero. b. The order in which polynomials are subtracted does not matter.

SOLUTION: a. If a binomial has two terms that are each a degree of 0, then those terms can be combined and the binomial becomes a monomial. For example, 18 + 7 = 25. If one of the terms of the binomial does not have a degree of 0, then the binomial cannot have a degree of 0, since the degree of a polynomial is the greatest degree of any term in the polynomial. b. Subtraction is not commutative. While 2 + 5 = 5 + 2, 2 – 5 ≠ 5 – 2. This is also true for polynomials. Sample answer: (2x – 3) – (4x – 3) = –2x, but (4x – 3) – (2x – 3) = 2x

63. CHALLENGE Write a polynomial that represents the sum of an odd integer 2n + 1 and the next two consecutive odd integers.

SOLUTION:

64. WRITING IN MATH Why would you add or subtract equations that represent real-world situations? Explain.

SOLUTION:

65. WRITING IN MATH Describe how to add and subtract polynomials using both the vertical and horizontal formats.

SOLUTION: To add polynomials in a horizontal format, you combine like terms. For the vertical format, you write the polynomials in standard form, align like terms in columns, and combine like terms.

To subtract polynomials in a horizontal format you find the additive inverse of the polynomial you are subtracting, andthen combine like terms. For the vertical format, you write the polynomials in standard form, align like terms in columns, and subtract by adding the additive inverse.

66. Three consecutive integers can be represented by x, x + 1, and x + 2. What is the sum of these three integers? A x(x + 1)(x + 2)

B x3 + 3

C 3x + 3 D x + 3

SOLUTION:

The correct choice is C.

67. SHORT RESPONSE What is the perimeter of a square with sides that measure 2x + 3 units?

SOLUTION:

The perimeter of the square is 8x + 12 units.

68. Jim cuts a board in the shape of a regular hexagon and pounds in a nail at each vertex, as shown. How many rubber bands will he need to stretch a rubber band across every possible pair of nails?

F 15 G 14 H 12 J 9

SOLUTION: The first nail would connect to 5 others, the second to 4 others, the third to 3 others, etc.5 + 4 + 3 + 2 + 1 = 15 The correct choice is F.

69. Which ordered pair is in the solution set of the system of inequalities shown in the graph?

A (−3, 0) B (0, −3) C (5, 0) D (0, 5)

SOLUTION: Choice A is outside the shaded area for both inequalities. Choices B and D are inside the shaded area for only one inequality. Choice C is the only point in the solution for both inequalities. So, the correct choice is C.

70. COMPUTERS A computer technician charges by the hour to fix and repair computer equipment. The total cost of the technician for one hour is $75, for two hours is $125, for three hours is $175, for four hours is $225, and so on. Write a recursive formula for the sequence.

SOLUTION: Write out the terms. $75, $125, $175, $225, ... The first term is 75, and 50 is added to form each following term. Therefore, we have a1 = 75, an = an – 1 + 50, n ≥ 2.

Determine whether each sequence is arithmetic, geometric, or neither. Explain.71. 8, –32, 128, –512, ...

SOLUTION: Check for a common difference. –32 – 8 = –40 128 – (–32) = 160 There is no common difference. Check for a common ratio. –32 ÷ 8 = –4 128 ÷ (–32) = –4 Geometric; the common ratio is –4.

72. 25, 8, –9, –26, ...

SOLUTION: Check for a common difference. 8 – 25 = –17 –9 – 8 = –17 Arithmetic; the common difference is –17.

73.

SOLUTION: Check for a common difference.

There is no common difference. Check for a common ratio.

There is no common ratio, so the sequence is not arithmetic or geometric.

74. 43, 52, 61, 70, ...

SOLUTION: Check for a common difference. 52 – 43 = 9 61 – 52 = 9 Arithmetic; the common difference is 9.

75. –27, –16, –5, 6, ...

SOLUTION: Check for a common difference. –16 – (–27) = 11 –5 – (–16) = 11 Arithmetic; the common difference is 11.

76. 200, 100, 50, 25, …

SOLUTION: Check for a common difference. 100 – 200 = –100 50 – 100 = –50 There is no common difference. Check for a common ratio. 100 ÷ 200 = 0.5 50 ÷ 100 = 0.5

Geometric; the common ratio is 0.5 or .

77. JOBS Kimi received an offer for a new job. She wants to compare the offer with her current job. What is total amount of sales that Kimi must get each month to make the same income at either job?

SOLUTION: Let s be Kimi’s monthly sales.

Kimi must sell $80,000 each month to make the same income at either job.

Determine whether each sequence is an arithmetic sequence. If it is, state the common difference.78. 24, 16, 8, 0, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is –8, because 16 – 24 = –8; 8 –16 = –8; etc.

79. , 13, 26, …

SOLUTION: Find the difference between the terms.

– =

13 – =

26 – 13 = 13 There is not a common different. The sequence is not an arithmetic sequence

80. 7, 6, 5, 4, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is –1, because 6 – 7 = –1; 5 – 6 = –1; 4 – 5 = –1;etc.

81. 10, 12, 15, 18, …

SOLUTION: Find the difference between the terms. 12 – 10 = 2 15 – 12 = 3 18 – 15 = 3 There is no common difference. The sequence is not an arithmetic sequence.

82. −15, −11, −7, −3, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is 4, because –11 – (–15) = 4; –7 – (–11) = 4; –3 – (–7) = 4;etc.

83. −0.3, 0.2, 0.7, 1.2, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is 0.5, because 0.2 – (–0.3) = 0.5; 0.7 – 0.2 = 0.5; 1.2 – 0.7 = 0.5; etc.

Simplify.

84. t(t5)(t

7)

SOLUTION:

85. n3(n

2)(−2n

3)

SOLUTION:

86. (5t5v

2)(10t

3v

4)

SOLUTION:

87. (−8u4z

5)(5uz

4)

SOLUTION:

88. [(3)2]3

SOLUTION:

89. [(2)3]2

SOLUTION:

90. (2m4k

3)2(−3mk

2)3

SOLUTION:

91. (6xy2)2(2x

2y

2z

2)3

SOLUTION:

eSolutions Manual - Powered by Cognero Page 22

8-1 Adding and Subtracting Polynomials

Determine whether each expression is a polynomial. If it is a polynomial, find the degree and determine whether it is a monomial, binomial, or trinomial.

1. 7ab + 6b2 – 2a

3

SOLUTION:

A polynomial is a monomial or the sum of monomials. 7ab + 6b2 – 2a

3 is the sum of 3 monomials, so it is a

polynomial. The degree of a polynomial is the greatest degree of any term in the polynomial.The degree of each term is 2, 2, and

3, so the degree of 7ab + 6b2 – 2a

3 is 3. The polynomial has three terms, so it is a trinomial.

2. 2y – 5 + 3y2

SOLUTION:

2y – 5 + 3y2

is the sum of monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 1, 0, and

2, so the degree of 2y – 5 + 3y2 is 2. The polynomial has three terms, so it is a trinomial.

3. 3x2

SOLUTION:

A polynomial is a monomial or the sum of monomials, so 3x2 is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of 3x2 is 2. The

polynomial has one term, so it is a monomial.

4.

SOLUTION: A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer exponents. It has

only one term. is a division of two monomials, so it is not a monomial.

5. 5m2p

3 + 6

SOLUTION:

A polynomial is a monomial or the sum of monomials. 5m2p

3 + 6

is the sum of 2 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 5 and 0,

so the degree of 5m2p

3 + 6 is 5. The polynomial has two terms, so it is a binomial.

6. 5q–4

+ 6q

SOLUTION: A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer exponents. It has

only one term. 5q-4 is equal to , which is a division of two monomials, so it is not a monomial.

Write each polynomial in standard form. Identify the leading coefficient.

7. –4d4 + 1 – d

2

SOLUTION: Find the degree of each term.

–4d4→ 4

1 → 0

–d2 → 2

The greatest degree is 4, from the term –4d4, so the leading coefficient of –4d

4 + 1 – d

2 is –4.

Rewrite the polynomial with each monomial in descending order according to degree.

–4d4 – d

2 + 1

8. 2x5 – 12 + 3x

SOLUTION: Find the degree of each term.

2x5 → 5

–12 → 0 3x → 1

The greatest degree is 5, from the term 2x5, so the leading coefficient of 2x

5 – 12 + 3x is 2.

Rewrite the polynomial with each monomial in descending order according to degree.

2x5 + 3x – 12

9. 4z – 2z2 – 5z

4

SOLUTION: Find the degree of each term. 4z → 1

– 2z2 → 2

– 5z4→ 4

The greatest degree is 4, from the term – 5z4, so the leading coefficient of 4z – 2z

2 – 5z

4 is –5.

Rewrite the polynomial with each monomial in descending order according to degree.

–5z4 – 2z

2 + 4z

10. 2a + 4a3 – 5a

2 – 1

SOLUTION: Find the degree of each term. 2a → 1

4a3→ 3

– 5a2 → 2

– 1 → 0

The greatest degree is 3, from the term 4a3, so the leading coefficient of 2a + 4a

3 – 5a

2 – 1 is 4.

Rewrite the polynomial with each monomial in descending order according to degree.

4a3– 5a

2 + 2a – 1

Find each sum or difference.

11. (6x3 − 4) + (−2x

3 + 9)

SOLUTION:

12. (g3 − 2g2 + 5g + 6) − (g2 + 2g)

SOLUTION:

13. (4 + 2a2 − 2a) − (3a

2 − 8a + 7)

SOLUTION:

14. (8y − 4y2) + (3y − 9y

2)

SOLUTION:

15. (−4z3 − 2z + 8) − (4z

3 + 3z2 − 5)

SOLUTION:

16. (−3d2 − 8 + 2d) + (4d − 12 + d2

)

SOLUTION:

17. (y + 5) + (2y + 4y2 – 2)

SOLUTION:

18. (3n3 − 5n + n2

) − (−8n2 + 3n

3)

SOLUTION:

19. CCSS SENSE-MAKING The total number of students T who traveled for spring break consists of two groups: students who flew to their destinations F and students who drove to their destination D. The number (in thousands) of students who flew and the total number of students who flew or drove can be modeled by the following equations,where n is the number of years since 1995. T = 14n + 21

F = 8n + 7 a. Write an equation that models the number of students who drove to their destination for this time period. b. Predict the number of students who will drive to their destination in 2012. c. How many students will drive or fly to their destination in 2015?

SOLUTION: a.

D = 6n + 14 b. n = 2012 – 1995 = 17

The number of students who will drive to their destination in 2012 is 116,000 students. c. n = 2015 – 1995 = 20

The number of students who will drive or fly to their destination in 2015 is 301,000 students.

Determine whether each expression is a polynomial. If it is a polynomial, find the degree and determine whether it is a monomial, binomial, or trinomial.

20.

SOLUTION: A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer exponents. It has

only one term. is a division of two monomials, so it is not a monomial.

21.

SOLUTION: A polynomial is a monomial or the sum of monomials. 21 is a monomial, so it is also a polynomial. The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of 21 is 0. The polynomial has only one term, so it is a monomial.

22. c4 – 2c

2 + 1

SOLUTION:

A polynomial is a monomial or the sum of monomials. c4 – 2c

2 + 1

is the sum of 3 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 4, 2, and

0, so the degree of c4 – 2c2 + 1 is 4. The polynomial has three terms, so it is a trinomial.

23. d + 3dc

SOLUTION: A polynomial is a monomial or the sum of monomials. A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer

exponents. 3dc

has a variable in the exponent, so it is not a monomial.

24. a – a2

SOLUTION:

A polynomial is a monomial or the sum of monomials. a – a2 is the sum of 2 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 1 and 2,

so the degree of a – a2 is 2. The polynomial has two terms, so it is a binomial.

25. 5n3 + nq

3

SOLUTION:

A polynomial is a monomial or the sum of monomials. 5n3 + nq3 is the sum of 2 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 3 and 4,

so the degree of 5n3 + nq3 is 4. The polynomial has two terms, so it is a binomial.

Write each polynomial in standard form. Identify the leading coefficient.

26. 5x2 – 2 + 3x

SOLUTION: Find the degree of each term.

5x2 → 2

– 2 → 0

3x → 1

The greatest degree is 2, from the term 5x2, so the leading coefficient of 5x

2 – 2 + 3x is 5.

Rewrite the polynomial with each monomial in descending order according to degree.

5x2 + 3x – 2

27. 8y + 7y3

SOLUTION: Find the degree of each term.

7y3 → 3

8y → 1

The greatest degree is 3, from the term 7y3, so the leading coefficient of 8y + 7y

3 is 7.

Rewrite the polynomial with each monomial in descending order according to degree.

7y3 + 8y

28. 4 – 3c – 5c2

SOLUTION: Find the degree of each term. 4 → 0 3c → 1

– 5c2 → 2

The greatest degree is 2, from the term – 5c2, so the leading coefficient of 4 – 3c – 5c

2 is –5.

Rewrite the polynomial with each monomial in descending order according to degree.

–5c2 – 3c + 4

29. –y3 + 3y – 3y

2 + 2

SOLUTION: Find the degree of each term.

–y3 → 3

3y → 1

3y2 → 2

2 → 0

The greatest degree is 3, from the term –y3, so the leading coefficient of –y

3 + 3y – 3y

2 + 2 is –1.

Rewrite the polynomial with each monomial in descending order according to degree. –y 3 – 3y 2 + 3y + 2

30. 11t + 2t2 – 3 + t

5

SOLUTION: Find the degree of each term. 11t → 1

2t2 → 2

–3 → 0

t5 → 5

The greatest degree is 5, from the term t5, so the leading coefficient of 11t + 2t2 – 3 + t

5 is 1.

Rewrite the polynomial with each monomial in descending order according to degree.

t5 + 2t

2 + 11t – 3

31. 2 + r – r3

SOLUTION: Find the degree of each term. 2 → 0 r → 1

– r3→ 3

The greatest degree is 3, from the term – r3, so the leading coefficient of 2 + r – r3

is –1. Rewrite the polynomial with each monomial in descending order according to degree.

–r3 + r + 2

32.

SOLUTION: Find the degree of each term.

→ 0

–3x4 → 4

7 → 0

The greatest degree is 4, from the term –3x4, so the leading coefficient of is –3.

Rewrite the polynomial with each monomial in descending order according to degree.

33. –9b2 + 10b – b

6

SOLUTION: Find the degree of each term.

–9b2 → 2

10b → 1

–b6→ 6

The greatest degree is 6, from the term –b6, so the leading coefficient of –9b2 + 10b – b6 is –1. Rewrite the polynomial with each monomial in descending order according to degree.

–b6 – 9b

2 + 10b

Find each sum or difference.

34. (2c2 + 6c + 4) + (5c – 7)

SOLUTION:

35. (2x + 3x2) − (7 − 8x

2)

SOLUTION:

36. (3c3 − c + 11) − (c2 + 2c + 8)

SOLUTION:

37. (z2 + z) + (z

2 − 11)

SOLUTION:

38. (2x − 2y + 1) − (3y + 4x)

SOLUTION:

39. (4a − 5b2 + 3) + (6 − 2a + 3b

2)

SOLUTION:

40. (x2y − 3x

2 + y) + (3y − 2x2y)

SOLUTION:

41. (−8xy + 3x2 − 5y) + (4x

2 − 2y + 6xy)

SOLUTION:

42. (5n − 2p2 + 2np) − (4p

2 + 4n)

SOLUTION:

43. (4rxt − 8r2x + x2

) − (6rx2 + 5rxt − 2x

2)

SOLUTION:

44. PETS From 1999 through 2009, the number of dogs D and the number of cats C (in hundreds) adopted from animal

shelters in the United States are modeled by the equations D = 2n + 3 and C = n + 4, where n is the number of yearssince 1999. a. Write an equation that models the total number T of dogs and cats adopted in hundreds for this time period. b. If this trend continues, how many dogs and cats will be adopted in 2013?

SOLUTION: a.

So, an equation that models the total number of dogs and cats adopted is T = 3n + 7. b. Evaluate the equation for the total number of dogs and cats for n = 2013 – 1999 = 14.

The number of cats and dogs adopted in 2013 will be 49 × 100 or 4900 cats and dogs.

Classify each polynomial according to its degree and number of terms.

45. 4x – 3x2 + 5

SOLUTION:

Find the degree of each term of 4x – 3x2 + 5.

4x → 1

– 3x2 → 2

5 → 0

The greatest degree is 2 and there are 3 terms, so 4x – 3x2 + 5 is a quadratic trinomial.

46. 11z3

SOLUTION: Find the degree of each term. 11z3

→ 3

The greatest degree is 3 and there is one term, so 11z3 is a cubic monomial.

47. 9 + y4

SOLUTION:

Find the degree of each term of 9 + y 4.

9 → 1

y4→ 4

The greatest degree is 4 and there are 2 terms, so 9 + y 4 is a quartic binomial.

48. 3x3 – 7

SOLUTION:

Find the degree of each term of 3x3 – 7.

3x3 → 3

–7 → 0

The greatest degree is 3 and there are 2 terms, so 3x3 – 7 is a cubic binomial.

49. –2x5 – x

2 + 5x – 8

SOLUTION:

Find the degree of each term of –2x5 – x2 + 5x – 8.

–2x5 → 5

–x2 → 2

5x → 1 – 8 → 0

The greatest degree is 5 and there are 4 terms, so –2x5 – x2 + 5x – 8 is a quintic polynomial.

50. 10t – 4t2 + 6t

3

SOLUTION:

Find the degree of each term of 10t – 4t2 + 6t

3.

10t → 1

4t2 → 2

6t3 → 3

The greatest degree is 3 and there are 3 terms, so 10t – 4t2 + 6t

3 is a cubic trinomial.

51. ENROLLMENT In a rapidly growing school system, the numbers (in hundreds) of total students N and

K-5 students P enrolled from 2000 to 2009 are modeled by the equations N = 1.25t2 – t + 7.5 and P =

0.7t2 – 0.95t + 3.8, where t is the number of years since 2000.

a. Write an equation modeling the number of 6-12 students S enrolled for this time period. b. How many 6-12 students were enrolled in the school system in 2007?

SOLUTION: a. To write an equation that represents the number of 6-12 students enrolled, subtract the equations that represent the total number of students and the number of K-5 students.

b. Replace t with 7 in the equation for S to determine the number of students enrolled in 6-12 in 2007.

In 2007 there were 30.3 hundreds or 3030 students enrolled in 6-12.

52. CCSS REASONING The perimeter of the figure shown is represented by the expression 3x2 − 7x + 2. Write a

polynomial that represents the measure of the third side.

SOLUTION:

53. GEOMETRY Consider the rectangle.

a. What does (4x2 + 2x – 1)(2x

2 – x + 3) represent?

b. What does 2(4x2 + 2x – 1) + 2(2x

2 – x + 3) represent?

SOLUTION:

a. (4x2 + 2x – 1)(2x

2 – x + 3) is a multiplication of the length and the width of the rectangle, which is the formula for

the area of a rectangle.

b. 2(4x2 + 2x – 1) + 2(2x

2 – x + 3) is the sum of twice the length and twice the width of a rectangle, which is the

formula for the perimeter of the rectangle

Find each sum or difference.

54. (4x + 2y − 6z) + (5y − 2z + 7x) + (−9z − 2x − 3y)

SOLUTION:

55. (5a2 − 4) + (a

2 − 2a + 12) + (4a2 − 6a + 8)

SOLUTION:

56. (3c2 − 7) + (4c + 7) − (c

2 + 5c − 8)

SOLUTION:

57. (3n3 + 3n − 10) − (4n

2 − 5n) + (4n3 − 3n

2 − 9n + 4)

SOLUTION:

58. FOOTBALL The National Football League is divided into two conferences, the American A and the National N. From 2002 through 2009, the total attendance T (in thousands) for both conferences and for the American Conference games are modeled by the following equations, where x is the number of years since 2002.

T = –0.69x3 + 55.83x

2 + 643.31x + 10,538

A = –3.78x3 + 58.96x

2 + 265.96x + 5257

Determine how many people attended National Conference football games in 2009.

SOLUTION:

Let x = 7 represent 2009, then find how many people attended a National Conference football game in 2009.

In 2009 the number of people who attended National Conference football games was about 8829 thousand. Multiply by 1000 to find the attendance in standard form. 8829 ×1000 = 8,829,000. So, about 8,829,000 people attended National Conference football games in 2009.

59. CAR RENTAL The cost to rent a car for a day is $15 plus $0.15 for each mile driven. a. Write a polynomial that represents the cost of renting a car for m miles. b. If a car is driven 145 miles, how much would it cost to rent? c. If a car is driven 105 miles each day for four days, how much would it cost to rent a car? d. If a car is driven 220 miles each day for seven days, how much would it cost to rent a car?

SOLUTION: a. The cost to rent a car is the daily rate and the mileage cost or 15 + 0.15m. b. Substitute 145 for m to find the daily cost to drive 145 miles.

The cost to rent the car would be $36.75. c. The expression represents the cost per day. For 4 days, multiply the entire expression by 4 to find the cost for a 4 day trip. Substitute 105 for m.

The cost to rent the car would be $123. d. The expression represents the cost per day. For 7 days, multiply the entire expression by 7 to find the cost for a 7 day trip. Substitute 220 for m.

The cost to rent the car would be $336.

60. MULTIPLE REPRESENTATIONS In this problem, you will explore perimeter and area. a. Geometric Draw three rectangles that each have a perimeter of 400 feet. b. Tabular Record the width and length of each rectangle in a table like the one shown below. Find the area of eachrectangle.

c. Graphical On a coordinate system, graph the area of rectangle 4 in terms of the length, x. Use the graph to determine the largest area possible. d. Analytical Determine the length and width that produce the largest area.

SOLUTION: a. For the perimeters to be 400, create lengths and widths that sum to 200. Note that some lengths are already provided in the table in part b.

b. The area is length multiplied by width. Make sure the units are squared for area. The sum of the length and width must be 200, so if the length is x, the width must be 200 – x.

c. The length will be the x-values on the horizontal and the area will be the y-values on the vertical. The length cannot pass 200 since the sum of the length and width is 200. Set the intervals for the x-axis to 25 feet. Extend the table of values to find more points to plot on the graph. It appears that 10,000 is the greatest area, so set the intervals

for the y-axis to 1000 ft2. After graphing, it appears that the highest point on the graph is at an area of 10,000 ft

2.

d. The associated x-value with the maximum area is x = 100, so the length must be 100 and the width must be 200 – 100, or 100. The length and width of the rectangle must be 100 feet each to have the largest area.

61. CCSS CRITIQUE Cheyenne and Sebastian are finding (2x2 − x) − (3x + 3x

2 − 2). Is either of them correct? Explain your reasoning.

SOLUTION:

Neither is correct. Cheyenne, did not distribute the negative to the 2nd and 3rd terms when she found the additive inverse. Sebastian did not distribute the negate to the 3rd terms when he found the additive inverse. To find the additive inverse, all terms should be multiplied by −1.

62. REASONING Determine whether each of the following statements is true or false . Explain your reasoning. a. A binomial can have a degree of zero. b. The order in which polynomials are subtracted does not matter.

SOLUTION: a. If a binomial has two terms that are each a degree of 0, then those terms can be combined and the binomial becomes a monomial. For example, 18 + 7 = 25. If one of the terms of the binomial does not have a degree of 0, then the binomial cannot have a degree of 0, since the degree of a polynomial is the greatest degree of any term in the polynomial. b. Subtraction is not commutative. While 2 + 5 = 5 + 2, 2 – 5 ≠ 5 – 2. This is also true for polynomials. Sample answer: (2x – 3) – (4x – 3) = –2x, but (4x – 3) – (2x – 3) = 2x

63. CHALLENGE Write a polynomial that represents the sum of an odd integer 2n + 1 and the next two consecutive odd integers.

SOLUTION:

64. WRITING IN MATH Why would you add or subtract equations that represent real-world situations? Explain.

SOLUTION:

65. WRITING IN MATH Describe how to add and subtract polynomials using both the vertical and horizontal formats.

SOLUTION: To add polynomials in a horizontal format, you combine like terms. For the vertical format, you write the polynomials in standard form, align like terms in columns, and combine like terms.

To subtract polynomials in a horizontal format you find the additive inverse of the polynomial you are subtracting, andthen combine like terms. For the vertical format, you write the polynomials in standard form, align like terms in columns, and subtract by adding the additive inverse.

66. Three consecutive integers can be represented by x, x + 1, and x + 2. What is the sum of these three integers? A x(x + 1)(x + 2)

B x3 + 3

C 3x + 3 D x + 3

SOLUTION:

The correct choice is C.

67. SHORT RESPONSE What is the perimeter of a square with sides that measure 2x + 3 units?

SOLUTION:

The perimeter of the square is 8x + 12 units.

68. Jim cuts a board in the shape of a regular hexagon and pounds in a nail at each vertex, as shown. How many rubber bands will he need to stretch a rubber band across every possible pair of nails?

F 15 G 14 H 12 J 9

SOLUTION: The first nail would connect to 5 others, the second to 4 others, the third to 3 others, etc.5 + 4 + 3 + 2 + 1 = 15 The correct choice is F.

69. Which ordered pair is in the solution set of the system of inequalities shown in the graph?

A (−3, 0) B (0, −3) C (5, 0) D (0, 5)

SOLUTION: Choice A is outside the shaded area for both inequalities. Choices B and D are inside the shaded area for only one inequality. Choice C is the only point in the solution for both inequalities. So, the correct choice is C.

70. COMPUTERS A computer technician charges by the hour to fix and repair computer equipment. The total cost of the technician for one hour is $75, for two hours is $125, for three hours is $175, for four hours is $225, and so on. Write a recursive formula for the sequence.

SOLUTION: Write out the terms. $75, $125, $175, $225, ... The first term is 75, and 50 is added to form each following term. Therefore, we have a1 = 75, an = an – 1 + 50, n ≥ 2.

Determine whether each sequence is arithmetic, geometric, or neither. Explain.71. 8, –32, 128, –512, ...

SOLUTION: Check for a common difference. –32 – 8 = –40 128 – (–32) = 160 There is no common difference. Check for a common ratio. –32 ÷ 8 = –4 128 ÷ (–32) = –4 Geometric; the common ratio is –4.

72. 25, 8, –9, –26, ...

SOLUTION: Check for a common difference. 8 – 25 = –17 –9 – 8 = –17 Arithmetic; the common difference is –17.

73.

SOLUTION: Check for a common difference.

There is no common difference. Check for a common ratio.

There is no common ratio, so the sequence is not arithmetic or geometric.

74. 43, 52, 61, 70, ...

SOLUTION: Check for a common difference. 52 – 43 = 9 61 – 52 = 9 Arithmetic; the common difference is 9.

75. –27, –16, –5, 6, ...

SOLUTION: Check for a common difference. –16 – (–27) = 11 –5 – (–16) = 11 Arithmetic; the common difference is 11.

76. 200, 100, 50, 25, …

SOLUTION: Check for a common difference. 100 – 200 = –100 50 – 100 = –50 There is no common difference. Check for a common ratio. 100 ÷ 200 = 0.5 50 ÷ 100 = 0.5

Geometric; the common ratio is 0.5 or .

77. JOBS Kimi received an offer for a new job. She wants to compare the offer with her current job. What is total amount of sales that Kimi must get each month to make the same income at either job?

SOLUTION: Let s be Kimi’s monthly sales.

Kimi must sell $80,000 each month to make the same income at either job.

Determine whether each sequence is an arithmetic sequence. If it is, state the common difference.78. 24, 16, 8, 0, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is –8, because 16 – 24 = –8; 8 –16 = –8; etc.

79. , 13, 26, …

SOLUTION: Find the difference between the terms.

– =

13 – =

26 – 13 = 13 There is not a common different. The sequence is not an arithmetic sequence

80. 7, 6, 5, 4, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is –1, because 6 – 7 = –1; 5 – 6 = –1; 4 – 5 = –1;etc.

81. 10, 12, 15, 18, …

SOLUTION: Find the difference between the terms. 12 – 10 = 2 15 – 12 = 3 18 – 15 = 3 There is no common difference. The sequence is not an arithmetic sequence.

82. −15, −11, −7, −3, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is 4, because –11 – (–15) = 4; –7 – (–11) = 4; –3 – (–7) = 4;etc.

83. −0.3, 0.2, 0.7, 1.2, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is 0.5, because 0.2 – (–0.3) = 0.5; 0.7 – 0.2 = 0.5; 1.2 – 0.7 = 0.5; etc.

Simplify.

84. t(t5)(t

7)

SOLUTION:

85. n3(n

2)(−2n

3)

SOLUTION:

86. (5t5v

2)(10t

3v

4)

SOLUTION:

87. (−8u4z

5)(5uz

4)

SOLUTION:

88. [(3)2]3

SOLUTION:

89. [(2)3]2

SOLUTION:

90. (2m4k

3)2(−3mk

2)3

SOLUTION:

91. (6xy2)2(2x

2y

2z

2)3

SOLUTION:

eSolutions Manual - Powered by Cognero Page 23

8-1 Adding and Subtracting Polynomials

Determine whether each expression is a polynomial. If it is a polynomial, find the degree and determine whether it is a monomial, binomial, or trinomial.

1. 7ab + 6b2 – 2a

3

SOLUTION:

A polynomial is a monomial or the sum of monomials. 7ab + 6b2 – 2a

3 is the sum of 3 monomials, so it is a

polynomial. The degree of a polynomial is the greatest degree of any term in the polynomial.The degree of each term is 2, 2, and

3, so the degree of 7ab + 6b2 – 2a

3 is 3. The polynomial has three terms, so it is a trinomial.

2. 2y – 5 + 3y2

SOLUTION:

2y – 5 + 3y2

is the sum of monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 1, 0, and

2, so the degree of 2y – 5 + 3y2 is 2. The polynomial has three terms, so it is a trinomial.

3. 3x2

SOLUTION:

A polynomial is a monomial or the sum of monomials, so 3x2 is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of 3x2 is 2. The

polynomial has one term, so it is a monomial.

4.

SOLUTION: A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer exponents. It has

only one term. is a division of two monomials, so it is not a monomial.

5. 5m2p

3 + 6

SOLUTION:

A polynomial is a monomial or the sum of monomials. 5m2p

3 + 6

is the sum of 2 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 5 and 0,

so the degree of 5m2p

3 + 6 is 5. The polynomial has two terms, so it is a binomial.

6. 5q–4

+ 6q

SOLUTION: A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer exponents. It has

only one term. 5q-4 is equal to , which is a division of two monomials, so it is not a monomial.

Write each polynomial in standard form. Identify the leading coefficient.

7. –4d4 + 1 – d

2

SOLUTION: Find the degree of each term.

–4d4→ 4

1 → 0

–d2 → 2

The greatest degree is 4, from the term –4d4, so the leading coefficient of –4d

4 + 1 – d

2 is –4.

Rewrite the polynomial with each monomial in descending order according to degree.

–4d4 – d

2 + 1

8. 2x5 – 12 + 3x

SOLUTION: Find the degree of each term.

2x5 → 5

–12 → 0 3x → 1

The greatest degree is 5, from the term 2x5, so the leading coefficient of 2x

5 – 12 + 3x is 2.

Rewrite the polynomial with each monomial in descending order according to degree.

2x5 + 3x – 12

9. 4z – 2z2 – 5z

4

SOLUTION: Find the degree of each term. 4z → 1

– 2z2 → 2

– 5z4→ 4

The greatest degree is 4, from the term – 5z4, so the leading coefficient of 4z – 2z

2 – 5z

4 is –5.

Rewrite the polynomial with each monomial in descending order according to degree.

–5z4 – 2z

2 + 4z

10. 2a + 4a3 – 5a

2 – 1

SOLUTION: Find the degree of each term. 2a → 1

4a3→ 3

– 5a2 → 2

– 1 → 0

The greatest degree is 3, from the term 4a3, so the leading coefficient of 2a + 4a

3 – 5a

2 – 1 is 4.

Rewrite the polynomial with each monomial in descending order according to degree.

4a3– 5a

2 + 2a – 1

Find each sum or difference.

11. (6x3 − 4) + (−2x

3 + 9)

SOLUTION:

12. (g3 − 2g2 + 5g + 6) − (g2 + 2g)

SOLUTION:

13. (4 + 2a2 − 2a) − (3a

2 − 8a + 7)

SOLUTION:

14. (8y − 4y2) + (3y − 9y

2)

SOLUTION:

15. (−4z3 − 2z + 8) − (4z

3 + 3z2 − 5)

SOLUTION:

16. (−3d2 − 8 + 2d) + (4d − 12 + d2

)

SOLUTION:

17. (y + 5) + (2y + 4y2 – 2)

SOLUTION:

18. (3n3 − 5n + n2

) − (−8n2 + 3n

3)

SOLUTION:

19. CCSS SENSE-MAKING The total number of students T who traveled for spring break consists of two groups: students who flew to their destinations F and students who drove to their destination D. The number (in thousands) of students who flew and the total number of students who flew or drove can be modeled by the following equations,where n is the number of years since 1995. T = 14n + 21

F = 8n + 7 a. Write an equation that models the number of students who drove to their destination for this time period. b. Predict the number of students who will drive to their destination in 2012. c. How many students will drive or fly to their destination in 2015?

SOLUTION: a.

D = 6n + 14 b. n = 2012 – 1995 = 17

The number of students who will drive to their destination in 2012 is 116,000 students. c. n = 2015 – 1995 = 20

The number of students who will drive or fly to their destination in 2015 is 301,000 students.

Determine whether each expression is a polynomial. If it is a polynomial, find the degree and determine whether it is a monomial, binomial, or trinomial.

20.

SOLUTION: A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer exponents. It has

only one term. is a division of two monomials, so it is not a monomial.

21.

SOLUTION: A polynomial is a monomial or the sum of monomials. 21 is a monomial, so it is also a polynomial. The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of 21 is 0. The polynomial has only one term, so it is a monomial.

22. c4 – 2c

2 + 1

SOLUTION:

A polynomial is a monomial or the sum of monomials. c4 – 2c

2 + 1

is the sum of 3 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 4, 2, and

0, so the degree of c4 – 2c2 + 1 is 4. The polynomial has three terms, so it is a trinomial.

23. d + 3dc

SOLUTION: A polynomial is a monomial or the sum of monomials. A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer

exponents. 3dc

has a variable in the exponent, so it is not a monomial.

24. a – a2

SOLUTION:

A polynomial is a monomial or the sum of monomials. a – a2 is the sum of 2 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 1 and 2,

so the degree of a – a2 is 2. The polynomial has two terms, so it is a binomial.

25. 5n3 + nq

3

SOLUTION:

A polynomial is a monomial or the sum of monomials. 5n3 + nq3 is the sum of 2 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 3 and 4,

so the degree of 5n3 + nq3 is 4. The polynomial has two terms, so it is a binomial.

Write each polynomial in standard form. Identify the leading coefficient.

26. 5x2 – 2 + 3x

SOLUTION: Find the degree of each term.

5x2 → 2

– 2 → 0

3x → 1

The greatest degree is 2, from the term 5x2, so the leading coefficient of 5x

2 – 2 + 3x is 5.

Rewrite the polynomial with each monomial in descending order according to degree.

5x2 + 3x – 2

27. 8y + 7y3

SOLUTION: Find the degree of each term.

7y3 → 3

8y → 1

The greatest degree is 3, from the term 7y3, so the leading coefficient of 8y + 7y

3 is 7.

Rewrite the polynomial with each monomial in descending order according to degree.

7y3 + 8y

28. 4 – 3c – 5c2

SOLUTION: Find the degree of each term. 4 → 0 3c → 1

– 5c2 → 2

The greatest degree is 2, from the term – 5c2, so the leading coefficient of 4 – 3c – 5c

2 is –5.

Rewrite the polynomial with each monomial in descending order according to degree.

–5c2 – 3c + 4

29. –y3 + 3y – 3y

2 + 2

SOLUTION: Find the degree of each term.

–y3 → 3

3y → 1

3y2 → 2

2 → 0

The greatest degree is 3, from the term –y3, so the leading coefficient of –y

3 + 3y – 3y

2 + 2 is –1.

Rewrite the polynomial with each monomial in descending order according to degree. –y 3 – 3y 2 + 3y + 2

30. 11t + 2t2 – 3 + t

5

SOLUTION: Find the degree of each term. 11t → 1

2t2 → 2

–3 → 0

t5 → 5

The greatest degree is 5, from the term t5, so the leading coefficient of 11t + 2t2 – 3 + t

5 is 1.

Rewrite the polynomial with each monomial in descending order according to degree.

t5 + 2t

2 + 11t – 3

31. 2 + r – r3

SOLUTION: Find the degree of each term. 2 → 0 r → 1

– r3→ 3

The greatest degree is 3, from the term – r3, so the leading coefficient of 2 + r – r3

is –1. Rewrite the polynomial with each monomial in descending order according to degree.

–r3 + r + 2

32.

SOLUTION: Find the degree of each term.

→ 0

–3x4 → 4

7 → 0

The greatest degree is 4, from the term –3x4, so the leading coefficient of is –3.

Rewrite the polynomial with each monomial in descending order according to degree.

33. –9b2 + 10b – b

6

SOLUTION: Find the degree of each term.

–9b2 → 2

10b → 1

–b6→ 6

The greatest degree is 6, from the term –b6, so the leading coefficient of –9b2 + 10b – b6 is –1. Rewrite the polynomial with each monomial in descending order according to degree.

–b6 – 9b

2 + 10b

Find each sum or difference.

34. (2c2 + 6c + 4) + (5c – 7)

SOLUTION:

35. (2x + 3x2) − (7 − 8x

2)

SOLUTION:

36. (3c3 − c + 11) − (c2 + 2c + 8)

SOLUTION:

37. (z2 + z) + (z

2 − 11)

SOLUTION:

38. (2x − 2y + 1) − (3y + 4x)

SOLUTION:

39. (4a − 5b2 + 3) + (6 − 2a + 3b

2)

SOLUTION:

40. (x2y − 3x

2 + y) + (3y − 2x2y)

SOLUTION:

41. (−8xy + 3x2 − 5y) + (4x

2 − 2y + 6xy)

SOLUTION:

42. (5n − 2p2 + 2np) − (4p

2 + 4n)

SOLUTION:

43. (4rxt − 8r2x + x2

) − (6rx2 + 5rxt − 2x

2)

SOLUTION:

44. PETS From 1999 through 2009, the number of dogs D and the number of cats C (in hundreds) adopted from animal

shelters in the United States are modeled by the equations D = 2n + 3 and C = n + 4, where n is the number of yearssince 1999. a. Write an equation that models the total number T of dogs and cats adopted in hundreds for this time period. b. If this trend continues, how many dogs and cats will be adopted in 2013?

SOLUTION: a.

So, an equation that models the total number of dogs and cats adopted is T = 3n + 7. b. Evaluate the equation for the total number of dogs and cats for n = 2013 – 1999 = 14.

The number of cats and dogs adopted in 2013 will be 49 × 100 or 4900 cats and dogs.

Classify each polynomial according to its degree and number of terms.

45. 4x – 3x2 + 5

SOLUTION:

Find the degree of each term of 4x – 3x2 + 5.

4x → 1

– 3x2 → 2

5 → 0

The greatest degree is 2 and there are 3 terms, so 4x – 3x2 + 5 is a quadratic trinomial.

46. 11z3

SOLUTION: Find the degree of each term. 11z3

→ 3

The greatest degree is 3 and there is one term, so 11z3 is a cubic monomial.

47. 9 + y4

SOLUTION:

Find the degree of each term of 9 + y 4.

9 → 1

y4→ 4

The greatest degree is 4 and there are 2 terms, so 9 + y 4 is a quartic binomial.

48. 3x3 – 7

SOLUTION:

Find the degree of each term of 3x3 – 7.

3x3 → 3

–7 → 0

The greatest degree is 3 and there are 2 terms, so 3x3 – 7 is a cubic binomial.

49. –2x5 – x

2 + 5x – 8

SOLUTION:

Find the degree of each term of –2x5 – x2 + 5x – 8.

–2x5 → 5

–x2 → 2

5x → 1 – 8 → 0

The greatest degree is 5 and there are 4 terms, so –2x5 – x2 + 5x – 8 is a quintic polynomial.

50. 10t – 4t2 + 6t

3

SOLUTION:

Find the degree of each term of 10t – 4t2 + 6t

3.

10t → 1

4t2 → 2

6t3 → 3

The greatest degree is 3 and there are 3 terms, so 10t – 4t2 + 6t

3 is a cubic trinomial.

51. ENROLLMENT In a rapidly growing school system, the numbers (in hundreds) of total students N and

K-5 students P enrolled from 2000 to 2009 are modeled by the equations N = 1.25t2 – t + 7.5 and P =

0.7t2 – 0.95t + 3.8, where t is the number of years since 2000.

a. Write an equation modeling the number of 6-12 students S enrolled for this time period. b. How many 6-12 students were enrolled in the school system in 2007?

SOLUTION: a. To write an equation that represents the number of 6-12 students enrolled, subtract the equations that represent the total number of students and the number of K-5 students.

b. Replace t with 7 in the equation for S to determine the number of students enrolled in 6-12 in 2007.

In 2007 there were 30.3 hundreds or 3030 students enrolled in 6-12.

52. CCSS REASONING The perimeter of the figure shown is represented by the expression 3x2 − 7x + 2. Write a

polynomial that represents the measure of the third side.

SOLUTION:

53. GEOMETRY Consider the rectangle.

a. What does (4x2 + 2x – 1)(2x

2 – x + 3) represent?

b. What does 2(4x2 + 2x – 1) + 2(2x

2 – x + 3) represent?

SOLUTION:

a. (4x2 + 2x – 1)(2x

2 – x + 3) is a multiplication of the length and the width of the rectangle, which is the formula for

the area of a rectangle.

b. 2(4x2 + 2x – 1) + 2(2x

2 – x + 3) is the sum of twice the length and twice the width of a rectangle, which is the

formula for the perimeter of the rectangle

Find each sum or difference.

54. (4x + 2y − 6z) + (5y − 2z + 7x) + (−9z − 2x − 3y)

SOLUTION:

55. (5a2 − 4) + (a

2 − 2a + 12) + (4a2 − 6a + 8)

SOLUTION:

56. (3c2 − 7) + (4c + 7) − (c

2 + 5c − 8)

SOLUTION:

57. (3n3 + 3n − 10) − (4n

2 − 5n) + (4n3 − 3n

2 − 9n + 4)

SOLUTION:

58. FOOTBALL The National Football League is divided into two conferences, the American A and the National N. From 2002 through 2009, the total attendance T (in thousands) for both conferences and for the American Conference games are modeled by the following equations, where x is the number of years since 2002.

T = –0.69x3 + 55.83x

2 + 643.31x + 10,538

A = –3.78x3 + 58.96x

2 + 265.96x + 5257

Determine how many people attended National Conference football games in 2009.

SOLUTION:

Let x = 7 represent 2009, then find how many people attended a National Conference football game in 2009.

In 2009 the number of people who attended National Conference football games was about 8829 thousand. Multiply by 1000 to find the attendance in standard form. 8829 ×1000 = 8,829,000. So, about 8,829,000 people attended National Conference football games in 2009.

59. CAR RENTAL The cost to rent a car for a day is $15 plus $0.15 for each mile driven. a. Write a polynomial that represents the cost of renting a car for m miles. b. If a car is driven 145 miles, how much would it cost to rent? c. If a car is driven 105 miles each day for four days, how much would it cost to rent a car? d. If a car is driven 220 miles each day for seven days, how much would it cost to rent a car?

SOLUTION: a. The cost to rent a car is the daily rate and the mileage cost or 15 + 0.15m. b. Substitute 145 for m to find the daily cost to drive 145 miles.

The cost to rent the car would be $36.75. c. The expression represents the cost per day. For 4 days, multiply the entire expression by 4 to find the cost for a 4 day trip. Substitute 105 for m.

The cost to rent the car would be $123. d. The expression represents the cost per day. For 7 days, multiply the entire expression by 7 to find the cost for a 7 day trip. Substitute 220 for m.

The cost to rent the car would be $336.

60. MULTIPLE REPRESENTATIONS In this problem, you will explore perimeter and area. a. Geometric Draw three rectangles that each have a perimeter of 400 feet. b. Tabular Record the width and length of each rectangle in a table like the one shown below. Find the area of eachrectangle.

c. Graphical On a coordinate system, graph the area of rectangle 4 in terms of the length, x. Use the graph to determine the largest area possible. d. Analytical Determine the length and width that produce the largest area.

SOLUTION: a. For the perimeters to be 400, create lengths and widths that sum to 200. Note that some lengths are already provided in the table in part b.

b. The area is length multiplied by width. Make sure the units are squared for area. The sum of the length and width must be 200, so if the length is x, the width must be 200 – x.

c. The length will be the x-values on the horizontal and the area will be the y-values on the vertical. The length cannot pass 200 since the sum of the length and width is 200. Set the intervals for the x-axis to 25 feet. Extend the table of values to find more points to plot on the graph. It appears that 10,000 is the greatest area, so set the intervals

for the y-axis to 1000 ft2. After graphing, it appears that the highest point on the graph is at an area of 10,000 ft

2.

d. The associated x-value with the maximum area is x = 100, so the length must be 100 and the width must be 200 – 100, or 100. The length and width of the rectangle must be 100 feet each to have the largest area.

61. CCSS CRITIQUE Cheyenne and Sebastian are finding (2x2 − x) − (3x + 3x

2 − 2). Is either of them correct? Explain your reasoning.

SOLUTION:

Neither is correct. Cheyenne, did not distribute the negative to the 2nd and 3rd terms when she found the additive inverse. Sebastian did not distribute the negate to the 3rd terms when he found the additive inverse. To find the additive inverse, all terms should be multiplied by −1.

62. REASONING Determine whether each of the following statements is true or false . Explain your reasoning. a. A binomial can have a degree of zero. b. The order in which polynomials are subtracted does not matter.

SOLUTION: a. If a binomial has two terms that are each a degree of 0, then those terms can be combined and the binomial becomes a monomial. For example, 18 + 7 = 25. If one of the terms of the binomial does not have a degree of 0, then the binomial cannot have a degree of 0, since the degree of a polynomial is the greatest degree of any term in the polynomial. b. Subtraction is not commutative. While 2 + 5 = 5 + 2, 2 – 5 ≠ 5 – 2. This is also true for polynomials. Sample answer: (2x – 3) – (4x – 3) = –2x, but (4x – 3) – (2x – 3) = 2x

63. CHALLENGE Write a polynomial that represents the sum of an odd integer 2n + 1 and the next two consecutive odd integers.

SOLUTION:

64. WRITING IN MATH Why would you add or subtract equations that represent real-world situations? Explain.

SOLUTION:

65. WRITING IN MATH Describe how to add and subtract polynomials using both the vertical and horizontal formats.

SOLUTION: To add polynomials in a horizontal format, you combine like terms. For the vertical format, you write the polynomials in standard form, align like terms in columns, and combine like terms.

To subtract polynomials in a horizontal format you find the additive inverse of the polynomial you are subtracting, andthen combine like terms. For the vertical format, you write the polynomials in standard form, align like terms in columns, and subtract by adding the additive inverse.

66. Three consecutive integers can be represented by x, x + 1, and x + 2. What is the sum of these three integers? A x(x + 1)(x + 2)

B x3 + 3

C 3x + 3 D x + 3

SOLUTION:

The correct choice is C.

67. SHORT RESPONSE What is the perimeter of a square with sides that measure 2x + 3 units?

SOLUTION:

The perimeter of the square is 8x + 12 units.

68. Jim cuts a board in the shape of a regular hexagon and pounds in a nail at each vertex, as shown. How many rubber bands will he need to stretch a rubber band across every possible pair of nails?

F 15 G 14 H 12 J 9

SOLUTION: The first nail would connect to 5 others, the second to 4 others, the third to 3 others, etc.5 + 4 + 3 + 2 + 1 = 15 The correct choice is F.

69. Which ordered pair is in the solution set of the system of inequalities shown in the graph?

A (−3, 0) B (0, −3) C (5, 0) D (0, 5)

SOLUTION: Choice A is outside the shaded area for both inequalities. Choices B and D are inside the shaded area for only one inequality. Choice C is the only point in the solution for both inequalities. So, the correct choice is C.

70. COMPUTERS A computer technician charges by the hour to fix and repair computer equipment. The total cost of the technician for one hour is $75, for two hours is $125, for three hours is $175, for four hours is $225, and so on. Write a recursive formula for the sequence.

SOLUTION: Write out the terms. $75, $125, $175, $225, ... The first term is 75, and 50 is added to form each following term. Therefore, we have a1 = 75, an = an – 1 + 50, n ≥ 2.

Determine whether each sequence is arithmetic, geometric, or neither. Explain.71. 8, –32, 128, –512, ...

SOLUTION: Check for a common difference. –32 – 8 = –40 128 – (–32) = 160 There is no common difference. Check for a common ratio. –32 ÷ 8 = –4 128 ÷ (–32) = –4 Geometric; the common ratio is –4.

72. 25, 8, –9, –26, ...

SOLUTION: Check for a common difference. 8 – 25 = –17 –9 – 8 = –17 Arithmetic; the common difference is –17.

73.

SOLUTION: Check for a common difference.

There is no common difference. Check for a common ratio.

There is no common ratio, so the sequence is not arithmetic or geometric.

74. 43, 52, 61, 70, ...

SOLUTION: Check for a common difference. 52 – 43 = 9 61 – 52 = 9 Arithmetic; the common difference is 9.

75. –27, –16, –5, 6, ...

SOLUTION: Check for a common difference. –16 – (–27) = 11 –5 – (–16) = 11 Arithmetic; the common difference is 11.

76. 200, 100, 50, 25, …

SOLUTION: Check for a common difference. 100 – 200 = –100 50 – 100 = –50 There is no common difference. Check for a common ratio. 100 ÷ 200 = 0.5 50 ÷ 100 = 0.5

Geometric; the common ratio is 0.5 or .

77. JOBS Kimi received an offer for a new job. She wants to compare the offer with her current job. What is total amount of sales that Kimi must get each month to make the same income at either job?

SOLUTION: Let s be Kimi’s monthly sales.

Kimi must sell $80,000 each month to make the same income at either job.

Determine whether each sequence is an arithmetic sequence. If it is, state the common difference.78. 24, 16, 8, 0, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is –8, because 16 – 24 = –8; 8 –16 = –8; etc.

79. , 13, 26, …

SOLUTION: Find the difference between the terms.

– =

13 – =

26 – 13 = 13 There is not a common different. The sequence is not an arithmetic sequence

80. 7, 6, 5, 4, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is –1, because 6 – 7 = –1; 5 – 6 = –1; 4 – 5 = –1;etc.

81. 10, 12, 15, 18, …

SOLUTION: Find the difference between the terms. 12 – 10 = 2 15 – 12 = 3 18 – 15 = 3 There is no common difference. The sequence is not an arithmetic sequence.

82. −15, −11, −7, −3, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is 4, because –11 – (–15) = 4; –7 – (–11) = 4; –3 – (–7) = 4;etc.

83. −0.3, 0.2, 0.7, 1.2, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is 0.5, because 0.2 – (–0.3) = 0.5; 0.7 – 0.2 = 0.5; 1.2 – 0.7 = 0.5; etc.

Simplify.

84. t(t5)(t

7)

SOLUTION:

85. n3(n

2)(−2n

3)

SOLUTION:

86. (5t5v

2)(10t

3v

4)

SOLUTION:

87. (−8u4z

5)(5uz

4)

SOLUTION:

88. [(3)2]3

SOLUTION:

89. [(2)3]2

SOLUTION:

90. (2m4k

3)2(−3mk

2)3

SOLUTION:

91. (6xy2)2(2x

2y

2z

2)3

SOLUTION:

eSolutions Manual - Powered by Cognero Page 24

8-1 Adding and Subtracting Polynomials

Determine whether each expression is a polynomial. If it is a polynomial, find the degree and determine whether it is a monomial, binomial, or trinomial.

1. 7ab + 6b2 – 2a

3

SOLUTION:

A polynomial is a monomial or the sum of monomials. 7ab + 6b2 – 2a

3 is the sum of 3 monomials, so it is a

polynomial. The degree of a polynomial is the greatest degree of any term in the polynomial.The degree of each term is 2, 2, and

3, so the degree of 7ab + 6b2 – 2a

3 is 3. The polynomial has three terms, so it is a trinomial.

2. 2y – 5 + 3y2

SOLUTION:

2y – 5 + 3y2

is the sum of monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 1, 0, and

2, so the degree of 2y – 5 + 3y2 is 2. The polynomial has three terms, so it is a trinomial.

3. 3x2

SOLUTION:

A polynomial is a monomial or the sum of monomials, so 3x2 is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of 3x2 is 2. The

polynomial has one term, so it is a monomial.

4.

SOLUTION: A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer exponents. It has

only one term. is a division of two monomials, so it is not a monomial.

5. 5m2p

3 + 6

SOLUTION:

A polynomial is a monomial or the sum of monomials. 5m2p

3 + 6

is the sum of 2 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 5 and 0,

so the degree of 5m2p

3 + 6 is 5. The polynomial has two terms, so it is a binomial.

6. 5q–4

+ 6q

SOLUTION: A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer exponents. It has

only one term. 5q-4 is equal to , which is a division of two monomials, so it is not a monomial.

Write each polynomial in standard form. Identify the leading coefficient.

7. –4d4 + 1 – d

2

SOLUTION: Find the degree of each term.

–4d4→ 4

1 → 0

–d2 → 2

The greatest degree is 4, from the term –4d4, so the leading coefficient of –4d

4 + 1 – d

2 is –4.

Rewrite the polynomial with each monomial in descending order according to degree.

–4d4 – d

2 + 1

8. 2x5 – 12 + 3x

SOLUTION: Find the degree of each term.

2x5 → 5

–12 → 0 3x → 1

The greatest degree is 5, from the term 2x5, so the leading coefficient of 2x

5 – 12 + 3x is 2.

Rewrite the polynomial with each monomial in descending order according to degree.

2x5 + 3x – 12

9. 4z – 2z2 – 5z

4

SOLUTION: Find the degree of each term. 4z → 1

– 2z2 → 2

– 5z4→ 4

The greatest degree is 4, from the term – 5z4, so the leading coefficient of 4z – 2z

2 – 5z

4 is –5.

Rewrite the polynomial with each monomial in descending order according to degree.

–5z4 – 2z

2 + 4z

10. 2a + 4a3 – 5a

2 – 1

SOLUTION: Find the degree of each term. 2a → 1

4a3→ 3

– 5a2 → 2

– 1 → 0

The greatest degree is 3, from the term 4a3, so the leading coefficient of 2a + 4a

3 – 5a

2 – 1 is 4.

Rewrite the polynomial with each monomial in descending order according to degree.

4a3– 5a

2 + 2a – 1

Find each sum or difference.

11. (6x3 − 4) + (−2x

3 + 9)

SOLUTION:

12. (g3 − 2g2 + 5g + 6) − (g2 + 2g)

SOLUTION:

13. (4 + 2a2 − 2a) − (3a

2 − 8a + 7)

SOLUTION:

14. (8y − 4y2) + (3y − 9y

2)

SOLUTION:

15. (−4z3 − 2z + 8) − (4z

3 + 3z2 − 5)

SOLUTION:

16. (−3d2 − 8 + 2d) + (4d − 12 + d2

)

SOLUTION:

17. (y + 5) + (2y + 4y2 – 2)

SOLUTION:

18. (3n3 − 5n + n2

) − (−8n2 + 3n

3)

SOLUTION:

19. CCSS SENSE-MAKING The total number of students T who traveled for spring break consists of two groups: students who flew to their destinations F and students who drove to their destination D. The number (in thousands) of students who flew and the total number of students who flew or drove can be modeled by the following equations,where n is the number of years since 1995. T = 14n + 21

F = 8n + 7 a. Write an equation that models the number of students who drove to their destination for this time period. b. Predict the number of students who will drive to their destination in 2012. c. How many students will drive or fly to their destination in 2015?

SOLUTION: a.

D = 6n + 14 b. n = 2012 – 1995 = 17

The number of students who will drive to their destination in 2012 is 116,000 students. c. n = 2015 – 1995 = 20

The number of students who will drive or fly to their destination in 2015 is 301,000 students.

Determine whether each expression is a polynomial. If it is a polynomial, find the degree and determine whether it is a monomial, binomial, or trinomial.

20.

SOLUTION: A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer exponents. It has

only one term. is a division of two monomials, so it is not a monomial.

21.

SOLUTION: A polynomial is a monomial or the sum of monomials. 21 is a monomial, so it is also a polynomial. The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of 21 is 0. The polynomial has only one term, so it is a monomial.

22. c4 – 2c

2 + 1

SOLUTION:

A polynomial is a monomial or the sum of monomials. c4 – 2c

2 + 1

is the sum of 3 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 4, 2, and

0, so the degree of c4 – 2c2 + 1 is 4. The polynomial has three terms, so it is a trinomial.

23. d + 3dc

SOLUTION: A polynomial is a monomial or the sum of monomials. A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer

exponents. 3dc

has a variable in the exponent, so it is not a monomial.

24. a – a2

SOLUTION:

A polynomial is a monomial or the sum of monomials. a – a2 is the sum of 2 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 1 and 2,

so the degree of a – a2 is 2. The polynomial has two terms, so it is a binomial.

25. 5n3 + nq

3

SOLUTION:

A polynomial is a monomial or the sum of monomials. 5n3 + nq3 is the sum of 2 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 3 and 4,

so the degree of 5n3 + nq3 is 4. The polynomial has two terms, so it is a binomial.

Write each polynomial in standard form. Identify the leading coefficient.

26. 5x2 – 2 + 3x

SOLUTION: Find the degree of each term.

5x2 → 2

– 2 → 0

3x → 1

The greatest degree is 2, from the term 5x2, so the leading coefficient of 5x

2 – 2 + 3x is 5.

Rewrite the polynomial with each monomial in descending order according to degree.

5x2 + 3x – 2

27. 8y + 7y3

SOLUTION: Find the degree of each term.

7y3 → 3

8y → 1

The greatest degree is 3, from the term 7y3, so the leading coefficient of 8y + 7y

3 is 7.

Rewrite the polynomial with each monomial in descending order according to degree.

7y3 + 8y

28. 4 – 3c – 5c2

SOLUTION: Find the degree of each term. 4 → 0 3c → 1

– 5c2 → 2

The greatest degree is 2, from the term – 5c2, so the leading coefficient of 4 – 3c – 5c

2 is –5.

Rewrite the polynomial with each monomial in descending order according to degree.

–5c2 – 3c + 4

29. –y3 + 3y – 3y

2 + 2

SOLUTION: Find the degree of each term.

–y3 → 3

3y → 1

3y2 → 2

2 → 0

The greatest degree is 3, from the term –y3, so the leading coefficient of –y

3 + 3y – 3y

2 + 2 is –1.

Rewrite the polynomial with each monomial in descending order according to degree. –y 3 – 3y 2 + 3y + 2

30. 11t + 2t2 – 3 + t

5

SOLUTION: Find the degree of each term. 11t → 1

2t2 → 2

–3 → 0

t5 → 5

The greatest degree is 5, from the term t5, so the leading coefficient of 11t + 2t2 – 3 + t

5 is 1.

Rewrite the polynomial with each monomial in descending order according to degree.

t5 + 2t

2 + 11t – 3

31. 2 + r – r3

SOLUTION: Find the degree of each term. 2 → 0 r → 1

– r3→ 3

The greatest degree is 3, from the term – r3, so the leading coefficient of 2 + r – r3

is –1. Rewrite the polynomial with each monomial in descending order according to degree.

–r3 + r + 2

32.

SOLUTION: Find the degree of each term.

→ 0

–3x4 → 4

7 → 0

The greatest degree is 4, from the term –3x4, so the leading coefficient of is –3.

Rewrite the polynomial with each monomial in descending order according to degree.

33. –9b2 + 10b – b

6

SOLUTION: Find the degree of each term.

–9b2 → 2

10b → 1

–b6→ 6

The greatest degree is 6, from the term –b6, so the leading coefficient of –9b2 + 10b – b6 is –1. Rewrite the polynomial with each monomial in descending order according to degree.

–b6 – 9b

2 + 10b

Find each sum or difference.

34. (2c2 + 6c + 4) + (5c – 7)

SOLUTION:

35. (2x + 3x2) − (7 − 8x

2)

SOLUTION:

36. (3c3 − c + 11) − (c2 + 2c + 8)

SOLUTION:

37. (z2 + z) + (z

2 − 11)

SOLUTION:

38. (2x − 2y + 1) − (3y + 4x)

SOLUTION:

39. (4a − 5b2 + 3) + (6 − 2a + 3b

2)

SOLUTION:

40. (x2y − 3x

2 + y) + (3y − 2x2y)

SOLUTION:

41. (−8xy + 3x2 − 5y) + (4x

2 − 2y + 6xy)

SOLUTION:

42. (5n − 2p2 + 2np) − (4p

2 + 4n)

SOLUTION:

43. (4rxt − 8r2x + x2

) − (6rx2 + 5rxt − 2x

2)

SOLUTION:

44. PETS From 1999 through 2009, the number of dogs D and the number of cats C (in hundreds) adopted from animal

shelters in the United States are modeled by the equations D = 2n + 3 and C = n + 4, where n is the number of yearssince 1999. a. Write an equation that models the total number T of dogs and cats adopted in hundreds for this time period. b. If this trend continues, how many dogs and cats will be adopted in 2013?

SOLUTION: a.

So, an equation that models the total number of dogs and cats adopted is T = 3n + 7. b. Evaluate the equation for the total number of dogs and cats for n = 2013 – 1999 = 14.

The number of cats and dogs adopted in 2013 will be 49 × 100 or 4900 cats and dogs.

Classify each polynomial according to its degree and number of terms.

45. 4x – 3x2 + 5

SOLUTION:

Find the degree of each term of 4x – 3x2 + 5.

4x → 1

– 3x2 → 2

5 → 0

The greatest degree is 2 and there are 3 terms, so 4x – 3x2 + 5 is a quadratic trinomial.

46. 11z3

SOLUTION: Find the degree of each term. 11z3

→ 3

The greatest degree is 3 and there is one term, so 11z3 is a cubic monomial.

47. 9 + y4

SOLUTION:

Find the degree of each term of 9 + y 4.

9 → 1

y4→ 4

The greatest degree is 4 and there are 2 terms, so 9 + y 4 is a quartic binomial.

48. 3x3 – 7

SOLUTION:

Find the degree of each term of 3x3 – 7.

3x3 → 3

–7 → 0

The greatest degree is 3 and there are 2 terms, so 3x3 – 7 is a cubic binomial.

49. –2x5 – x

2 + 5x – 8

SOLUTION:

Find the degree of each term of –2x5 – x2 + 5x – 8.

–2x5 → 5

–x2 → 2

5x → 1 – 8 → 0

The greatest degree is 5 and there are 4 terms, so –2x5 – x2 + 5x – 8 is a quintic polynomial.

50. 10t – 4t2 + 6t

3

SOLUTION:

Find the degree of each term of 10t – 4t2 + 6t

3.

10t → 1

4t2 → 2

6t3 → 3

The greatest degree is 3 and there are 3 terms, so 10t – 4t2 + 6t

3 is a cubic trinomial.

51. ENROLLMENT In a rapidly growing school system, the numbers (in hundreds) of total students N and

K-5 students P enrolled from 2000 to 2009 are modeled by the equations N = 1.25t2 – t + 7.5 and P =

0.7t2 – 0.95t + 3.8, where t is the number of years since 2000.

a. Write an equation modeling the number of 6-12 students S enrolled for this time period. b. How many 6-12 students were enrolled in the school system in 2007?

SOLUTION: a. To write an equation that represents the number of 6-12 students enrolled, subtract the equations that represent the total number of students and the number of K-5 students.

b. Replace t with 7 in the equation for S to determine the number of students enrolled in 6-12 in 2007.

In 2007 there were 30.3 hundreds or 3030 students enrolled in 6-12.

52. CCSS REASONING The perimeter of the figure shown is represented by the expression 3x2 − 7x + 2. Write a

polynomial that represents the measure of the third side.

SOLUTION:

53. GEOMETRY Consider the rectangle.

a. What does (4x2 + 2x – 1)(2x

2 – x + 3) represent?

b. What does 2(4x2 + 2x – 1) + 2(2x

2 – x + 3) represent?

SOLUTION:

a. (4x2 + 2x – 1)(2x

2 – x + 3) is a multiplication of the length and the width of the rectangle, which is the formula for

the area of a rectangle.

b. 2(4x2 + 2x – 1) + 2(2x

2 – x + 3) is the sum of twice the length and twice the width of a rectangle, which is the

formula for the perimeter of the rectangle

Find each sum or difference.

54. (4x + 2y − 6z) + (5y − 2z + 7x) + (−9z − 2x − 3y)

SOLUTION:

55. (5a2 − 4) + (a

2 − 2a + 12) + (4a2 − 6a + 8)

SOLUTION:

56. (3c2 − 7) + (4c + 7) − (c

2 + 5c − 8)

SOLUTION:

57. (3n3 + 3n − 10) − (4n

2 − 5n) + (4n3 − 3n

2 − 9n + 4)

SOLUTION:

58. FOOTBALL The National Football League is divided into two conferences, the American A and the National N. From 2002 through 2009, the total attendance T (in thousands) for both conferences and for the American Conference games are modeled by the following equations, where x is the number of years since 2002.

T = –0.69x3 + 55.83x

2 + 643.31x + 10,538

A = –3.78x3 + 58.96x

2 + 265.96x + 5257

Determine how many people attended National Conference football games in 2009.

SOLUTION:

Let x = 7 represent 2009, then find how many people attended a National Conference football game in 2009.

In 2009 the number of people who attended National Conference football games was about 8829 thousand. Multiply by 1000 to find the attendance in standard form. 8829 ×1000 = 8,829,000. So, about 8,829,000 people attended National Conference football games in 2009.

59. CAR RENTAL The cost to rent a car for a day is $15 plus $0.15 for each mile driven. a. Write a polynomial that represents the cost of renting a car for m miles. b. If a car is driven 145 miles, how much would it cost to rent? c. If a car is driven 105 miles each day for four days, how much would it cost to rent a car? d. If a car is driven 220 miles each day for seven days, how much would it cost to rent a car?

SOLUTION: a. The cost to rent a car is the daily rate and the mileage cost or 15 + 0.15m. b. Substitute 145 for m to find the daily cost to drive 145 miles.

The cost to rent the car would be $36.75. c. The expression represents the cost per day. For 4 days, multiply the entire expression by 4 to find the cost for a 4 day trip. Substitute 105 for m.

The cost to rent the car would be $123. d. The expression represents the cost per day. For 7 days, multiply the entire expression by 7 to find the cost for a 7 day trip. Substitute 220 for m.

The cost to rent the car would be $336.

60. MULTIPLE REPRESENTATIONS In this problem, you will explore perimeter and area. a. Geometric Draw three rectangles that each have a perimeter of 400 feet. b. Tabular Record the width and length of each rectangle in a table like the one shown below. Find the area of eachrectangle.

c. Graphical On a coordinate system, graph the area of rectangle 4 in terms of the length, x. Use the graph to determine the largest area possible. d. Analytical Determine the length and width that produce the largest area.

SOLUTION: a. For the perimeters to be 400, create lengths and widths that sum to 200. Note that some lengths are already provided in the table in part b.

b. The area is length multiplied by width. Make sure the units are squared for area. The sum of the length and width must be 200, so if the length is x, the width must be 200 – x.

c. The length will be the x-values on the horizontal and the area will be the y-values on the vertical. The length cannot pass 200 since the sum of the length and width is 200. Set the intervals for the x-axis to 25 feet. Extend the table of values to find more points to plot on the graph. It appears that 10,000 is the greatest area, so set the intervals

for the y-axis to 1000 ft2. After graphing, it appears that the highest point on the graph is at an area of 10,000 ft

2.

d. The associated x-value with the maximum area is x = 100, so the length must be 100 and the width must be 200 – 100, or 100. The length and width of the rectangle must be 100 feet each to have the largest area.

61. CCSS CRITIQUE Cheyenne and Sebastian are finding (2x2 − x) − (3x + 3x

2 − 2). Is either of them correct? Explain your reasoning.

SOLUTION:

Neither is correct. Cheyenne, did not distribute the negative to the 2nd and 3rd terms when she found the additive inverse. Sebastian did not distribute the negate to the 3rd terms when he found the additive inverse. To find the additive inverse, all terms should be multiplied by −1.

62. REASONING Determine whether each of the following statements is true or false . Explain your reasoning. a. A binomial can have a degree of zero. b. The order in which polynomials are subtracted does not matter.

SOLUTION: a. If a binomial has two terms that are each a degree of 0, then those terms can be combined and the binomial becomes a monomial. For example, 18 + 7 = 25. If one of the terms of the binomial does not have a degree of 0, then the binomial cannot have a degree of 0, since the degree of a polynomial is the greatest degree of any term in the polynomial. b. Subtraction is not commutative. While 2 + 5 = 5 + 2, 2 – 5 ≠ 5 – 2. This is also true for polynomials. Sample answer: (2x – 3) – (4x – 3) = –2x, but (4x – 3) – (2x – 3) = 2x

63. CHALLENGE Write a polynomial that represents the sum of an odd integer 2n + 1 and the next two consecutive odd integers.

SOLUTION:

64. WRITING IN MATH Why would you add or subtract equations that represent real-world situations? Explain.

SOLUTION:

65. WRITING IN MATH Describe how to add and subtract polynomials using both the vertical and horizontal formats.

SOLUTION: To add polynomials in a horizontal format, you combine like terms. For the vertical format, you write the polynomials in standard form, align like terms in columns, and combine like terms.

To subtract polynomials in a horizontal format you find the additive inverse of the polynomial you are subtracting, andthen combine like terms. For the vertical format, you write the polynomials in standard form, align like terms in columns, and subtract by adding the additive inverse.

66. Three consecutive integers can be represented by x, x + 1, and x + 2. What is the sum of these three integers? A x(x + 1)(x + 2)

B x3 + 3

C 3x + 3 D x + 3

SOLUTION:

The correct choice is C.

67. SHORT RESPONSE What is the perimeter of a square with sides that measure 2x + 3 units?

SOLUTION:

The perimeter of the square is 8x + 12 units.

68. Jim cuts a board in the shape of a regular hexagon and pounds in a nail at each vertex, as shown. How many rubber bands will he need to stretch a rubber band across every possible pair of nails?

F 15 G 14 H 12 J 9

SOLUTION: The first nail would connect to 5 others, the second to 4 others, the third to 3 others, etc.5 + 4 + 3 + 2 + 1 = 15 The correct choice is F.

69. Which ordered pair is in the solution set of the system of inequalities shown in the graph?

A (−3, 0) B (0, −3) C (5, 0) D (0, 5)

SOLUTION: Choice A is outside the shaded area for both inequalities. Choices B and D are inside the shaded area for only one inequality. Choice C is the only point in the solution for both inequalities. So, the correct choice is C.

70. COMPUTERS A computer technician charges by the hour to fix and repair computer equipment. The total cost of the technician for one hour is $75, for two hours is $125, for three hours is $175, for four hours is $225, and so on. Write a recursive formula for the sequence.

SOLUTION: Write out the terms. $75, $125, $175, $225, ... The first term is 75, and 50 is added to form each following term. Therefore, we have a1 = 75, an = an – 1 + 50, n ≥ 2.

Determine whether each sequence is arithmetic, geometric, or neither. Explain.71. 8, –32, 128, –512, ...

SOLUTION: Check for a common difference. –32 – 8 = –40 128 – (–32) = 160 There is no common difference. Check for a common ratio. –32 ÷ 8 = –4 128 ÷ (–32) = –4 Geometric; the common ratio is –4.

72. 25, 8, –9, –26, ...

SOLUTION: Check for a common difference. 8 – 25 = –17 –9 – 8 = –17 Arithmetic; the common difference is –17.

73.

SOLUTION: Check for a common difference.

There is no common difference. Check for a common ratio.

There is no common ratio, so the sequence is not arithmetic or geometric.

74. 43, 52, 61, 70, ...

SOLUTION: Check for a common difference. 52 – 43 = 9 61 – 52 = 9 Arithmetic; the common difference is 9.

75. –27, –16, –5, 6, ...

SOLUTION: Check for a common difference. –16 – (–27) = 11 –5 – (–16) = 11 Arithmetic; the common difference is 11.

76. 200, 100, 50, 25, …

SOLUTION: Check for a common difference. 100 – 200 = –100 50 – 100 = –50 There is no common difference. Check for a common ratio. 100 ÷ 200 = 0.5 50 ÷ 100 = 0.5

Geometric; the common ratio is 0.5 or .

77. JOBS Kimi received an offer for a new job. She wants to compare the offer with her current job. What is total amount of sales that Kimi must get each month to make the same income at either job?

SOLUTION: Let s be Kimi’s monthly sales.

Kimi must sell $80,000 each month to make the same income at either job.

Determine whether each sequence is an arithmetic sequence. If it is, state the common difference.78. 24, 16, 8, 0, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is –8, because 16 – 24 = –8; 8 –16 = –8; etc.

79. , 13, 26, …

SOLUTION: Find the difference between the terms.

– =

13 – =

26 – 13 = 13 There is not a common different. The sequence is not an arithmetic sequence

80. 7, 6, 5, 4, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is –1, because 6 – 7 = –1; 5 – 6 = –1; 4 – 5 = –1;etc.

81. 10, 12, 15, 18, …

SOLUTION: Find the difference between the terms. 12 – 10 = 2 15 – 12 = 3 18 – 15 = 3 There is no common difference. The sequence is not an arithmetic sequence.

82. −15, −11, −7, −3, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is 4, because –11 – (–15) = 4; –7 – (–11) = 4; –3 – (–7) = 4;etc.

83. −0.3, 0.2, 0.7, 1.2, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is 0.5, because 0.2 – (–0.3) = 0.5; 0.7 – 0.2 = 0.5; 1.2 – 0.7 = 0.5; etc.

Simplify.

84. t(t5)(t

7)

SOLUTION:

85. n3(n

2)(−2n

3)

SOLUTION:

86. (5t5v

2)(10t

3v

4)

SOLUTION:

87. (−8u4z

5)(5uz

4)

SOLUTION:

88. [(3)2]3

SOLUTION:

89. [(2)3]2

SOLUTION:

90. (2m4k

3)2(−3mk

2)3

SOLUTION:

91. (6xy2)2(2x

2y

2z

2)3

SOLUTION:

eSolutions Manual - Powered by Cognero Page 25

8-1 Adding and Subtracting Polynomials

Determine whether each expression is a polynomial. If it is a polynomial, find the degree and determine whether it is a monomial, binomial, or trinomial.

1. 7ab + 6b2 – 2a

3

SOLUTION:

A polynomial is a monomial or the sum of monomials. 7ab + 6b2 – 2a

3 is the sum of 3 monomials, so it is a

polynomial. The degree of a polynomial is the greatest degree of any term in the polynomial.The degree of each term is 2, 2, and

3, so the degree of 7ab + 6b2 – 2a

3 is 3. The polynomial has three terms, so it is a trinomial.

2. 2y – 5 + 3y2

SOLUTION:

2y – 5 + 3y2

is the sum of monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 1, 0, and

2, so the degree of 2y – 5 + 3y2 is 2. The polynomial has three terms, so it is a trinomial.

3. 3x2

SOLUTION:

A polynomial is a monomial or the sum of monomials, so 3x2 is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of 3x2 is 2. The

polynomial has one term, so it is a monomial.

4.

SOLUTION: A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer exponents. It has

only one term. is a division of two monomials, so it is not a monomial.

5. 5m2p

3 + 6

SOLUTION:

A polynomial is a monomial or the sum of monomials. 5m2p

3 + 6

is the sum of 2 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 5 and 0,

so the degree of 5m2p

3 + 6 is 5. The polynomial has two terms, so it is a binomial.

6. 5q–4

+ 6q

SOLUTION: A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer exponents. It has

only one term. 5q-4 is equal to , which is a division of two monomials, so it is not a monomial.

Write each polynomial in standard form. Identify the leading coefficient.

7. –4d4 + 1 – d

2

SOLUTION: Find the degree of each term.

–4d4→ 4

1 → 0

–d2 → 2

The greatest degree is 4, from the term –4d4, so the leading coefficient of –4d

4 + 1 – d

2 is –4.

Rewrite the polynomial with each monomial in descending order according to degree.

–4d4 – d

2 + 1

8. 2x5 – 12 + 3x

SOLUTION: Find the degree of each term.

2x5 → 5

–12 → 0 3x → 1

The greatest degree is 5, from the term 2x5, so the leading coefficient of 2x

5 – 12 + 3x is 2.

Rewrite the polynomial with each monomial in descending order according to degree.

2x5 + 3x – 12

9. 4z – 2z2 – 5z

4

SOLUTION: Find the degree of each term. 4z → 1

– 2z2 → 2

– 5z4→ 4

The greatest degree is 4, from the term – 5z4, so the leading coefficient of 4z – 2z

2 – 5z

4 is –5.

Rewrite the polynomial with each monomial in descending order according to degree.

–5z4 – 2z

2 + 4z

10. 2a + 4a3 – 5a

2 – 1

SOLUTION: Find the degree of each term. 2a → 1

4a3→ 3

– 5a2 → 2

– 1 → 0

The greatest degree is 3, from the term 4a3, so the leading coefficient of 2a + 4a

3 – 5a

2 – 1 is 4.

Rewrite the polynomial with each monomial in descending order according to degree.

4a3– 5a

2 + 2a – 1

Find each sum or difference.

11. (6x3 − 4) + (−2x

3 + 9)

SOLUTION:

12. (g3 − 2g2 + 5g + 6) − (g2 + 2g)

SOLUTION:

13. (4 + 2a2 − 2a) − (3a

2 − 8a + 7)

SOLUTION:

14. (8y − 4y2) + (3y − 9y

2)

SOLUTION:

15. (−4z3 − 2z + 8) − (4z

3 + 3z2 − 5)

SOLUTION:

16. (−3d2 − 8 + 2d) + (4d − 12 + d2

)

SOLUTION:

17. (y + 5) + (2y + 4y2 – 2)

SOLUTION:

18. (3n3 − 5n + n2

) − (−8n2 + 3n

3)

SOLUTION:

19. CCSS SENSE-MAKING The total number of students T who traveled for spring break consists of two groups: students who flew to their destinations F and students who drove to their destination D. The number (in thousands) of students who flew and the total number of students who flew or drove can be modeled by the following equations,where n is the number of years since 1995. T = 14n + 21

F = 8n + 7 a. Write an equation that models the number of students who drove to their destination for this time period. b. Predict the number of students who will drive to their destination in 2012. c. How many students will drive or fly to their destination in 2015?

SOLUTION: a.

D = 6n + 14 b. n = 2012 – 1995 = 17

The number of students who will drive to their destination in 2012 is 116,000 students. c. n = 2015 – 1995 = 20

The number of students who will drive or fly to their destination in 2015 is 301,000 students.

Determine whether each expression is a polynomial. If it is a polynomial, find the degree and determine whether it is a monomial, binomial, or trinomial.

20.

SOLUTION: A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer exponents. It has

only one term. is a division of two monomials, so it is not a monomial.

21.

SOLUTION: A polynomial is a monomial or the sum of monomials. 21 is a monomial, so it is also a polynomial. The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of 21 is 0. The polynomial has only one term, so it is a monomial.

22. c4 – 2c

2 + 1

SOLUTION:

A polynomial is a monomial or the sum of monomials. c4 – 2c

2 + 1

is the sum of 3 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 4, 2, and

0, so the degree of c4 – 2c2 + 1 is 4. The polynomial has three terms, so it is a trinomial.

23. d + 3dc

SOLUTION: A polynomial is a monomial or the sum of monomials. A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer

exponents. 3dc

has a variable in the exponent, so it is not a monomial.

24. a – a2

SOLUTION:

A polynomial is a monomial or the sum of monomials. a – a2 is the sum of 2 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 1 and 2,

so the degree of a – a2 is 2. The polynomial has two terms, so it is a binomial.

25. 5n3 + nq

3

SOLUTION:

A polynomial is a monomial or the sum of monomials. 5n3 + nq3 is the sum of 2 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 3 and 4,

so the degree of 5n3 + nq3 is 4. The polynomial has two terms, so it is a binomial.

Write each polynomial in standard form. Identify the leading coefficient.

26. 5x2 – 2 + 3x

SOLUTION: Find the degree of each term.

5x2 → 2

– 2 → 0

3x → 1

The greatest degree is 2, from the term 5x2, so the leading coefficient of 5x

2 – 2 + 3x is 5.

Rewrite the polynomial with each monomial in descending order according to degree.

5x2 + 3x – 2

27. 8y + 7y3

SOLUTION: Find the degree of each term.

7y3 → 3

8y → 1

The greatest degree is 3, from the term 7y3, so the leading coefficient of 8y + 7y

3 is 7.

Rewrite the polynomial with each monomial in descending order according to degree.

7y3 + 8y

28. 4 – 3c – 5c2

SOLUTION: Find the degree of each term. 4 → 0 3c → 1

– 5c2 → 2

The greatest degree is 2, from the term – 5c2, so the leading coefficient of 4 – 3c – 5c

2 is –5.

Rewrite the polynomial with each monomial in descending order according to degree.

–5c2 – 3c + 4

29. –y3 + 3y – 3y

2 + 2

SOLUTION: Find the degree of each term.

–y3 → 3

3y → 1

3y2 → 2

2 → 0

The greatest degree is 3, from the term –y3, so the leading coefficient of –y

3 + 3y – 3y

2 + 2 is –1.

Rewrite the polynomial with each monomial in descending order according to degree. –y 3 – 3y 2 + 3y + 2

30. 11t + 2t2 – 3 + t

5

SOLUTION: Find the degree of each term. 11t → 1

2t2 → 2

–3 → 0

t5 → 5

The greatest degree is 5, from the term t5, so the leading coefficient of 11t + 2t2 – 3 + t

5 is 1.

Rewrite the polynomial with each monomial in descending order according to degree.

t5 + 2t

2 + 11t – 3

31. 2 + r – r3

SOLUTION: Find the degree of each term. 2 → 0 r → 1

– r3→ 3

The greatest degree is 3, from the term – r3, so the leading coefficient of 2 + r – r3

is –1. Rewrite the polynomial with each monomial in descending order according to degree.

–r3 + r + 2

32.

SOLUTION: Find the degree of each term.

→ 0

–3x4 → 4

7 → 0

The greatest degree is 4, from the term –3x4, so the leading coefficient of is –3.

Rewrite the polynomial with each monomial in descending order according to degree.

33. –9b2 + 10b – b

6

SOLUTION: Find the degree of each term.

–9b2 → 2

10b → 1

–b6→ 6

The greatest degree is 6, from the term –b6, so the leading coefficient of –9b2 + 10b – b6 is –1. Rewrite the polynomial with each monomial in descending order according to degree.

–b6 – 9b

2 + 10b

Find each sum or difference.

34. (2c2 + 6c + 4) + (5c – 7)

SOLUTION:

35. (2x + 3x2) − (7 − 8x

2)

SOLUTION:

36. (3c3 − c + 11) − (c2 + 2c + 8)

SOLUTION:

37. (z2 + z) + (z

2 − 11)

SOLUTION:

38. (2x − 2y + 1) − (3y + 4x)

SOLUTION:

39. (4a − 5b2 + 3) + (6 − 2a + 3b

2)

SOLUTION:

40. (x2y − 3x

2 + y) + (3y − 2x2y)

SOLUTION:

41. (−8xy + 3x2 − 5y) + (4x

2 − 2y + 6xy)

SOLUTION:

42. (5n − 2p2 + 2np) − (4p

2 + 4n)

SOLUTION:

43. (4rxt − 8r2x + x2

) − (6rx2 + 5rxt − 2x

2)

SOLUTION:

44. PETS From 1999 through 2009, the number of dogs D and the number of cats C (in hundreds) adopted from animal

shelters in the United States are modeled by the equations D = 2n + 3 and C = n + 4, where n is the number of yearssince 1999. a. Write an equation that models the total number T of dogs and cats adopted in hundreds for this time period. b. If this trend continues, how many dogs and cats will be adopted in 2013?

SOLUTION: a.

So, an equation that models the total number of dogs and cats adopted is T = 3n + 7. b. Evaluate the equation for the total number of dogs and cats for n = 2013 – 1999 = 14.

The number of cats and dogs adopted in 2013 will be 49 × 100 or 4900 cats and dogs.

Classify each polynomial according to its degree and number of terms.

45. 4x – 3x2 + 5

SOLUTION:

Find the degree of each term of 4x – 3x2 + 5.

4x → 1

– 3x2 → 2

5 → 0

The greatest degree is 2 and there are 3 terms, so 4x – 3x2 + 5 is a quadratic trinomial.

46. 11z3

SOLUTION: Find the degree of each term. 11z3

→ 3

The greatest degree is 3 and there is one term, so 11z3 is a cubic monomial.

47. 9 + y4

SOLUTION:

Find the degree of each term of 9 + y 4.

9 → 1

y4→ 4

The greatest degree is 4 and there are 2 terms, so 9 + y 4 is a quartic binomial.

48. 3x3 – 7

SOLUTION:

Find the degree of each term of 3x3 – 7.

3x3 → 3

–7 → 0

The greatest degree is 3 and there are 2 terms, so 3x3 – 7 is a cubic binomial.

49. –2x5 – x

2 + 5x – 8

SOLUTION:

Find the degree of each term of –2x5 – x2 + 5x – 8.

–2x5 → 5

–x2 → 2

5x → 1 – 8 → 0

The greatest degree is 5 and there are 4 terms, so –2x5 – x2 + 5x – 8 is a quintic polynomial.

50. 10t – 4t2 + 6t

3

SOLUTION:

Find the degree of each term of 10t – 4t2 + 6t

3.

10t → 1

4t2 → 2

6t3 → 3

The greatest degree is 3 and there are 3 terms, so 10t – 4t2 + 6t

3 is a cubic trinomial.

51. ENROLLMENT In a rapidly growing school system, the numbers (in hundreds) of total students N and

K-5 students P enrolled from 2000 to 2009 are modeled by the equations N = 1.25t2 – t + 7.5 and P =

0.7t2 – 0.95t + 3.8, where t is the number of years since 2000.

a. Write an equation modeling the number of 6-12 students S enrolled for this time period. b. How many 6-12 students were enrolled in the school system in 2007?

SOLUTION: a. To write an equation that represents the number of 6-12 students enrolled, subtract the equations that represent the total number of students and the number of K-5 students.

b. Replace t with 7 in the equation for S to determine the number of students enrolled in 6-12 in 2007.

In 2007 there were 30.3 hundreds or 3030 students enrolled in 6-12.

52. CCSS REASONING The perimeter of the figure shown is represented by the expression 3x2 − 7x + 2. Write a

polynomial that represents the measure of the third side.

SOLUTION:

53. GEOMETRY Consider the rectangle.

a. What does (4x2 + 2x – 1)(2x

2 – x + 3) represent?

b. What does 2(4x2 + 2x – 1) + 2(2x

2 – x + 3) represent?

SOLUTION:

a. (4x2 + 2x – 1)(2x

2 – x + 3) is a multiplication of the length and the width of the rectangle, which is the formula for

the area of a rectangle.

b. 2(4x2 + 2x – 1) + 2(2x

2 – x + 3) is the sum of twice the length and twice the width of a rectangle, which is the

formula for the perimeter of the rectangle

Find each sum or difference.

54. (4x + 2y − 6z) + (5y − 2z + 7x) + (−9z − 2x − 3y)

SOLUTION:

55. (5a2 − 4) + (a

2 − 2a + 12) + (4a2 − 6a + 8)

SOLUTION:

56. (3c2 − 7) + (4c + 7) − (c

2 + 5c − 8)

SOLUTION:

57. (3n3 + 3n − 10) − (4n

2 − 5n) + (4n3 − 3n

2 − 9n + 4)

SOLUTION:

58. FOOTBALL The National Football League is divided into two conferences, the American A and the National N. From 2002 through 2009, the total attendance T (in thousands) for both conferences and for the American Conference games are modeled by the following equations, where x is the number of years since 2002.

T = –0.69x3 + 55.83x

2 + 643.31x + 10,538

A = –3.78x3 + 58.96x

2 + 265.96x + 5257

Determine how many people attended National Conference football games in 2009.

SOLUTION:

Let x = 7 represent 2009, then find how many people attended a National Conference football game in 2009.

In 2009 the number of people who attended National Conference football games was about 8829 thousand. Multiply by 1000 to find the attendance in standard form. 8829 ×1000 = 8,829,000. So, about 8,829,000 people attended National Conference football games in 2009.

59. CAR RENTAL The cost to rent a car for a day is $15 plus $0.15 for each mile driven. a. Write a polynomial that represents the cost of renting a car for m miles. b. If a car is driven 145 miles, how much would it cost to rent? c. If a car is driven 105 miles each day for four days, how much would it cost to rent a car? d. If a car is driven 220 miles each day for seven days, how much would it cost to rent a car?

SOLUTION: a. The cost to rent a car is the daily rate and the mileage cost or 15 + 0.15m. b. Substitute 145 for m to find the daily cost to drive 145 miles.

The cost to rent the car would be $36.75. c. The expression represents the cost per day. For 4 days, multiply the entire expression by 4 to find the cost for a 4 day trip. Substitute 105 for m.

The cost to rent the car would be $123. d. The expression represents the cost per day. For 7 days, multiply the entire expression by 7 to find the cost for a 7 day trip. Substitute 220 for m.

The cost to rent the car would be $336.

60. MULTIPLE REPRESENTATIONS In this problem, you will explore perimeter and area. a. Geometric Draw three rectangles that each have a perimeter of 400 feet. b. Tabular Record the width and length of each rectangle in a table like the one shown below. Find the area of eachrectangle.

c. Graphical On a coordinate system, graph the area of rectangle 4 in terms of the length, x. Use the graph to determine the largest area possible. d. Analytical Determine the length and width that produce the largest area.

SOLUTION: a. For the perimeters to be 400, create lengths and widths that sum to 200. Note that some lengths are already provided in the table in part b.

b. The area is length multiplied by width. Make sure the units are squared for area. The sum of the length and width must be 200, so if the length is x, the width must be 200 – x.

c. The length will be the x-values on the horizontal and the area will be the y-values on the vertical. The length cannot pass 200 since the sum of the length and width is 200. Set the intervals for the x-axis to 25 feet. Extend the table of values to find more points to plot on the graph. It appears that 10,000 is the greatest area, so set the intervals

for the y-axis to 1000 ft2. After graphing, it appears that the highest point on the graph is at an area of 10,000 ft

2.

d. The associated x-value with the maximum area is x = 100, so the length must be 100 and the width must be 200 – 100, or 100. The length and width of the rectangle must be 100 feet each to have the largest area.

61. CCSS CRITIQUE Cheyenne and Sebastian are finding (2x2 − x) − (3x + 3x

2 − 2). Is either of them correct? Explain your reasoning.

SOLUTION:

Neither is correct. Cheyenne, did not distribute the negative to the 2nd and 3rd terms when she found the additive inverse. Sebastian did not distribute the negate to the 3rd terms when he found the additive inverse. To find the additive inverse, all terms should be multiplied by −1.

62. REASONING Determine whether each of the following statements is true or false . Explain your reasoning. a. A binomial can have a degree of zero. b. The order in which polynomials are subtracted does not matter.

SOLUTION: a. If a binomial has two terms that are each a degree of 0, then those terms can be combined and the binomial becomes a monomial. For example, 18 + 7 = 25. If one of the terms of the binomial does not have a degree of 0, then the binomial cannot have a degree of 0, since the degree of a polynomial is the greatest degree of any term in the polynomial. b. Subtraction is not commutative. While 2 + 5 = 5 + 2, 2 – 5 ≠ 5 – 2. This is also true for polynomials. Sample answer: (2x – 3) – (4x – 3) = –2x, but (4x – 3) – (2x – 3) = 2x

63. CHALLENGE Write a polynomial that represents the sum of an odd integer 2n + 1 and the next two consecutive odd integers.

SOLUTION:

64. WRITING IN MATH Why would you add or subtract equations that represent real-world situations? Explain.

SOLUTION:

65. WRITING IN MATH Describe how to add and subtract polynomials using both the vertical and horizontal formats.

SOLUTION: To add polynomials in a horizontal format, you combine like terms. For the vertical format, you write the polynomials in standard form, align like terms in columns, and combine like terms.

To subtract polynomials in a horizontal format you find the additive inverse of the polynomial you are subtracting, andthen combine like terms. For the vertical format, you write the polynomials in standard form, align like terms in columns, and subtract by adding the additive inverse.

66. Three consecutive integers can be represented by x, x + 1, and x + 2. What is the sum of these three integers? A x(x + 1)(x + 2)

B x3 + 3

C 3x + 3 D x + 3

SOLUTION:

The correct choice is C.

67. SHORT RESPONSE What is the perimeter of a square with sides that measure 2x + 3 units?

SOLUTION:

The perimeter of the square is 8x + 12 units.

68. Jim cuts a board in the shape of a regular hexagon and pounds in a nail at each vertex, as shown. How many rubber bands will he need to stretch a rubber band across every possible pair of nails?

F 15 G 14 H 12 J 9

SOLUTION: The first nail would connect to 5 others, the second to 4 others, the third to 3 others, etc.5 + 4 + 3 + 2 + 1 = 15 The correct choice is F.

69. Which ordered pair is in the solution set of the system of inequalities shown in the graph?

A (−3, 0) B (0, −3) C (5, 0) D (0, 5)

SOLUTION: Choice A is outside the shaded area for both inequalities. Choices B and D are inside the shaded area for only one inequality. Choice C is the only point in the solution for both inequalities. So, the correct choice is C.

70. COMPUTERS A computer technician charges by the hour to fix and repair computer equipment. The total cost of the technician for one hour is $75, for two hours is $125, for three hours is $175, for four hours is $225, and so on. Write a recursive formula for the sequence.

SOLUTION: Write out the terms. $75, $125, $175, $225, ... The first term is 75, and 50 is added to form each following term. Therefore, we have a1 = 75, an = an – 1 + 50, n ≥ 2.

Determine whether each sequence is arithmetic, geometric, or neither. Explain.71. 8, –32, 128, –512, ...

SOLUTION: Check for a common difference. –32 – 8 = –40 128 – (–32) = 160 There is no common difference. Check for a common ratio. –32 ÷ 8 = –4 128 ÷ (–32) = –4 Geometric; the common ratio is –4.

72. 25, 8, –9, –26, ...

SOLUTION: Check for a common difference. 8 – 25 = –17 –9 – 8 = –17 Arithmetic; the common difference is –17.

73.

SOLUTION: Check for a common difference.

There is no common difference. Check for a common ratio.

There is no common ratio, so the sequence is not arithmetic or geometric.

74. 43, 52, 61, 70, ...

SOLUTION: Check for a common difference. 52 – 43 = 9 61 – 52 = 9 Arithmetic; the common difference is 9.

75. –27, –16, –5, 6, ...

SOLUTION: Check for a common difference. –16 – (–27) = 11 –5 – (–16) = 11 Arithmetic; the common difference is 11.

76. 200, 100, 50, 25, …

SOLUTION: Check for a common difference. 100 – 200 = –100 50 – 100 = –50 There is no common difference. Check for a common ratio. 100 ÷ 200 = 0.5 50 ÷ 100 = 0.5

Geometric; the common ratio is 0.5 or .

77. JOBS Kimi received an offer for a new job. She wants to compare the offer with her current job. What is total amount of sales that Kimi must get each month to make the same income at either job?

SOLUTION: Let s be Kimi’s monthly sales.

Kimi must sell $80,000 each month to make the same income at either job.

Determine whether each sequence is an arithmetic sequence. If it is, state the common difference.78. 24, 16, 8, 0, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is –8, because 16 – 24 = –8; 8 –16 = –8; etc.

79. , 13, 26, …

SOLUTION: Find the difference between the terms.

– =

13 – =

26 – 13 = 13 There is not a common different. The sequence is not an arithmetic sequence

80. 7, 6, 5, 4, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is –1, because 6 – 7 = –1; 5 – 6 = –1; 4 – 5 = –1;etc.

81. 10, 12, 15, 18, …

SOLUTION: Find the difference between the terms. 12 – 10 = 2 15 – 12 = 3 18 – 15 = 3 There is no common difference. The sequence is not an arithmetic sequence.

82. −15, −11, −7, −3, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is 4, because –11 – (–15) = 4; –7 – (–11) = 4; –3 – (–7) = 4;etc.

83. −0.3, 0.2, 0.7, 1.2, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is 0.5, because 0.2 – (–0.3) = 0.5; 0.7 – 0.2 = 0.5; 1.2 – 0.7 = 0.5; etc.

Simplify.

84. t(t5)(t

7)

SOLUTION:

85. n3(n

2)(−2n

3)

SOLUTION:

86. (5t5v

2)(10t

3v

4)

SOLUTION:

87. (−8u4z

5)(5uz

4)

SOLUTION:

88. [(3)2]3

SOLUTION:

89. [(2)3]2

SOLUTION:

90. (2m4k

3)2(−3mk

2)3

SOLUTION:

91. (6xy2)2(2x

2y

2z

2)3

SOLUTION:

eSolutions Manual - Powered by Cognero Page 26

8-1 Adding and Subtracting Polynomials

Determine whether each expression is a polynomial. If it is a polynomial, find the degree and determine whether it is a monomial, binomial, or trinomial.

1. 7ab + 6b2 – 2a

3

SOLUTION:

A polynomial is a monomial or the sum of monomials. 7ab + 6b2 – 2a

3 is the sum of 3 monomials, so it is a

polynomial. The degree of a polynomial is the greatest degree of any term in the polynomial.The degree of each term is 2, 2, and

3, so the degree of 7ab + 6b2 – 2a

3 is 3. The polynomial has three terms, so it is a trinomial.

2. 2y – 5 + 3y2

SOLUTION:

2y – 5 + 3y2

is the sum of monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 1, 0, and

2, so the degree of 2y – 5 + 3y2 is 2. The polynomial has three terms, so it is a trinomial.

3. 3x2

SOLUTION:

A polynomial is a monomial or the sum of monomials, so 3x2 is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of 3x2 is 2. The

polynomial has one term, so it is a monomial.

4.

SOLUTION: A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer exponents. It has

only one term. is a division of two monomials, so it is not a monomial.

5. 5m2p

3 + 6

SOLUTION:

A polynomial is a monomial or the sum of monomials. 5m2p

3 + 6

is the sum of 2 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 5 and 0,

so the degree of 5m2p

3 + 6 is 5. The polynomial has two terms, so it is a binomial.

6. 5q–4

+ 6q

SOLUTION: A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer exponents. It has

only one term. 5q-4 is equal to , which is a division of two monomials, so it is not a monomial.

Write each polynomial in standard form. Identify the leading coefficient.

7. –4d4 + 1 – d

2

SOLUTION: Find the degree of each term.

–4d4→ 4

1 → 0

–d2 → 2

The greatest degree is 4, from the term –4d4, so the leading coefficient of –4d

4 + 1 – d

2 is –4.

Rewrite the polynomial with each monomial in descending order according to degree.

–4d4 – d

2 + 1

8. 2x5 – 12 + 3x

SOLUTION: Find the degree of each term.

2x5 → 5

–12 → 0 3x → 1

The greatest degree is 5, from the term 2x5, so the leading coefficient of 2x

5 – 12 + 3x is 2.

Rewrite the polynomial with each monomial in descending order according to degree.

2x5 + 3x – 12

9. 4z – 2z2 – 5z

4

SOLUTION: Find the degree of each term. 4z → 1

– 2z2 → 2

– 5z4→ 4

The greatest degree is 4, from the term – 5z4, so the leading coefficient of 4z – 2z

2 – 5z

4 is –5.

Rewrite the polynomial with each monomial in descending order according to degree.

–5z4 – 2z

2 + 4z

10. 2a + 4a3 – 5a

2 – 1

SOLUTION: Find the degree of each term. 2a → 1

4a3→ 3

– 5a2 → 2

– 1 → 0

The greatest degree is 3, from the term 4a3, so the leading coefficient of 2a + 4a

3 – 5a

2 – 1 is 4.

Rewrite the polynomial with each monomial in descending order according to degree.

4a3– 5a

2 + 2a – 1

Find each sum or difference.

11. (6x3 − 4) + (−2x

3 + 9)

SOLUTION:

12. (g3 − 2g2 + 5g + 6) − (g2 + 2g)

SOLUTION:

13. (4 + 2a2 − 2a) − (3a

2 − 8a + 7)

SOLUTION:

14. (8y − 4y2) + (3y − 9y

2)

SOLUTION:

15. (−4z3 − 2z + 8) − (4z

3 + 3z2 − 5)

SOLUTION:

16. (−3d2 − 8 + 2d) + (4d − 12 + d2

)

SOLUTION:

17. (y + 5) + (2y + 4y2 – 2)

SOLUTION:

18. (3n3 − 5n + n2

) − (−8n2 + 3n

3)

SOLUTION:

19. CCSS SENSE-MAKING The total number of students T who traveled for spring break consists of two groups: students who flew to their destinations F and students who drove to their destination D. The number (in thousands) of students who flew and the total number of students who flew or drove can be modeled by the following equations,where n is the number of years since 1995. T = 14n + 21

F = 8n + 7 a. Write an equation that models the number of students who drove to their destination for this time period. b. Predict the number of students who will drive to their destination in 2012. c. How many students will drive or fly to their destination in 2015?

SOLUTION: a.

D = 6n + 14 b. n = 2012 – 1995 = 17

The number of students who will drive to their destination in 2012 is 116,000 students. c. n = 2015 – 1995 = 20

The number of students who will drive or fly to their destination in 2015 is 301,000 students.

Determine whether each expression is a polynomial. If it is a polynomial, find the degree and determine whether it is a monomial, binomial, or trinomial.

20.

SOLUTION: A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer exponents. It has

only one term. is a division of two monomials, so it is not a monomial.

21.

SOLUTION: A polynomial is a monomial or the sum of monomials. 21 is a monomial, so it is also a polynomial. The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of 21 is 0. The polynomial has only one term, so it is a monomial.

22. c4 – 2c

2 + 1

SOLUTION:

A polynomial is a monomial or the sum of monomials. c4 – 2c

2 + 1

is the sum of 3 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 4, 2, and

0, so the degree of c4 – 2c2 + 1 is 4. The polynomial has three terms, so it is a trinomial.

23. d + 3dc

SOLUTION: A polynomial is a monomial or the sum of monomials. A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer

exponents. 3dc

has a variable in the exponent, so it is not a monomial.

24. a – a2

SOLUTION:

A polynomial is a monomial or the sum of monomials. a – a2 is the sum of 2 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 1 and 2,

so the degree of a – a2 is 2. The polynomial has two terms, so it is a binomial.

25. 5n3 + nq

3

SOLUTION:

A polynomial is a monomial or the sum of monomials. 5n3 + nq3 is the sum of 2 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 3 and 4,

so the degree of 5n3 + nq3 is 4. The polynomial has two terms, so it is a binomial.

Write each polynomial in standard form. Identify the leading coefficient.

26. 5x2 – 2 + 3x

SOLUTION: Find the degree of each term.

5x2 → 2

– 2 → 0

3x → 1

The greatest degree is 2, from the term 5x2, so the leading coefficient of 5x

2 – 2 + 3x is 5.

Rewrite the polynomial with each monomial in descending order according to degree.

5x2 + 3x – 2

27. 8y + 7y3

SOLUTION: Find the degree of each term.

7y3 → 3

8y → 1

The greatest degree is 3, from the term 7y3, so the leading coefficient of 8y + 7y

3 is 7.

Rewrite the polynomial with each monomial in descending order according to degree.

7y3 + 8y

28. 4 – 3c – 5c2

SOLUTION: Find the degree of each term. 4 → 0 3c → 1

– 5c2 → 2

The greatest degree is 2, from the term – 5c2, so the leading coefficient of 4 – 3c – 5c

2 is –5.

Rewrite the polynomial with each monomial in descending order according to degree.

–5c2 – 3c + 4

29. –y3 + 3y – 3y

2 + 2

SOLUTION: Find the degree of each term.

–y3 → 3

3y → 1

3y2 → 2

2 → 0

The greatest degree is 3, from the term –y3, so the leading coefficient of –y

3 + 3y – 3y

2 + 2 is –1.

Rewrite the polynomial with each monomial in descending order according to degree. –y 3 – 3y 2 + 3y + 2

30. 11t + 2t2 – 3 + t

5

SOLUTION: Find the degree of each term. 11t → 1

2t2 → 2

–3 → 0

t5 → 5

The greatest degree is 5, from the term t5, so the leading coefficient of 11t + 2t2 – 3 + t

5 is 1.

Rewrite the polynomial with each monomial in descending order according to degree.

t5 + 2t

2 + 11t – 3

31. 2 + r – r3

SOLUTION: Find the degree of each term. 2 → 0 r → 1

– r3→ 3

The greatest degree is 3, from the term – r3, so the leading coefficient of 2 + r – r3

is –1. Rewrite the polynomial with each monomial in descending order according to degree.

–r3 + r + 2

32.

SOLUTION: Find the degree of each term.

→ 0

–3x4 → 4

7 → 0

The greatest degree is 4, from the term –3x4, so the leading coefficient of is –3.

Rewrite the polynomial with each monomial in descending order according to degree.

33. –9b2 + 10b – b

6

SOLUTION: Find the degree of each term.

–9b2 → 2

10b → 1

–b6→ 6

The greatest degree is 6, from the term –b6, so the leading coefficient of –9b2 + 10b – b6 is –1. Rewrite the polynomial with each monomial in descending order according to degree.

–b6 – 9b

2 + 10b

Find each sum or difference.

34. (2c2 + 6c + 4) + (5c – 7)

SOLUTION:

35. (2x + 3x2) − (7 − 8x

2)

SOLUTION:

36. (3c3 − c + 11) − (c2 + 2c + 8)

SOLUTION:

37. (z2 + z) + (z

2 − 11)

SOLUTION:

38. (2x − 2y + 1) − (3y + 4x)

SOLUTION:

39. (4a − 5b2 + 3) + (6 − 2a + 3b

2)

SOLUTION:

40. (x2y − 3x

2 + y) + (3y − 2x2y)

SOLUTION:

41. (−8xy + 3x2 − 5y) + (4x

2 − 2y + 6xy)

SOLUTION:

42. (5n − 2p2 + 2np) − (4p

2 + 4n)

SOLUTION:

43. (4rxt − 8r2x + x2

) − (6rx2 + 5rxt − 2x

2)

SOLUTION:

44. PETS From 1999 through 2009, the number of dogs D and the number of cats C (in hundreds) adopted from animal

shelters in the United States are modeled by the equations D = 2n + 3 and C = n + 4, where n is the number of yearssince 1999. a. Write an equation that models the total number T of dogs and cats adopted in hundreds for this time period. b. If this trend continues, how many dogs and cats will be adopted in 2013?

SOLUTION: a.

So, an equation that models the total number of dogs and cats adopted is T = 3n + 7. b. Evaluate the equation for the total number of dogs and cats for n = 2013 – 1999 = 14.

The number of cats and dogs adopted in 2013 will be 49 × 100 or 4900 cats and dogs.

Classify each polynomial according to its degree and number of terms.

45. 4x – 3x2 + 5

SOLUTION:

Find the degree of each term of 4x – 3x2 + 5.

4x → 1

– 3x2 → 2

5 → 0

The greatest degree is 2 and there are 3 terms, so 4x – 3x2 + 5 is a quadratic trinomial.

46. 11z3

SOLUTION: Find the degree of each term. 11z3

→ 3

The greatest degree is 3 and there is one term, so 11z3 is a cubic monomial.

47. 9 + y4

SOLUTION:

Find the degree of each term of 9 + y 4.

9 → 1

y4→ 4

The greatest degree is 4 and there are 2 terms, so 9 + y 4 is a quartic binomial.

48. 3x3 – 7

SOLUTION:

Find the degree of each term of 3x3 – 7.

3x3 → 3

–7 → 0

The greatest degree is 3 and there are 2 terms, so 3x3 – 7 is a cubic binomial.

49. –2x5 – x

2 + 5x – 8

SOLUTION:

Find the degree of each term of –2x5 – x2 + 5x – 8.

–2x5 → 5

–x2 → 2

5x → 1 – 8 → 0

The greatest degree is 5 and there are 4 terms, so –2x5 – x2 + 5x – 8 is a quintic polynomial.

50. 10t – 4t2 + 6t

3

SOLUTION:

Find the degree of each term of 10t – 4t2 + 6t

3.

10t → 1

4t2 → 2

6t3 → 3

The greatest degree is 3 and there are 3 terms, so 10t – 4t2 + 6t

3 is a cubic trinomial.

51. ENROLLMENT In a rapidly growing school system, the numbers (in hundreds) of total students N and

K-5 students P enrolled from 2000 to 2009 are modeled by the equations N = 1.25t2 – t + 7.5 and P =

0.7t2 – 0.95t + 3.8, where t is the number of years since 2000.

a. Write an equation modeling the number of 6-12 students S enrolled for this time period. b. How many 6-12 students were enrolled in the school system in 2007?

SOLUTION: a. To write an equation that represents the number of 6-12 students enrolled, subtract the equations that represent the total number of students and the number of K-5 students.

b. Replace t with 7 in the equation for S to determine the number of students enrolled in 6-12 in 2007.

In 2007 there were 30.3 hundreds or 3030 students enrolled in 6-12.

52. CCSS REASONING The perimeter of the figure shown is represented by the expression 3x2 − 7x + 2. Write a

polynomial that represents the measure of the third side.

SOLUTION:

53. GEOMETRY Consider the rectangle.

a. What does (4x2 + 2x – 1)(2x

2 – x + 3) represent?

b. What does 2(4x2 + 2x – 1) + 2(2x

2 – x + 3) represent?

SOLUTION:

a. (4x2 + 2x – 1)(2x

2 – x + 3) is a multiplication of the length and the width of the rectangle, which is the formula for

the area of a rectangle.

b. 2(4x2 + 2x – 1) + 2(2x

2 – x + 3) is the sum of twice the length and twice the width of a rectangle, which is the

formula for the perimeter of the rectangle

Find each sum or difference.

54. (4x + 2y − 6z) + (5y − 2z + 7x) + (−9z − 2x − 3y)

SOLUTION:

55. (5a2 − 4) + (a

2 − 2a + 12) + (4a2 − 6a + 8)

SOLUTION:

56. (3c2 − 7) + (4c + 7) − (c

2 + 5c − 8)

SOLUTION:

57. (3n3 + 3n − 10) − (4n

2 − 5n) + (4n3 − 3n

2 − 9n + 4)

SOLUTION:

58. FOOTBALL The National Football League is divided into two conferences, the American A and the National N. From 2002 through 2009, the total attendance T (in thousands) for both conferences and for the American Conference games are modeled by the following equations, where x is the number of years since 2002.

T = –0.69x3 + 55.83x

2 + 643.31x + 10,538

A = –3.78x3 + 58.96x

2 + 265.96x + 5257

Determine how many people attended National Conference football games in 2009.

SOLUTION:

Let x = 7 represent 2009, then find how many people attended a National Conference football game in 2009.

In 2009 the number of people who attended National Conference football games was about 8829 thousand. Multiply by 1000 to find the attendance in standard form. 8829 ×1000 = 8,829,000. So, about 8,829,000 people attended National Conference football games in 2009.

59. CAR RENTAL The cost to rent a car for a day is $15 plus $0.15 for each mile driven. a. Write a polynomial that represents the cost of renting a car for m miles. b. If a car is driven 145 miles, how much would it cost to rent? c. If a car is driven 105 miles each day for four days, how much would it cost to rent a car? d. If a car is driven 220 miles each day for seven days, how much would it cost to rent a car?

SOLUTION: a. The cost to rent a car is the daily rate and the mileage cost or 15 + 0.15m. b. Substitute 145 for m to find the daily cost to drive 145 miles.

The cost to rent the car would be $36.75. c. The expression represents the cost per day. For 4 days, multiply the entire expression by 4 to find the cost for a 4 day trip. Substitute 105 for m.

The cost to rent the car would be $123. d. The expression represents the cost per day. For 7 days, multiply the entire expression by 7 to find the cost for a 7 day trip. Substitute 220 for m.

The cost to rent the car would be $336.

60. MULTIPLE REPRESENTATIONS In this problem, you will explore perimeter and area. a. Geometric Draw three rectangles that each have a perimeter of 400 feet. b. Tabular Record the width and length of each rectangle in a table like the one shown below. Find the area of eachrectangle.

c. Graphical On a coordinate system, graph the area of rectangle 4 in terms of the length, x. Use the graph to determine the largest area possible. d. Analytical Determine the length and width that produce the largest area.

SOLUTION: a. For the perimeters to be 400, create lengths and widths that sum to 200. Note that some lengths are already provided in the table in part b.

b. The area is length multiplied by width. Make sure the units are squared for area. The sum of the length and width must be 200, so if the length is x, the width must be 200 – x.

c. The length will be the x-values on the horizontal and the area will be the y-values on the vertical. The length cannot pass 200 since the sum of the length and width is 200. Set the intervals for the x-axis to 25 feet. Extend the table of values to find more points to plot on the graph. It appears that 10,000 is the greatest area, so set the intervals

for the y-axis to 1000 ft2. After graphing, it appears that the highest point on the graph is at an area of 10,000 ft

2.

d. The associated x-value with the maximum area is x = 100, so the length must be 100 and the width must be 200 – 100, or 100. The length and width of the rectangle must be 100 feet each to have the largest area.

61. CCSS CRITIQUE Cheyenne and Sebastian are finding (2x2 − x) − (3x + 3x

2 − 2). Is either of them correct? Explain your reasoning.

SOLUTION:

Neither is correct. Cheyenne, did not distribute the negative to the 2nd and 3rd terms when she found the additive inverse. Sebastian did not distribute the negate to the 3rd terms when he found the additive inverse. To find the additive inverse, all terms should be multiplied by −1.

62. REASONING Determine whether each of the following statements is true or false . Explain your reasoning. a. A binomial can have a degree of zero. b. The order in which polynomials are subtracted does not matter.

SOLUTION: a. If a binomial has two terms that are each a degree of 0, then those terms can be combined and the binomial becomes a monomial. For example, 18 + 7 = 25. If one of the terms of the binomial does not have a degree of 0, then the binomial cannot have a degree of 0, since the degree of a polynomial is the greatest degree of any term in the polynomial. b. Subtraction is not commutative. While 2 + 5 = 5 + 2, 2 – 5 ≠ 5 – 2. This is also true for polynomials. Sample answer: (2x – 3) – (4x – 3) = –2x, but (4x – 3) – (2x – 3) = 2x

63. CHALLENGE Write a polynomial that represents the sum of an odd integer 2n + 1 and the next two consecutive odd integers.

SOLUTION:

64. WRITING IN MATH Why would you add or subtract equations that represent real-world situations? Explain.

SOLUTION:

65. WRITING IN MATH Describe how to add and subtract polynomials using both the vertical and horizontal formats.

SOLUTION: To add polynomials in a horizontal format, you combine like terms. For the vertical format, you write the polynomials in standard form, align like terms in columns, and combine like terms.

To subtract polynomials in a horizontal format you find the additive inverse of the polynomial you are subtracting, andthen combine like terms. For the vertical format, you write the polynomials in standard form, align like terms in columns, and subtract by adding the additive inverse.

66. Three consecutive integers can be represented by x, x + 1, and x + 2. What is the sum of these three integers? A x(x + 1)(x + 2)

B x3 + 3

C 3x + 3 D x + 3

SOLUTION:

The correct choice is C.

67. SHORT RESPONSE What is the perimeter of a square with sides that measure 2x + 3 units?

SOLUTION:

The perimeter of the square is 8x + 12 units.

68. Jim cuts a board in the shape of a regular hexagon and pounds in a nail at each vertex, as shown. How many rubber bands will he need to stretch a rubber band across every possible pair of nails?

F 15 G 14 H 12 J 9

SOLUTION: The first nail would connect to 5 others, the second to 4 others, the third to 3 others, etc.5 + 4 + 3 + 2 + 1 = 15 The correct choice is F.

69. Which ordered pair is in the solution set of the system of inequalities shown in the graph?

A (−3, 0) B (0, −3) C (5, 0) D (0, 5)

SOLUTION: Choice A is outside the shaded area for both inequalities. Choices B and D are inside the shaded area for only one inequality. Choice C is the only point in the solution for both inequalities. So, the correct choice is C.

70. COMPUTERS A computer technician charges by the hour to fix and repair computer equipment. The total cost of the technician for one hour is $75, for two hours is $125, for three hours is $175, for four hours is $225, and so on. Write a recursive formula for the sequence.

SOLUTION: Write out the terms. $75, $125, $175, $225, ... The first term is 75, and 50 is added to form each following term. Therefore, we have a1 = 75, an = an – 1 + 50, n ≥ 2.

Determine whether each sequence is arithmetic, geometric, or neither. Explain.71. 8, –32, 128, –512, ...

SOLUTION: Check for a common difference. –32 – 8 = –40 128 – (–32) = 160 There is no common difference. Check for a common ratio. –32 ÷ 8 = –4 128 ÷ (–32) = –4 Geometric; the common ratio is –4.

72. 25, 8, –9, –26, ...

SOLUTION: Check for a common difference. 8 – 25 = –17 –9 – 8 = –17 Arithmetic; the common difference is –17.

73.

SOLUTION: Check for a common difference.

There is no common difference. Check for a common ratio.

There is no common ratio, so the sequence is not arithmetic or geometric.

74. 43, 52, 61, 70, ...

SOLUTION: Check for a common difference. 52 – 43 = 9 61 – 52 = 9 Arithmetic; the common difference is 9.

75. –27, –16, –5, 6, ...

SOLUTION: Check for a common difference. –16 – (–27) = 11 –5 – (–16) = 11 Arithmetic; the common difference is 11.

76. 200, 100, 50, 25, …

SOLUTION: Check for a common difference. 100 – 200 = –100 50 – 100 = –50 There is no common difference. Check for a common ratio. 100 ÷ 200 = 0.5 50 ÷ 100 = 0.5

Geometric; the common ratio is 0.5 or .

77. JOBS Kimi received an offer for a new job. She wants to compare the offer with her current job. What is total amount of sales that Kimi must get each month to make the same income at either job?

SOLUTION: Let s be Kimi’s monthly sales.

Kimi must sell $80,000 each month to make the same income at either job.

Determine whether each sequence is an arithmetic sequence. If it is, state the common difference.78. 24, 16, 8, 0, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is –8, because 16 – 24 = –8; 8 –16 = –8; etc.

79. , 13, 26, …

SOLUTION: Find the difference between the terms.

– =

13 – =

26 – 13 = 13 There is not a common different. The sequence is not an arithmetic sequence

80. 7, 6, 5, 4, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is –1, because 6 – 7 = –1; 5 – 6 = –1; 4 – 5 = –1;etc.

81. 10, 12, 15, 18, …

SOLUTION: Find the difference between the terms. 12 – 10 = 2 15 – 12 = 3 18 – 15 = 3 There is no common difference. The sequence is not an arithmetic sequence.

82. −15, −11, −7, −3, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is 4, because –11 – (–15) = 4; –7 – (–11) = 4; –3 – (–7) = 4;etc.

83. −0.3, 0.2, 0.7, 1.2, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is 0.5, because 0.2 – (–0.3) = 0.5; 0.7 – 0.2 = 0.5; 1.2 – 0.7 = 0.5; etc.

Simplify.

84. t(t5)(t

7)

SOLUTION:

85. n3(n

2)(−2n

3)

SOLUTION:

86. (5t5v

2)(10t

3v

4)

SOLUTION:

87. (−8u4z

5)(5uz

4)

SOLUTION:

88. [(3)2]3

SOLUTION:

89. [(2)3]2

SOLUTION:

90. (2m4k

3)2(−3mk

2)3

SOLUTION:

91. (6xy2)2(2x

2y

2z

2)3

SOLUTION:

eSolutions Manual - Powered by Cognero Page 27

8-1 Adding and Subtracting Polynomials

Determine whether each expression is a polynomial. If it is a polynomial, find the degree and determine whether it is a monomial, binomial, or trinomial.

1. 7ab + 6b2 – 2a

3

SOLUTION:

A polynomial is a monomial or the sum of monomials. 7ab + 6b2 – 2a

3 is the sum of 3 monomials, so it is a

polynomial. The degree of a polynomial is the greatest degree of any term in the polynomial.The degree of each term is 2, 2, and

3, so the degree of 7ab + 6b2 – 2a

3 is 3. The polynomial has three terms, so it is a trinomial.

2. 2y – 5 + 3y2

SOLUTION:

2y – 5 + 3y2

is the sum of monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 1, 0, and

2, so the degree of 2y – 5 + 3y2 is 2. The polynomial has three terms, so it is a trinomial.

3. 3x2

SOLUTION:

A polynomial is a monomial or the sum of monomials, so 3x2 is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of 3x2 is 2. The

polynomial has one term, so it is a monomial.

4.

SOLUTION: A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer exponents. It has

only one term. is a division of two monomials, so it is not a monomial.

5. 5m2p

3 + 6

SOLUTION:

A polynomial is a monomial or the sum of monomials. 5m2p

3 + 6

is the sum of 2 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 5 and 0,

so the degree of 5m2p

3 + 6 is 5. The polynomial has two terms, so it is a binomial.

6. 5q–4

+ 6q

SOLUTION: A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer exponents. It has

only one term. 5q-4 is equal to , which is a division of two monomials, so it is not a monomial.

Write each polynomial in standard form. Identify the leading coefficient.

7. –4d4 + 1 – d

2

SOLUTION: Find the degree of each term.

–4d4→ 4

1 → 0

–d2 → 2

The greatest degree is 4, from the term –4d4, so the leading coefficient of –4d

4 + 1 – d

2 is –4.

Rewrite the polynomial with each monomial in descending order according to degree.

–4d4 – d

2 + 1

8. 2x5 – 12 + 3x

SOLUTION: Find the degree of each term.

2x5 → 5

–12 → 0 3x → 1

The greatest degree is 5, from the term 2x5, so the leading coefficient of 2x

5 – 12 + 3x is 2.

Rewrite the polynomial with each monomial in descending order according to degree.

2x5 + 3x – 12

9. 4z – 2z2 – 5z

4

SOLUTION: Find the degree of each term. 4z → 1

– 2z2 → 2

– 5z4→ 4

The greatest degree is 4, from the term – 5z4, so the leading coefficient of 4z – 2z

2 – 5z

4 is –5.

Rewrite the polynomial with each monomial in descending order according to degree.

–5z4 – 2z

2 + 4z

10. 2a + 4a3 – 5a

2 – 1

SOLUTION: Find the degree of each term. 2a → 1

4a3→ 3

– 5a2 → 2

– 1 → 0

The greatest degree is 3, from the term 4a3, so the leading coefficient of 2a + 4a

3 – 5a

2 – 1 is 4.

Rewrite the polynomial with each monomial in descending order according to degree.

4a3– 5a

2 + 2a – 1

Find each sum or difference.

11. (6x3 − 4) + (−2x

3 + 9)

SOLUTION:

12. (g3 − 2g2 + 5g + 6) − (g2 + 2g)

SOLUTION:

13. (4 + 2a2 − 2a) − (3a

2 − 8a + 7)

SOLUTION:

14. (8y − 4y2) + (3y − 9y

2)

SOLUTION:

15. (−4z3 − 2z + 8) − (4z

3 + 3z2 − 5)

SOLUTION:

16. (−3d2 − 8 + 2d) + (4d − 12 + d2

)

SOLUTION:

17. (y + 5) + (2y + 4y2 – 2)

SOLUTION:

18. (3n3 − 5n + n2

) − (−8n2 + 3n

3)

SOLUTION:

19. CCSS SENSE-MAKING The total number of students T who traveled for spring break consists of two groups: students who flew to their destinations F and students who drove to their destination D. The number (in thousands) of students who flew and the total number of students who flew or drove can be modeled by the following equations,where n is the number of years since 1995. T = 14n + 21

F = 8n + 7 a. Write an equation that models the number of students who drove to their destination for this time period. b. Predict the number of students who will drive to their destination in 2012. c. How many students will drive or fly to their destination in 2015?

SOLUTION: a.

D = 6n + 14 b. n = 2012 – 1995 = 17

The number of students who will drive to their destination in 2012 is 116,000 students. c. n = 2015 – 1995 = 20

The number of students who will drive or fly to their destination in 2015 is 301,000 students.

Determine whether each expression is a polynomial. If it is a polynomial, find the degree and determine whether it is a monomial, binomial, or trinomial.

20.

SOLUTION: A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer exponents. It has

only one term. is a division of two monomials, so it is not a monomial.

21.

SOLUTION: A polynomial is a monomial or the sum of monomials. 21 is a monomial, so it is also a polynomial. The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of 21 is 0. The polynomial has only one term, so it is a monomial.

22. c4 – 2c

2 + 1

SOLUTION:

A polynomial is a monomial or the sum of monomials. c4 – 2c

2 + 1

is the sum of 3 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 4, 2, and

0, so the degree of c4 – 2c2 + 1 is 4. The polynomial has three terms, so it is a trinomial.

23. d + 3dc

SOLUTION: A polynomial is a monomial or the sum of monomials. A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer

exponents. 3dc

has a variable in the exponent, so it is not a monomial.

24. a – a2

SOLUTION:

A polynomial is a monomial or the sum of monomials. a – a2 is the sum of 2 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 1 and 2,

so the degree of a – a2 is 2. The polynomial has two terms, so it is a binomial.

25. 5n3 + nq

3

SOLUTION:

A polynomial is a monomial or the sum of monomials. 5n3 + nq3 is the sum of 2 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 3 and 4,

so the degree of 5n3 + nq3 is 4. The polynomial has two terms, so it is a binomial.

Write each polynomial in standard form. Identify the leading coefficient.

26. 5x2 – 2 + 3x

SOLUTION: Find the degree of each term.

5x2 → 2

– 2 → 0

3x → 1

The greatest degree is 2, from the term 5x2, so the leading coefficient of 5x

2 – 2 + 3x is 5.

Rewrite the polynomial with each monomial in descending order according to degree.

5x2 + 3x – 2

27. 8y + 7y3

SOLUTION: Find the degree of each term.

7y3 → 3

8y → 1

The greatest degree is 3, from the term 7y3, so the leading coefficient of 8y + 7y

3 is 7.

Rewrite the polynomial with each monomial in descending order according to degree.

7y3 + 8y

28. 4 – 3c – 5c2

SOLUTION: Find the degree of each term. 4 → 0 3c → 1

– 5c2 → 2

The greatest degree is 2, from the term – 5c2, so the leading coefficient of 4 – 3c – 5c

2 is –5.

Rewrite the polynomial with each monomial in descending order according to degree.

–5c2 – 3c + 4

29. –y3 + 3y – 3y

2 + 2

SOLUTION: Find the degree of each term.

–y3 → 3

3y → 1

3y2 → 2

2 → 0

The greatest degree is 3, from the term –y3, so the leading coefficient of –y

3 + 3y – 3y

2 + 2 is –1.

Rewrite the polynomial with each monomial in descending order according to degree. –y 3 – 3y 2 + 3y + 2

30. 11t + 2t2 – 3 + t

5

SOLUTION: Find the degree of each term. 11t → 1

2t2 → 2

–3 → 0

t5 → 5

The greatest degree is 5, from the term t5, so the leading coefficient of 11t + 2t2 – 3 + t

5 is 1.

Rewrite the polynomial with each monomial in descending order according to degree.

t5 + 2t

2 + 11t – 3

31. 2 + r – r3

SOLUTION: Find the degree of each term. 2 → 0 r → 1

– r3→ 3

The greatest degree is 3, from the term – r3, so the leading coefficient of 2 + r – r3

is –1. Rewrite the polynomial with each monomial in descending order according to degree.

–r3 + r + 2

32.

SOLUTION: Find the degree of each term.

→ 0

–3x4 → 4

7 → 0

The greatest degree is 4, from the term –3x4, so the leading coefficient of is –3.

Rewrite the polynomial with each monomial in descending order according to degree.

33. –9b2 + 10b – b

6

SOLUTION: Find the degree of each term.

–9b2 → 2

10b → 1

–b6→ 6

The greatest degree is 6, from the term –b6, so the leading coefficient of –9b2 + 10b – b6 is –1. Rewrite the polynomial with each monomial in descending order according to degree.

–b6 – 9b

2 + 10b

Find each sum or difference.

34. (2c2 + 6c + 4) + (5c – 7)

SOLUTION:

35. (2x + 3x2) − (7 − 8x

2)

SOLUTION:

36. (3c3 − c + 11) − (c2 + 2c + 8)

SOLUTION:

37. (z2 + z) + (z

2 − 11)

SOLUTION:

38. (2x − 2y + 1) − (3y + 4x)

SOLUTION:

39. (4a − 5b2 + 3) + (6 − 2a + 3b

2)

SOLUTION:

40. (x2y − 3x

2 + y) + (3y − 2x2y)

SOLUTION:

41. (−8xy + 3x2 − 5y) + (4x

2 − 2y + 6xy)

SOLUTION:

42. (5n − 2p2 + 2np) − (4p

2 + 4n)

SOLUTION:

43. (4rxt − 8r2x + x2

) − (6rx2 + 5rxt − 2x

2)

SOLUTION:

44. PETS From 1999 through 2009, the number of dogs D and the number of cats C (in hundreds) adopted from animal

shelters in the United States are modeled by the equations D = 2n + 3 and C = n + 4, where n is the number of yearssince 1999. a. Write an equation that models the total number T of dogs and cats adopted in hundreds for this time period. b. If this trend continues, how many dogs and cats will be adopted in 2013?

SOLUTION: a.

So, an equation that models the total number of dogs and cats adopted is T = 3n + 7. b. Evaluate the equation for the total number of dogs and cats for n = 2013 – 1999 = 14.

The number of cats and dogs adopted in 2013 will be 49 × 100 or 4900 cats and dogs.

Classify each polynomial according to its degree and number of terms.

45. 4x – 3x2 + 5

SOLUTION:

Find the degree of each term of 4x – 3x2 + 5.

4x → 1

– 3x2 → 2

5 → 0

The greatest degree is 2 and there are 3 terms, so 4x – 3x2 + 5 is a quadratic trinomial.

46. 11z3

SOLUTION: Find the degree of each term. 11z3

→ 3

The greatest degree is 3 and there is one term, so 11z3 is a cubic monomial.

47. 9 + y4

SOLUTION:

Find the degree of each term of 9 + y 4.

9 → 1

y4→ 4

The greatest degree is 4 and there are 2 terms, so 9 + y 4 is a quartic binomial.

48. 3x3 – 7

SOLUTION:

Find the degree of each term of 3x3 – 7.

3x3 → 3

–7 → 0

The greatest degree is 3 and there are 2 terms, so 3x3 – 7 is a cubic binomial.

49. –2x5 – x

2 + 5x – 8

SOLUTION:

Find the degree of each term of –2x5 – x2 + 5x – 8.

–2x5 → 5

–x2 → 2

5x → 1 – 8 → 0

The greatest degree is 5 and there are 4 terms, so –2x5 – x2 + 5x – 8 is a quintic polynomial.

50. 10t – 4t2 + 6t

3

SOLUTION:

Find the degree of each term of 10t – 4t2 + 6t

3.

10t → 1

4t2 → 2

6t3 → 3

The greatest degree is 3 and there are 3 terms, so 10t – 4t2 + 6t

3 is a cubic trinomial.

51. ENROLLMENT In a rapidly growing school system, the numbers (in hundreds) of total students N and

K-5 students P enrolled from 2000 to 2009 are modeled by the equations N = 1.25t2 – t + 7.5 and P =

0.7t2 – 0.95t + 3.8, where t is the number of years since 2000.

a. Write an equation modeling the number of 6-12 students S enrolled for this time period. b. How many 6-12 students were enrolled in the school system in 2007?

SOLUTION: a. To write an equation that represents the number of 6-12 students enrolled, subtract the equations that represent the total number of students and the number of K-5 students.

b. Replace t with 7 in the equation for S to determine the number of students enrolled in 6-12 in 2007.

In 2007 there were 30.3 hundreds or 3030 students enrolled in 6-12.

52. CCSS REASONING The perimeter of the figure shown is represented by the expression 3x2 − 7x + 2. Write a

polynomial that represents the measure of the third side.

SOLUTION:

53. GEOMETRY Consider the rectangle.

a. What does (4x2 + 2x – 1)(2x

2 – x + 3) represent?

b. What does 2(4x2 + 2x – 1) + 2(2x

2 – x + 3) represent?

SOLUTION:

a. (4x2 + 2x – 1)(2x

2 – x + 3) is a multiplication of the length and the width of the rectangle, which is the formula for

the area of a rectangle.

b. 2(4x2 + 2x – 1) + 2(2x

2 – x + 3) is the sum of twice the length and twice the width of a rectangle, which is the

formula for the perimeter of the rectangle

Find each sum or difference.

54. (4x + 2y − 6z) + (5y − 2z + 7x) + (−9z − 2x − 3y)

SOLUTION:

55. (5a2 − 4) + (a

2 − 2a + 12) + (4a2 − 6a + 8)

SOLUTION:

56. (3c2 − 7) + (4c + 7) − (c

2 + 5c − 8)

SOLUTION:

57. (3n3 + 3n − 10) − (4n

2 − 5n) + (4n3 − 3n

2 − 9n + 4)

SOLUTION:

58. FOOTBALL The National Football League is divided into two conferences, the American A and the National N. From 2002 through 2009, the total attendance T (in thousands) for both conferences and for the American Conference games are modeled by the following equations, where x is the number of years since 2002.

T = –0.69x3 + 55.83x

2 + 643.31x + 10,538

A = –3.78x3 + 58.96x

2 + 265.96x + 5257

Determine how many people attended National Conference football games in 2009.

SOLUTION:

Let x = 7 represent 2009, then find how many people attended a National Conference football game in 2009.

In 2009 the number of people who attended National Conference football games was about 8829 thousand. Multiply by 1000 to find the attendance in standard form. 8829 ×1000 = 8,829,000. So, about 8,829,000 people attended National Conference football games in 2009.

59. CAR RENTAL The cost to rent a car for a day is $15 plus $0.15 for each mile driven. a. Write a polynomial that represents the cost of renting a car for m miles. b. If a car is driven 145 miles, how much would it cost to rent? c. If a car is driven 105 miles each day for four days, how much would it cost to rent a car? d. If a car is driven 220 miles each day for seven days, how much would it cost to rent a car?

SOLUTION: a. The cost to rent a car is the daily rate and the mileage cost or 15 + 0.15m. b. Substitute 145 for m to find the daily cost to drive 145 miles.

The cost to rent the car would be $36.75. c. The expression represents the cost per day. For 4 days, multiply the entire expression by 4 to find the cost for a 4 day trip. Substitute 105 for m.

The cost to rent the car would be $123. d. The expression represents the cost per day. For 7 days, multiply the entire expression by 7 to find the cost for a 7 day trip. Substitute 220 for m.

The cost to rent the car would be $336.

60. MULTIPLE REPRESENTATIONS In this problem, you will explore perimeter and area. a. Geometric Draw three rectangles that each have a perimeter of 400 feet. b. Tabular Record the width and length of each rectangle in a table like the one shown below. Find the area of eachrectangle.

c. Graphical On a coordinate system, graph the area of rectangle 4 in terms of the length, x. Use the graph to determine the largest area possible. d. Analytical Determine the length and width that produce the largest area.

SOLUTION: a. For the perimeters to be 400, create lengths and widths that sum to 200. Note that some lengths are already provided in the table in part b.

b. The area is length multiplied by width. Make sure the units are squared for area. The sum of the length and width must be 200, so if the length is x, the width must be 200 – x.

c. The length will be the x-values on the horizontal and the area will be the y-values on the vertical. The length cannot pass 200 since the sum of the length and width is 200. Set the intervals for the x-axis to 25 feet. Extend the table of values to find more points to plot on the graph. It appears that 10,000 is the greatest area, so set the intervals

for the y-axis to 1000 ft2. After graphing, it appears that the highest point on the graph is at an area of 10,000 ft

2.

d. The associated x-value with the maximum area is x = 100, so the length must be 100 and the width must be 200 – 100, or 100. The length and width of the rectangle must be 100 feet each to have the largest area.

61. CCSS CRITIQUE Cheyenne and Sebastian are finding (2x2 − x) − (3x + 3x

2 − 2). Is either of them correct? Explain your reasoning.

SOLUTION:

Neither is correct. Cheyenne, did not distribute the negative to the 2nd and 3rd terms when she found the additive inverse. Sebastian did not distribute the negate to the 3rd terms when he found the additive inverse. To find the additive inverse, all terms should be multiplied by −1.

62. REASONING Determine whether each of the following statements is true or false . Explain your reasoning. a. A binomial can have a degree of zero. b. The order in which polynomials are subtracted does not matter.

SOLUTION: a. If a binomial has two terms that are each a degree of 0, then those terms can be combined and the binomial becomes a monomial. For example, 18 + 7 = 25. If one of the terms of the binomial does not have a degree of 0, then the binomial cannot have a degree of 0, since the degree of a polynomial is the greatest degree of any term in the polynomial. b. Subtraction is not commutative. While 2 + 5 = 5 + 2, 2 – 5 ≠ 5 – 2. This is also true for polynomials. Sample answer: (2x – 3) – (4x – 3) = –2x, but (4x – 3) – (2x – 3) = 2x

63. CHALLENGE Write a polynomial that represents the sum of an odd integer 2n + 1 and the next two consecutive odd integers.

SOLUTION:

64. WRITING IN MATH Why would you add or subtract equations that represent real-world situations? Explain.

SOLUTION:

65. WRITING IN MATH Describe how to add and subtract polynomials using both the vertical and horizontal formats.

SOLUTION: To add polynomials in a horizontal format, you combine like terms. For the vertical format, you write the polynomials in standard form, align like terms in columns, and combine like terms.

To subtract polynomials in a horizontal format you find the additive inverse of the polynomial you are subtracting, andthen combine like terms. For the vertical format, you write the polynomials in standard form, align like terms in columns, and subtract by adding the additive inverse.

66. Three consecutive integers can be represented by x, x + 1, and x + 2. What is the sum of these three integers? A x(x + 1)(x + 2)

B x3 + 3

C 3x + 3 D x + 3

SOLUTION:

The correct choice is C.

67. SHORT RESPONSE What is the perimeter of a square with sides that measure 2x + 3 units?

SOLUTION:

The perimeter of the square is 8x + 12 units.

68. Jim cuts a board in the shape of a regular hexagon and pounds in a nail at each vertex, as shown. How many rubber bands will he need to stretch a rubber band across every possible pair of nails?

F 15 G 14 H 12 J 9

SOLUTION: The first nail would connect to 5 others, the second to 4 others, the third to 3 others, etc.5 + 4 + 3 + 2 + 1 = 15 The correct choice is F.

69. Which ordered pair is in the solution set of the system of inequalities shown in the graph?

A (−3, 0) B (0, −3) C (5, 0) D (0, 5)

SOLUTION: Choice A is outside the shaded area for both inequalities. Choices B and D are inside the shaded area for only one inequality. Choice C is the only point in the solution for both inequalities. So, the correct choice is C.

70. COMPUTERS A computer technician charges by the hour to fix and repair computer equipment. The total cost of the technician for one hour is $75, for two hours is $125, for three hours is $175, for four hours is $225, and so on. Write a recursive formula for the sequence.

SOLUTION: Write out the terms. $75, $125, $175, $225, ... The first term is 75, and 50 is added to form each following term. Therefore, we have a1 = 75, an = an – 1 + 50, n ≥ 2.

Determine whether each sequence is arithmetic, geometric, or neither. Explain.71. 8, –32, 128, –512, ...

SOLUTION: Check for a common difference. –32 – 8 = –40 128 – (–32) = 160 There is no common difference. Check for a common ratio. –32 ÷ 8 = –4 128 ÷ (–32) = –4 Geometric; the common ratio is –4.

72. 25, 8, –9, –26, ...

SOLUTION: Check for a common difference. 8 – 25 = –17 –9 – 8 = –17 Arithmetic; the common difference is –17.

73.

SOLUTION: Check for a common difference.

There is no common difference. Check for a common ratio.

There is no common ratio, so the sequence is not arithmetic or geometric.

74. 43, 52, 61, 70, ...

SOLUTION: Check for a common difference. 52 – 43 = 9 61 – 52 = 9 Arithmetic; the common difference is 9.

75. –27, –16, –5, 6, ...

SOLUTION: Check for a common difference. –16 – (–27) = 11 –5 – (–16) = 11 Arithmetic; the common difference is 11.

76. 200, 100, 50, 25, …

SOLUTION: Check for a common difference. 100 – 200 = –100 50 – 100 = –50 There is no common difference. Check for a common ratio. 100 ÷ 200 = 0.5 50 ÷ 100 = 0.5

Geometric; the common ratio is 0.5 or .

77. JOBS Kimi received an offer for a new job. She wants to compare the offer with her current job. What is total amount of sales that Kimi must get each month to make the same income at either job?

SOLUTION: Let s be Kimi’s monthly sales.

Kimi must sell $80,000 each month to make the same income at either job.

Determine whether each sequence is an arithmetic sequence. If it is, state the common difference.78. 24, 16, 8, 0, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is –8, because 16 – 24 = –8; 8 –16 = –8; etc.

79. , 13, 26, …

SOLUTION: Find the difference between the terms.

– =

13 – =

26 – 13 = 13 There is not a common different. The sequence is not an arithmetic sequence

80. 7, 6, 5, 4, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is –1, because 6 – 7 = –1; 5 – 6 = –1; 4 – 5 = –1;etc.

81. 10, 12, 15, 18, …

SOLUTION: Find the difference between the terms. 12 – 10 = 2 15 – 12 = 3 18 – 15 = 3 There is no common difference. The sequence is not an arithmetic sequence.

82. −15, −11, −7, −3, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is 4, because –11 – (–15) = 4; –7 – (–11) = 4; –3 – (–7) = 4;etc.

83. −0.3, 0.2, 0.7, 1.2, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is 0.5, because 0.2 – (–0.3) = 0.5; 0.7 – 0.2 = 0.5; 1.2 – 0.7 = 0.5; etc.

Simplify.

84. t(t5)(t

7)

SOLUTION:

85. n3(n

2)(−2n

3)

SOLUTION:

86. (5t5v

2)(10t

3v

4)

SOLUTION:

87. (−8u4z

5)(5uz

4)

SOLUTION:

88. [(3)2]3

SOLUTION:

89. [(2)3]2

SOLUTION:

90. (2m4k

3)2(−3mk

2)3

SOLUTION:

91. (6xy2)2(2x

2y

2z

2)3

SOLUTION:

eSolutions Manual - Powered by Cognero Page 28

8-1 Adding and Subtracting Polynomials

Determine whether each expression is a polynomial. If it is a polynomial, find the degree and determine whether it is a monomial, binomial, or trinomial.

1. 7ab + 6b2 – 2a

3

SOLUTION:

A polynomial is a monomial or the sum of monomials. 7ab + 6b2 – 2a

3 is the sum of 3 monomials, so it is a

polynomial. The degree of a polynomial is the greatest degree of any term in the polynomial.The degree of each term is 2, 2, and

3, so the degree of 7ab + 6b2 – 2a

3 is 3. The polynomial has three terms, so it is a trinomial.

2. 2y – 5 + 3y2

SOLUTION:

2y – 5 + 3y2

is the sum of monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 1, 0, and

2, so the degree of 2y – 5 + 3y2 is 2. The polynomial has three terms, so it is a trinomial.

3. 3x2

SOLUTION:

A polynomial is a monomial or the sum of monomials, so 3x2 is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of 3x2 is 2. The

polynomial has one term, so it is a monomial.

4.

SOLUTION: A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer exponents. It has

only one term. is a division of two monomials, so it is not a monomial.

5. 5m2p

3 + 6

SOLUTION:

A polynomial is a monomial or the sum of monomials. 5m2p

3 + 6

is the sum of 2 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 5 and 0,

so the degree of 5m2p

3 + 6 is 5. The polynomial has two terms, so it is a binomial.

6. 5q–4

+ 6q

SOLUTION: A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer exponents. It has

only one term. 5q-4 is equal to , which is a division of two monomials, so it is not a monomial.

Write each polynomial in standard form. Identify the leading coefficient.

7. –4d4 + 1 – d

2

SOLUTION: Find the degree of each term.

–4d4→ 4

1 → 0

–d2 → 2

The greatest degree is 4, from the term –4d4, so the leading coefficient of –4d

4 + 1 – d

2 is –4.

Rewrite the polynomial with each monomial in descending order according to degree.

–4d4 – d

2 + 1

8. 2x5 – 12 + 3x

SOLUTION: Find the degree of each term.

2x5 → 5

–12 → 0 3x → 1

The greatest degree is 5, from the term 2x5, so the leading coefficient of 2x

5 – 12 + 3x is 2.

Rewrite the polynomial with each monomial in descending order according to degree.

2x5 + 3x – 12

9. 4z – 2z2 – 5z

4

SOLUTION: Find the degree of each term. 4z → 1

– 2z2 → 2

– 5z4→ 4

The greatest degree is 4, from the term – 5z4, so the leading coefficient of 4z – 2z

2 – 5z

4 is –5.

Rewrite the polynomial with each monomial in descending order according to degree.

–5z4 – 2z

2 + 4z

10. 2a + 4a3 – 5a

2 – 1

SOLUTION: Find the degree of each term. 2a → 1

4a3→ 3

– 5a2 → 2

– 1 → 0

The greatest degree is 3, from the term 4a3, so the leading coefficient of 2a + 4a

3 – 5a

2 – 1 is 4.

Rewrite the polynomial with each monomial in descending order according to degree.

4a3– 5a

2 + 2a – 1

Find each sum or difference.

11. (6x3 − 4) + (−2x

3 + 9)

SOLUTION:

12. (g3 − 2g2 + 5g + 6) − (g2 + 2g)

SOLUTION:

13. (4 + 2a2 − 2a) − (3a

2 − 8a + 7)

SOLUTION:

14. (8y − 4y2) + (3y − 9y

2)

SOLUTION:

15. (−4z3 − 2z + 8) − (4z

3 + 3z2 − 5)

SOLUTION:

16. (−3d2 − 8 + 2d) + (4d − 12 + d2

)

SOLUTION:

17. (y + 5) + (2y + 4y2 – 2)

SOLUTION:

18. (3n3 − 5n + n2

) − (−8n2 + 3n

3)

SOLUTION:

19. CCSS SENSE-MAKING The total number of students T who traveled for spring break consists of two groups: students who flew to their destinations F and students who drove to their destination D. The number (in thousands) of students who flew and the total number of students who flew or drove can be modeled by the following equations,where n is the number of years since 1995. T = 14n + 21

F = 8n + 7 a. Write an equation that models the number of students who drove to their destination for this time period. b. Predict the number of students who will drive to their destination in 2012. c. How many students will drive or fly to their destination in 2015?

SOLUTION: a.

D = 6n + 14 b. n = 2012 – 1995 = 17

The number of students who will drive to their destination in 2012 is 116,000 students. c. n = 2015 – 1995 = 20

The number of students who will drive or fly to their destination in 2015 is 301,000 students.

Determine whether each expression is a polynomial. If it is a polynomial, find the degree and determine whether it is a monomial, binomial, or trinomial.

20.

SOLUTION: A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer exponents. It has

only one term. is a division of two monomials, so it is not a monomial.

21.

SOLUTION: A polynomial is a monomial or the sum of monomials. 21 is a monomial, so it is also a polynomial. The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of 21 is 0. The polynomial has only one term, so it is a monomial.

22. c4 – 2c

2 + 1

SOLUTION:

A polynomial is a monomial or the sum of monomials. c4 – 2c

2 + 1

is the sum of 3 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 4, 2, and

0, so the degree of c4 – 2c2 + 1 is 4. The polynomial has three terms, so it is a trinomial.

23. d + 3dc

SOLUTION: A polynomial is a monomial or the sum of monomials. A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer

exponents. 3dc

has a variable in the exponent, so it is not a monomial.

24. a – a2

SOLUTION:

A polynomial is a monomial or the sum of monomials. a – a2 is the sum of 2 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 1 and 2,

so the degree of a – a2 is 2. The polynomial has two terms, so it is a binomial.

25. 5n3 + nq

3

SOLUTION:

A polynomial is a monomial or the sum of monomials. 5n3 + nq3 is the sum of 2 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 3 and 4,

so the degree of 5n3 + nq3 is 4. The polynomial has two terms, so it is a binomial.

Write each polynomial in standard form. Identify the leading coefficient.

26. 5x2 – 2 + 3x

SOLUTION: Find the degree of each term.

5x2 → 2

– 2 → 0

3x → 1

The greatest degree is 2, from the term 5x2, so the leading coefficient of 5x

2 – 2 + 3x is 5.

Rewrite the polynomial with each monomial in descending order according to degree.

5x2 + 3x – 2

27. 8y + 7y3

SOLUTION: Find the degree of each term.

7y3 → 3

8y → 1

The greatest degree is 3, from the term 7y3, so the leading coefficient of 8y + 7y

3 is 7.

Rewrite the polynomial with each monomial in descending order according to degree.

7y3 + 8y

28. 4 – 3c – 5c2

SOLUTION: Find the degree of each term. 4 → 0 3c → 1

– 5c2 → 2

The greatest degree is 2, from the term – 5c2, so the leading coefficient of 4 – 3c – 5c

2 is –5.

Rewrite the polynomial with each monomial in descending order according to degree.

–5c2 – 3c + 4

29. –y3 + 3y – 3y

2 + 2

SOLUTION: Find the degree of each term.

–y3 → 3

3y → 1

3y2 → 2

2 → 0

The greatest degree is 3, from the term –y3, so the leading coefficient of –y

3 + 3y – 3y

2 + 2 is –1.

Rewrite the polynomial with each monomial in descending order according to degree. –y 3 – 3y 2 + 3y + 2

30. 11t + 2t2 – 3 + t

5

SOLUTION: Find the degree of each term. 11t → 1

2t2 → 2

–3 → 0

t5 → 5

The greatest degree is 5, from the term t5, so the leading coefficient of 11t + 2t2 – 3 + t

5 is 1.

Rewrite the polynomial with each monomial in descending order according to degree.

t5 + 2t

2 + 11t – 3

31. 2 + r – r3

SOLUTION: Find the degree of each term. 2 → 0 r → 1

– r3→ 3

The greatest degree is 3, from the term – r3, so the leading coefficient of 2 + r – r3

is –1. Rewrite the polynomial with each monomial in descending order according to degree.

–r3 + r + 2

32.

SOLUTION: Find the degree of each term.

→ 0

–3x4 → 4

7 → 0

The greatest degree is 4, from the term –3x4, so the leading coefficient of is –3.

Rewrite the polynomial with each monomial in descending order according to degree.

33. –9b2 + 10b – b

6

SOLUTION: Find the degree of each term.

–9b2 → 2

10b → 1

–b6→ 6

The greatest degree is 6, from the term –b6, so the leading coefficient of –9b2 + 10b – b6 is –1. Rewrite the polynomial with each monomial in descending order according to degree.

–b6 – 9b

2 + 10b

Find each sum or difference.

34. (2c2 + 6c + 4) + (5c – 7)

SOLUTION:

35. (2x + 3x2) − (7 − 8x

2)

SOLUTION:

36. (3c3 − c + 11) − (c2 + 2c + 8)

SOLUTION:

37. (z2 + z) + (z

2 − 11)

SOLUTION:

38. (2x − 2y + 1) − (3y + 4x)

SOLUTION:

39. (4a − 5b2 + 3) + (6 − 2a + 3b

2)

SOLUTION:

40. (x2y − 3x

2 + y) + (3y − 2x2y)

SOLUTION:

41. (−8xy + 3x2 − 5y) + (4x

2 − 2y + 6xy)

SOLUTION:

42. (5n − 2p2 + 2np) − (4p

2 + 4n)

SOLUTION:

43. (4rxt − 8r2x + x2

) − (6rx2 + 5rxt − 2x

2)

SOLUTION:

44. PETS From 1999 through 2009, the number of dogs D and the number of cats C (in hundreds) adopted from animal

shelters in the United States are modeled by the equations D = 2n + 3 and C = n + 4, where n is the number of yearssince 1999. a. Write an equation that models the total number T of dogs and cats adopted in hundreds for this time period. b. If this trend continues, how many dogs and cats will be adopted in 2013?

SOLUTION: a.

So, an equation that models the total number of dogs and cats adopted is T = 3n + 7. b. Evaluate the equation for the total number of dogs and cats for n = 2013 – 1999 = 14.

The number of cats and dogs adopted in 2013 will be 49 × 100 or 4900 cats and dogs.

Classify each polynomial according to its degree and number of terms.

45. 4x – 3x2 + 5

SOLUTION:

Find the degree of each term of 4x – 3x2 + 5.

4x → 1

– 3x2 → 2

5 → 0

The greatest degree is 2 and there are 3 terms, so 4x – 3x2 + 5 is a quadratic trinomial.

46. 11z3

SOLUTION: Find the degree of each term. 11z3

→ 3

The greatest degree is 3 and there is one term, so 11z3 is a cubic monomial.

47. 9 + y4

SOLUTION:

Find the degree of each term of 9 + y 4.

9 → 1

y4→ 4

The greatest degree is 4 and there are 2 terms, so 9 + y 4 is a quartic binomial.

48. 3x3 – 7

SOLUTION:

Find the degree of each term of 3x3 – 7.

3x3 → 3

–7 → 0

The greatest degree is 3 and there are 2 terms, so 3x3 – 7 is a cubic binomial.

49. –2x5 – x

2 + 5x – 8

SOLUTION:

Find the degree of each term of –2x5 – x2 + 5x – 8.

–2x5 → 5

–x2 → 2

5x → 1 – 8 → 0

The greatest degree is 5 and there are 4 terms, so –2x5 – x2 + 5x – 8 is a quintic polynomial.

50. 10t – 4t2 + 6t

3

SOLUTION:

Find the degree of each term of 10t – 4t2 + 6t

3.

10t → 1

4t2 → 2

6t3 → 3

The greatest degree is 3 and there are 3 terms, so 10t – 4t2 + 6t

3 is a cubic trinomial.

51. ENROLLMENT In a rapidly growing school system, the numbers (in hundreds) of total students N and

K-5 students P enrolled from 2000 to 2009 are modeled by the equations N = 1.25t2 – t + 7.5 and P =

0.7t2 – 0.95t + 3.8, where t is the number of years since 2000.

a. Write an equation modeling the number of 6-12 students S enrolled for this time period. b. How many 6-12 students were enrolled in the school system in 2007?

SOLUTION: a. To write an equation that represents the number of 6-12 students enrolled, subtract the equations that represent the total number of students and the number of K-5 students.

b. Replace t with 7 in the equation for S to determine the number of students enrolled in 6-12 in 2007.

In 2007 there were 30.3 hundreds or 3030 students enrolled in 6-12.

52. CCSS REASONING The perimeter of the figure shown is represented by the expression 3x2 − 7x + 2. Write a

polynomial that represents the measure of the third side.

SOLUTION:

53. GEOMETRY Consider the rectangle.

a. What does (4x2 + 2x – 1)(2x

2 – x + 3) represent?

b. What does 2(4x2 + 2x – 1) + 2(2x

2 – x + 3) represent?

SOLUTION:

a. (4x2 + 2x – 1)(2x

2 – x + 3) is a multiplication of the length and the width of the rectangle, which is the formula for

the area of a rectangle.

b. 2(4x2 + 2x – 1) + 2(2x

2 – x + 3) is the sum of twice the length and twice the width of a rectangle, which is the

formula for the perimeter of the rectangle

Find each sum or difference.

54. (4x + 2y − 6z) + (5y − 2z + 7x) + (−9z − 2x − 3y)

SOLUTION:

55. (5a2 − 4) + (a

2 − 2a + 12) + (4a2 − 6a + 8)

SOLUTION:

56. (3c2 − 7) + (4c + 7) − (c

2 + 5c − 8)

SOLUTION:

57. (3n3 + 3n − 10) − (4n

2 − 5n) + (4n3 − 3n

2 − 9n + 4)

SOLUTION:

58. FOOTBALL The National Football League is divided into two conferences, the American A and the National N. From 2002 through 2009, the total attendance T (in thousands) for both conferences and for the American Conference games are modeled by the following equations, where x is the number of years since 2002.

T = –0.69x3 + 55.83x

2 + 643.31x + 10,538

A = –3.78x3 + 58.96x

2 + 265.96x + 5257

Determine how many people attended National Conference football games in 2009.

SOLUTION:

Let x = 7 represent 2009, then find how many people attended a National Conference football game in 2009.

In 2009 the number of people who attended National Conference football games was about 8829 thousand. Multiply by 1000 to find the attendance in standard form. 8829 ×1000 = 8,829,000. So, about 8,829,000 people attended National Conference football games in 2009.

59. CAR RENTAL The cost to rent a car for a day is $15 plus $0.15 for each mile driven. a. Write a polynomial that represents the cost of renting a car for m miles. b. If a car is driven 145 miles, how much would it cost to rent? c. If a car is driven 105 miles each day for four days, how much would it cost to rent a car? d. If a car is driven 220 miles each day for seven days, how much would it cost to rent a car?

SOLUTION: a. The cost to rent a car is the daily rate and the mileage cost or 15 + 0.15m. b. Substitute 145 for m to find the daily cost to drive 145 miles.

The cost to rent the car would be $36.75. c. The expression represents the cost per day. For 4 days, multiply the entire expression by 4 to find the cost for a 4 day trip. Substitute 105 for m.

The cost to rent the car would be $123. d. The expression represents the cost per day. For 7 days, multiply the entire expression by 7 to find the cost for a 7 day trip. Substitute 220 for m.

The cost to rent the car would be $336.

60. MULTIPLE REPRESENTATIONS In this problem, you will explore perimeter and area. a. Geometric Draw three rectangles that each have a perimeter of 400 feet. b. Tabular Record the width and length of each rectangle in a table like the one shown below. Find the area of eachrectangle.

c. Graphical On a coordinate system, graph the area of rectangle 4 in terms of the length, x. Use the graph to determine the largest area possible. d. Analytical Determine the length and width that produce the largest area.

SOLUTION: a. For the perimeters to be 400, create lengths and widths that sum to 200. Note that some lengths are already provided in the table in part b.

b. The area is length multiplied by width. Make sure the units are squared for area. The sum of the length and width must be 200, so if the length is x, the width must be 200 – x.

c. The length will be the x-values on the horizontal and the area will be the y-values on the vertical. The length cannot pass 200 since the sum of the length and width is 200. Set the intervals for the x-axis to 25 feet. Extend the table of values to find more points to plot on the graph. It appears that 10,000 is the greatest area, so set the intervals

for the y-axis to 1000 ft2. After graphing, it appears that the highest point on the graph is at an area of 10,000 ft

2.

d. The associated x-value with the maximum area is x = 100, so the length must be 100 and the width must be 200 – 100, or 100. The length and width of the rectangle must be 100 feet each to have the largest area.

61. CCSS CRITIQUE Cheyenne and Sebastian are finding (2x2 − x) − (3x + 3x

2 − 2). Is either of them correct? Explain your reasoning.

SOLUTION:

Neither is correct. Cheyenne, did not distribute the negative to the 2nd and 3rd terms when she found the additive inverse. Sebastian did not distribute the negate to the 3rd terms when he found the additive inverse. To find the additive inverse, all terms should be multiplied by −1.

62. REASONING Determine whether each of the following statements is true or false . Explain your reasoning. a. A binomial can have a degree of zero. b. The order in which polynomials are subtracted does not matter.

SOLUTION: a. If a binomial has two terms that are each a degree of 0, then those terms can be combined and the binomial becomes a monomial. For example, 18 + 7 = 25. If one of the terms of the binomial does not have a degree of 0, then the binomial cannot have a degree of 0, since the degree of a polynomial is the greatest degree of any term in the polynomial. b. Subtraction is not commutative. While 2 + 5 = 5 + 2, 2 – 5 ≠ 5 – 2. This is also true for polynomials. Sample answer: (2x – 3) – (4x – 3) = –2x, but (4x – 3) – (2x – 3) = 2x

63. CHALLENGE Write a polynomial that represents the sum of an odd integer 2n + 1 and the next two consecutive odd integers.

SOLUTION:

64. WRITING IN MATH Why would you add or subtract equations that represent real-world situations? Explain.

SOLUTION:

65. WRITING IN MATH Describe how to add and subtract polynomials using both the vertical and horizontal formats.

SOLUTION: To add polynomials in a horizontal format, you combine like terms. For the vertical format, you write the polynomials in standard form, align like terms in columns, and combine like terms.

To subtract polynomials in a horizontal format you find the additive inverse of the polynomial you are subtracting, andthen combine like terms. For the vertical format, you write the polynomials in standard form, align like terms in columns, and subtract by adding the additive inverse.

66. Three consecutive integers can be represented by x, x + 1, and x + 2. What is the sum of these three integers? A x(x + 1)(x + 2)

B x3 + 3

C 3x + 3 D x + 3

SOLUTION:

The correct choice is C.

67. SHORT RESPONSE What is the perimeter of a square with sides that measure 2x + 3 units?

SOLUTION:

The perimeter of the square is 8x + 12 units.

68. Jim cuts a board in the shape of a regular hexagon and pounds in a nail at each vertex, as shown. How many rubber bands will he need to stretch a rubber band across every possible pair of nails?

F 15 G 14 H 12 J 9

SOLUTION: The first nail would connect to 5 others, the second to 4 others, the third to 3 others, etc.5 + 4 + 3 + 2 + 1 = 15 The correct choice is F.

69. Which ordered pair is in the solution set of the system of inequalities shown in the graph?

A (−3, 0) B (0, −3) C (5, 0) D (0, 5)

SOLUTION: Choice A is outside the shaded area for both inequalities. Choices B and D are inside the shaded area for only one inequality. Choice C is the only point in the solution for both inequalities. So, the correct choice is C.

70. COMPUTERS A computer technician charges by the hour to fix and repair computer equipment. The total cost of the technician for one hour is $75, for two hours is $125, for three hours is $175, for four hours is $225, and so on. Write a recursive formula for the sequence.

SOLUTION: Write out the terms. $75, $125, $175, $225, ... The first term is 75, and 50 is added to form each following term. Therefore, we have a1 = 75, an = an – 1 + 50, n ≥ 2.

Determine whether each sequence is arithmetic, geometric, or neither. Explain.71. 8, –32, 128, –512, ...

SOLUTION: Check for a common difference. –32 – 8 = –40 128 – (–32) = 160 There is no common difference. Check for a common ratio. –32 ÷ 8 = –4 128 ÷ (–32) = –4 Geometric; the common ratio is –4.

72. 25, 8, –9, –26, ...

SOLUTION: Check for a common difference. 8 – 25 = –17 –9 – 8 = –17 Arithmetic; the common difference is –17.

73.

SOLUTION: Check for a common difference.

There is no common difference. Check for a common ratio.

There is no common ratio, so the sequence is not arithmetic or geometric.

74. 43, 52, 61, 70, ...

SOLUTION: Check for a common difference. 52 – 43 = 9 61 – 52 = 9 Arithmetic; the common difference is 9.

75. –27, –16, –5, 6, ...

SOLUTION: Check for a common difference. –16 – (–27) = 11 –5 – (–16) = 11 Arithmetic; the common difference is 11.

76. 200, 100, 50, 25, …

SOLUTION: Check for a common difference. 100 – 200 = –100 50 – 100 = –50 There is no common difference. Check for a common ratio. 100 ÷ 200 = 0.5 50 ÷ 100 = 0.5

Geometric; the common ratio is 0.5 or .

77. JOBS Kimi received an offer for a new job. She wants to compare the offer with her current job. What is total amount of sales that Kimi must get each month to make the same income at either job?

SOLUTION: Let s be Kimi’s monthly sales.

Kimi must sell $80,000 each month to make the same income at either job.

Determine whether each sequence is an arithmetic sequence. If it is, state the common difference.78. 24, 16, 8, 0, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is –8, because 16 – 24 = –8; 8 –16 = –8; etc.

79. , 13, 26, …

SOLUTION: Find the difference between the terms.

– =

13 – =

26 – 13 = 13 There is not a common different. The sequence is not an arithmetic sequence

80. 7, 6, 5, 4, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is –1, because 6 – 7 = –1; 5 – 6 = –1; 4 – 5 = –1;etc.

81. 10, 12, 15, 18, …

SOLUTION: Find the difference between the terms. 12 – 10 = 2 15 – 12 = 3 18 – 15 = 3 There is no common difference. The sequence is not an arithmetic sequence.

82. −15, −11, −7, −3, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is 4, because –11 – (–15) = 4; –7 – (–11) = 4; –3 – (–7) = 4;etc.

83. −0.3, 0.2, 0.7, 1.2, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is 0.5, because 0.2 – (–0.3) = 0.5; 0.7 – 0.2 = 0.5; 1.2 – 0.7 = 0.5; etc.

Simplify.

84. t(t5)(t

7)

SOLUTION:

85. n3(n

2)(−2n

3)

SOLUTION:

86. (5t5v

2)(10t

3v

4)

SOLUTION:

87. (−8u4z

5)(5uz

4)

SOLUTION:

88. [(3)2]3

SOLUTION:

89. [(2)3]2

SOLUTION:

90. (2m4k

3)2(−3mk

2)3

SOLUTION:

91. (6xy2)2(2x

2y

2z

2)3

SOLUTION:

eSolutions Manual - Powered by Cognero Page 29

8-1 Adding and Subtracting Polynomials

Determine whether each expression is a polynomial. If it is a polynomial, find the degree and determine whether it is a monomial, binomial, or trinomial.

1. 7ab + 6b2 – 2a

3

SOLUTION:

A polynomial is a monomial or the sum of monomials. 7ab + 6b2 – 2a

3 is the sum of 3 monomials, so it is a

polynomial. The degree of a polynomial is the greatest degree of any term in the polynomial.The degree of each term is 2, 2, and

3, so the degree of 7ab + 6b2 – 2a

3 is 3. The polynomial has three terms, so it is a trinomial.

2. 2y – 5 + 3y2

SOLUTION:

2y – 5 + 3y2

is the sum of monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 1, 0, and

2, so the degree of 2y – 5 + 3y2 is 2. The polynomial has three terms, so it is a trinomial.

3. 3x2

SOLUTION:

A polynomial is a monomial or the sum of monomials, so 3x2 is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of 3x2 is 2. The

polynomial has one term, so it is a monomial.

4.

SOLUTION: A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer exponents. It has

only one term. is a division of two monomials, so it is not a monomial.

5. 5m2p

3 + 6

SOLUTION:

A polynomial is a monomial or the sum of monomials. 5m2p

3 + 6

is the sum of 2 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 5 and 0,

so the degree of 5m2p

3 + 6 is 5. The polynomial has two terms, so it is a binomial.

6. 5q–4

+ 6q

SOLUTION: A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer exponents. It has

only one term. 5q-4 is equal to , which is a division of two monomials, so it is not a monomial.

Write each polynomial in standard form. Identify the leading coefficient.

7. –4d4 + 1 – d

2

SOLUTION: Find the degree of each term.

–4d4→ 4

1 → 0

–d2 → 2

The greatest degree is 4, from the term –4d4, so the leading coefficient of –4d

4 + 1 – d

2 is –4.

Rewrite the polynomial with each monomial in descending order according to degree.

–4d4 – d

2 + 1

8. 2x5 – 12 + 3x

SOLUTION: Find the degree of each term.

2x5 → 5

–12 → 0 3x → 1

The greatest degree is 5, from the term 2x5, so the leading coefficient of 2x

5 – 12 + 3x is 2.

Rewrite the polynomial with each monomial in descending order according to degree.

2x5 + 3x – 12

9. 4z – 2z2 – 5z

4

SOLUTION: Find the degree of each term. 4z → 1

– 2z2 → 2

– 5z4→ 4

The greatest degree is 4, from the term – 5z4, so the leading coefficient of 4z – 2z

2 – 5z

4 is –5.

Rewrite the polynomial with each monomial in descending order according to degree.

–5z4 – 2z

2 + 4z

10. 2a + 4a3 – 5a

2 – 1

SOLUTION: Find the degree of each term. 2a → 1

4a3→ 3

– 5a2 → 2

– 1 → 0

The greatest degree is 3, from the term 4a3, so the leading coefficient of 2a + 4a

3 – 5a

2 – 1 is 4.

Rewrite the polynomial with each monomial in descending order according to degree.

4a3– 5a

2 + 2a – 1

Find each sum or difference.

11. (6x3 − 4) + (−2x

3 + 9)

SOLUTION:

12. (g3 − 2g2 + 5g + 6) − (g2 + 2g)

SOLUTION:

13. (4 + 2a2 − 2a) − (3a

2 − 8a + 7)

SOLUTION:

14. (8y − 4y2) + (3y − 9y

2)

SOLUTION:

15. (−4z3 − 2z + 8) − (4z

3 + 3z2 − 5)

SOLUTION:

16. (−3d2 − 8 + 2d) + (4d − 12 + d2

)

SOLUTION:

17. (y + 5) + (2y + 4y2 – 2)

SOLUTION:

18. (3n3 − 5n + n2

) − (−8n2 + 3n

3)

SOLUTION:

19. CCSS SENSE-MAKING The total number of students T who traveled for spring break consists of two groups: students who flew to their destinations F and students who drove to their destination D. The number (in thousands) of students who flew and the total number of students who flew or drove can be modeled by the following equations,where n is the number of years since 1995. T = 14n + 21

F = 8n + 7 a. Write an equation that models the number of students who drove to their destination for this time period. b. Predict the number of students who will drive to their destination in 2012. c. How many students will drive or fly to their destination in 2015?

SOLUTION: a.

D = 6n + 14 b. n = 2012 – 1995 = 17

The number of students who will drive to their destination in 2012 is 116,000 students. c. n = 2015 – 1995 = 20

The number of students who will drive or fly to their destination in 2015 is 301,000 students.

Determine whether each expression is a polynomial. If it is a polynomial, find the degree and determine whether it is a monomial, binomial, or trinomial.

20.

SOLUTION: A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer exponents. It has

only one term. is a division of two monomials, so it is not a monomial.

21.

SOLUTION: A polynomial is a monomial or the sum of monomials. 21 is a monomial, so it is also a polynomial. The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of 21 is 0. The polynomial has only one term, so it is a monomial.

22. c4 – 2c

2 + 1

SOLUTION:

A polynomial is a monomial or the sum of monomials. c4 – 2c

2 + 1

is the sum of 3 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 4, 2, and

0, so the degree of c4 – 2c2 + 1 is 4. The polynomial has three terms, so it is a trinomial.

23. d + 3dc

SOLUTION: A polynomial is a monomial or the sum of monomials. A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer

exponents. 3dc

has a variable in the exponent, so it is not a monomial.

24. a – a2

SOLUTION:

A polynomial is a monomial or the sum of monomials. a – a2 is the sum of 2 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 1 and 2,

so the degree of a – a2 is 2. The polynomial has two terms, so it is a binomial.

25. 5n3 + nq

3

SOLUTION:

A polynomial is a monomial or the sum of monomials. 5n3 + nq3 is the sum of 2 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 3 and 4,

so the degree of 5n3 + nq3 is 4. The polynomial has two terms, so it is a binomial.

Write each polynomial in standard form. Identify the leading coefficient.

26. 5x2 – 2 + 3x

SOLUTION: Find the degree of each term.

5x2 → 2

– 2 → 0

3x → 1

The greatest degree is 2, from the term 5x2, so the leading coefficient of 5x

2 – 2 + 3x is 5.

Rewrite the polynomial with each monomial in descending order according to degree.

5x2 + 3x – 2

27. 8y + 7y3

SOLUTION: Find the degree of each term.

7y3 → 3

8y → 1

The greatest degree is 3, from the term 7y3, so the leading coefficient of 8y + 7y

3 is 7.

Rewrite the polynomial with each monomial in descending order according to degree.

7y3 + 8y

28. 4 – 3c – 5c2

SOLUTION: Find the degree of each term. 4 → 0 3c → 1

– 5c2 → 2

The greatest degree is 2, from the term – 5c2, so the leading coefficient of 4 – 3c – 5c

2 is –5.

Rewrite the polynomial with each monomial in descending order according to degree.

–5c2 – 3c + 4

29. –y3 + 3y – 3y

2 + 2

SOLUTION: Find the degree of each term.

–y3 → 3

3y → 1

3y2 → 2

2 → 0

The greatest degree is 3, from the term –y3, so the leading coefficient of –y

3 + 3y – 3y

2 + 2 is –1.

Rewrite the polynomial with each monomial in descending order according to degree. –y 3 – 3y 2 + 3y + 2

30. 11t + 2t2 – 3 + t

5

SOLUTION: Find the degree of each term. 11t → 1

2t2 → 2

–3 → 0

t5 → 5

The greatest degree is 5, from the term t5, so the leading coefficient of 11t + 2t2 – 3 + t

5 is 1.

Rewrite the polynomial with each monomial in descending order according to degree.

t5 + 2t

2 + 11t – 3

31. 2 + r – r3

SOLUTION: Find the degree of each term. 2 → 0 r → 1

– r3→ 3

The greatest degree is 3, from the term – r3, so the leading coefficient of 2 + r – r3

is –1. Rewrite the polynomial with each monomial in descending order according to degree.

–r3 + r + 2

32.

SOLUTION: Find the degree of each term.

→ 0

–3x4 → 4

7 → 0

The greatest degree is 4, from the term –3x4, so the leading coefficient of is –3.

Rewrite the polynomial with each monomial in descending order according to degree.

33. –9b2 + 10b – b

6

SOLUTION: Find the degree of each term.

–9b2 → 2

10b → 1

–b6→ 6

The greatest degree is 6, from the term –b6, so the leading coefficient of –9b2 + 10b – b6 is –1. Rewrite the polynomial with each monomial in descending order according to degree.

–b6 – 9b

2 + 10b

Find each sum or difference.

34. (2c2 + 6c + 4) + (5c – 7)

SOLUTION:

35. (2x + 3x2) − (7 − 8x

2)

SOLUTION:

36. (3c3 − c + 11) − (c2 + 2c + 8)

SOLUTION:

37. (z2 + z) + (z

2 − 11)

SOLUTION:

38. (2x − 2y + 1) − (3y + 4x)

SOLUTION:

39. (4a − 5b2 + 3) + (6 − 2a + 3b

2)

SOLUTION:

40. (x2y − 3x

2 + y) + (3y − 2x2y)

SOLUTION:

41. (−8xy + 3x2 − 5y) + (4x

2 − 2y + 6xy)

SOLUTION:

42. (5n − 2p2 + 2np) − (4p

2 + 4n)

SOLUTION:

43. (4rxt − 8r2x + x2

) − (6rx2 + 5rxt − 2x

2)

SOLUTION:

44. PETS From 1999 through 2009, the number of dogs D and the number of cats C (in hundreds) adopted from animal

shelters in the United States are modeled by the equations D = 2n + 3 and C = n + 4, where n is the number of yearssince 1999. a. Write an equation that models the total number T of dogs and cats adopted in hundreds for this time period. b. If this trend continues, how many dogs and cats will be adopted in 2013?

SOLUTION: a.

So, an equation that models the total number of dogs and cats adopted is T = 3n + 7. b. Evaluate the equation for the total number of dogs and cats for n = 2013 – 1999 = 14.

The number of cats and dogs adopted in 2013 will be 49 × 100 or 4900 cats and dogs.

Classify each polynomial according to its degree and number of terms.

45. 4x – 3x2 + 5

SOLUTION:

Find the degree of each term of 4x – 3x2 + 5.

4x → 1

– 3x2 → 2

5 → 0

The greatest degree is 2 and there are 3 terms, so 4x – 3x2 + 5 is a quadratic trinomial.

46. 11z3

SOLUTION: Find the degree of each term. 11z3

→ 3

The greatest degree is 3 and there is one term, so 11z3 is a cubic monomial.

47. 9 + y4

SOLUTION:

Find the degree of each term of 9 + y 4.

9 → 1

y4→ 4

The greatest degree is 4 and there are 2 terms, so 9 + y 4 is a quartic binomial.

48. 3x3 – 7

SOLUTION:

Find the degree of each term of 3x3 – 7.

3x3 → 3

–7 → 0

The greatest degree is 3 and there are 2 terms, so 3x3 – 7 is a cubic binomial.

49. –2x5 – x

2 + 5x – 8

SOLUTION:

Find the degree of each term of –2x5 – x2 + 5x – 8.

–2x5 → 5

–x2 → 2

5x → 1 – 8 → 0

The greatest degree is 5 and there are 4 terms, so –2x5 – x2 + 5x – 8 is a quintic polynomial.

50. 10t – 4t2 + 6t

3

SOLUTION:

Find the degree of each term of 10t – 4t2 + 6t

3.

10t → 1

4t2 → 2

6t3 → 3

The greatest degree is 3 and there are 3 terms, so 10t – 4t2 + 6t

3 is a cubic trinomial.

51. ENROLLMENT In a rapidly growing school system, the numbers (in hundreds) of total students N and

K-5 students P enrolled from 2000 to 2009 are modeled by the equations N = 1.25t2 – t + 7.5 and P =

0.7t2 – 0.95t + 3.8, where t is the number of years since 2000.

a. Write an equation modeling the number of 6-12 students S enrolled for this time period. b. How many 6-12 students were enrolled in the school system in 2007?

SOLUTION: a. To write an equation that represents the number of 6-12 students enrolled, subtract the equations that represent the total number of students and the number of K-5 students.

b. Replace t with 7 in the equation for S to determine the number of students enrolled in 6-12 in 2007.

In 2007 there were 30.3 hundreds or 3030 students enrolled in 6-12.

52. CCSS REASONING The perimeter of the figure shown is represented by the expression 3x2 − 7x + 2. Write a

polynomial that represents the measure of the third side.

SOLUTION:

53. GEOMETRY Consider the rectangle.

a. What does (4x2 + 2x – 1)(2x

2 – x + 3) represent?

b. What does 2(4x2 + 2x – 1) + 2(2x

2 – x + 3) represent?

SOLUTION:

a. (4x2 + 2x – 1)(2x

2 – x + 3) is a multiplication of the length and the width of the rectangle, which is the formula for

the area of a rectangle.

b. 2(4x2 + 2x – 1) + 2(2x

2 – x + 3) is the sum of twice the length and twice the width of a rectangle, which is the

formula for the perimeter of the rectangle

Find each sum or difference.

54. (4x + 2y − 6z) + (5y − 2z + 7x) + (−9z − 2x − 3y)

SOLUTION:

55. (5a2 − 4) + (a

2 − 2a + 12) + (4a2 − 6a + 8)

SOLUTION:

56. (3c2 − 7) + (4c + 7) − (c

2 + 5c − 8)

SOLUTION:

57. (3n3 + 3n − 10) − (4n

2 − 5n) + (4n3 − 3n

2 − 9n + 4)

SOLUTION:

58. FOOTBALL The National Football League is divided into two conferences, the American A and the National N. From 2002 through 2009, the total attendance T (in thousands) for both conferences and for the American Conference games are modeled by the following equations, where x is the number of years since 2002.

T = –0.69x3 + 55.83x

2 + 643.31x + 10,538

A = –3.78x3 + 58.96x

2 + 265.96x + 5257

Determine how many people attended National Conference football games in 2009.

SOLUTION:

Let x = 7 represent 2009, then find how many people attended a National Conference football game in 2009.

In 2009 the number of people who attended National Conference football games was about 8829 thousand. Multiply by 1000 to find the attendance in standard form. 8829 ×1000 = 8,829,000. So, about 8,829,000 people attended National Conference football games in 2009.

59. CAR RENTAL The cost to rent a car for a day is $15 plus $0.15 for each mile driven. a. Write a polynomial that represents the cost of renting a car for m miles. b. If a car is driven 145 miles, how much would it cost to rent? c. If a car is driven 105 miles each day for four days, how much would it cost to rent a car? d. If a car is driven 220 miles each day for seven days, how much would it cost to rent a car?

SOLUTION: a. The cost to rent a car is the daily rate and the mileage cost or 15 + 0.15m. b. Substitute 145 for m to find the daily cost to drive 145 miles.

The cost to rent the car would be $36.75. c. The expression represents the cost per day. For 4 days, multiply the entire expression by 4 to find the cost for a 4 day trip. Substitute 105 for m.

The cost to rent the car would be $123. d. The expression represents the cost per day. For 7 days, multiply the entire expression by 7 to find the cost for a 7 day trip. Substitute 220 for m.

The cost to rent the car would be $336.

60. MULTIPLE REPRESENTATIONS In this problem, you will explore perimeter and area. a. Geometric Draw three rectangles that each have a perimeter of 400 feet. b. Tabular Record the width and length of each rectangle in a table like the one shown below. Find the area of eachrectangle.

c. Graphical On a coordinate system, graph the area of rectangle 4 in terms of the length, x. Use the graph to determine the largest area possible. d. Analytical Determine the length and width that produce the largest area.

SOLUTION: a. For the perimeters to be 400, create lengths and widths that sum to 200. Note that some lengths are already provided in the table in part b.

b. The area is length multiplied by width. Make sure the units are squared for area. The sum of the length and width must be 200, so if the length is x, the width must be 200 – x.

c. The length will be the x-values on the horizontal and the area will be the y-values on the vertical. The length cannot pass 200 since the sum of the length and width is 200. Set the intervals for the x-axis to 25 feet. Extend the table of values to find more points to plot on the graph. It appears that 10,000 is the greatest area, so set the intervals

for the y-axis to 1000 ft2. After graphing, it appears that the highest point on the graph is at an area of 10,000 ft

2.

d. The associated x-value with the maximum area is x = 100, so the length must be 100 and the width must be 200 – 100, or 100. The length and width of the rectangle must be 100 feet each to have the largest area.

61. CCSS CRITIQUE Cheyenne and Sebastian are finding (2x2 − x) − (3x + 3x

2 − 2). Is either of them correct? Explain your reasoning.

SOLUTION:

Neither is correct. Cheyenne, did not distribute the negative to the 2nd and 3rd terms when she found the additive inverse. Sebastian did not distribute the negate to the 3rd terms when he found the additive inverse. To find the additive inverse, all terms should be multiplied by −1.

62. REASONING Determine whether each of the following statements is true or false . Explain your reasoning. a. A binomial can have a degree of zero. b. The order in which polynomials are subtracted does not matter.

SOLUTION: a. If a binomial has two terms that are each a degree of 0, then those terms can be combined and the binomial becomes a monomial. For example, 18 + 7 = 25. If one of the terms of the binomial does not have a degree of 0, then the binomial cannot have a degree of 0, since the degree of a polynomial is the greatest degree of any term in the polynomial. b. Subtraction is not commutative. While 2 + 5 = 5 + 2, 2 – 5 ≠ 5 – 2. This is also true for polynomials. Sample answer: (2x – 3) – (4x – 3) = –2x, but (4x – 3) – (2x – 3) = 2x

63. CHALLENGE Write a polynomial that represents the sum of an odd integer 2n + 1 and the next two consecutive odd integers.

SOLUTION:

64. WRITING IN MATH Why would you add or subtract equations that represent real-world situations? Explain.

SOLUTION:

65. WRITING IN MATH Describe how to add and subtract polynomials using both the vertical and horizontal formats.

SOLUTION: To add polynomials in a horizontal format, you combine like terms. For the vertical format, you write the polynomials in standard form, align like terms in columns, and combine like terms.

To subtract polynomials in a horizontal format you find the additive inverse of the polynomial you are subtracting, andthen combine like terms. For the vertical format, you write the polynomials in standard form, align like terms in columns, and subtract by adding the additive inverse.

66. Three consecutive integers can be represented by x, x + 1, and x + 2. What is the sum of these three integers? A x(x + 1)(x + 2)

B x3 + 3

C 3x + 3 D x + 3

SOLUTION:

The correct choice is C.

67. SHORT RESPONSE What is the perimeter of a square with sides that measure 2x + 3 units?

SOLUTION:

The perimeter of the square is 8x + 12 units.

68. Jim cuts a board in the shape of a regular hexagon and pounds in a nail at each vertex, as shown. How many rubber bands will he need to stretch a rubber band across every possible pair of nails?

F 15 G 14 H 12 J 9

SOLUTION: The first nail would connect to 5 others, the second to 4 others, the third to 3 others, etc.5 + 4 + 3 + 2 + 1 = 15 The correct choice is F.

69. Which ordered pair is in the solution set of the system of inequalities shown in the graph?

A (−3, 0) B (0, −3) C (5, 0) D (0, 5)

SOLUTION: Choice A is outside the shaded area for both inequalities. Choices B and D are inside the shaded area for only one inequality. Choice C is the only point in the solution for both inequalities. So, the correct choice is C.

70. COMPUTERS A computer technician charges by the hour to fix and repair computer equipment. The total cost of the technician for one hour is $75, for two hours is $125, for three hours is $175, for four hours is $225, and so on. Write a recursive formula for the sequence.

SOLUTION: Write out the terms. $75, $125, $175, $225, ... The first term is 75, and 50 is added to form each following term. Therefore, we have a1 = 75, an = an – 1 + 50, n ≥ 2.

Determine whether each sequence is arithmetic, geometric, or neither. Explain.71. 8, –32, 128, –512, ...

SOLUTION: Check for a common difference. –32 – 8 = –40 128 – (–32) = 160 There is no common difference. Check for a common ratio. –32 ÷ 8 = –4 128 ÷ (–32) = –4 Geometric; the common ratio is –4.

72. 25, 8, –9, –26, ...

SOLUTION: Check for a common difference. 8 – 25 = –17 –9 – 8 = –17 Arithmetic; the common difference is –17.

73.

SOLUTION: Check for a common difference.

There is no common difference. Check for a common ratio.

There is no common ratio, so the sequence is not arithmetic or geometric.

74. 43, 52, 61, 70, ...

SOLUTION: Check for a common difference. 52 – 43 = 9 61 – 52 = 9 Arithmetic; the common difference is 9.

75. –27, –16, –5, 6, ...

SOLUTION: Check for a common difference. –16 – (–27) = 11 –5 – (–16) = 11 Arithmetic; the common difference is 11.

76. 200, 100, 50, 25, …

SOLUTION: Check for a common difference. 100 – 200 = –100 50 – 100 = –50 There is no common difference. Check for a common ratio. 100 ÷ 200 = 0.5 50 ÷ 100 = 0.5

Geometric; the common ratio is 0.5 or .

77. JOBS Kimi received an offer for a new job. She wants to compare the offer with her current job. What is total amount of sales that Kimi must get each month to make the same income at either job?

SOLUTION: Let s be Kimi’s monthly sales.

Kimi must sell $80,000 each month to make the same income at either job.

Determine whether each sequence is an arithmetic sequence. If it is, state the common difference.78. 24, 16, 8, 0, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is –8, because 16 – 24 = –8; 8 –16 = –8; etc.

79. , 13, 26, …

SOLUTION: Find the difference between the terms.

– =

13 – =

26 – 13 = 13 There is not a common different. The sequence is not an arithmetic sequence

80. 7, 6, 5, 4, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is –1, because 6 – 7 = –1; 5 – 6 = –1; 4 – 5 = –1;etc.

81. 10, 12, 15, 18, …

SOLUTION: Find the difference between the terms. 12 – 10 = 2 15 – 12 = 3 18 – 15 = 3 There is no common difference. The sequence is not an arithmetic sequence.

82. −15, −11, −7, −3, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is 4, because –11 – (–15) = 4; –7 – (–11) = 4; –3 – (–7) = 4;etc.

83. −0.3, 0.2, 0.7, 1.2, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is 0.5, because 0.2 – (–0.3) = 0.5; 0.7 – 0.2 = 0.5; 1.2 – 0.7 = 0.5; etc.

Simplify.

84. t(t5)(t

7)

SOLUTION:

85. n3(n

2)(−2n

3)

SOLUTION:

86. (5t5v

2)(10t

3v

4)

SOLUTION:

87. (−8u4z

5)(5uz

4)

SOLUTION:

88. [(3)2]3

SOLUTION:

89. [(2)3]2

SOLUTION:

90. (2m4k

3)2(−3mk

2)3

SOLUTION:

91. (6xy2)2(2x

2y

2z

2)3

SOLUTION:

eSolutions Manual - Powered by Cognero Page 30

8-1 Adding and Subtracting Polynomials

Determine whether each expression is a polynomial. If it is a polynomial, find the degree and determine whether it is a monomial, binomial, or trinomial.

1. 7ab + 6b2 – 2a

3

SOLUTION:

A polynomial is a monomial or the sum of monomials. 7ab + 6b2 – 2a

3 is the sum of 3 monomials, so it is a

polynomial. The degree of a polynomial is the greatest degree of any term in the polynomial.The degree of each term is 2, 2, and

3, so the degree of 7ab + 6b2 – 2a

3 is 3. The polynomial has three terms, so it is a trinomial.

2. 2y – 5 + 3y2

SOLUTION:

2y – 5 + 3y2

is the sum of monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 1, 0, and

2, so the degree of 2y – 5 + 3y2 is 2. The polynomial has three terms, so it is a trinomial.

3. 3x2

SOLUTION:

A polynomial is a monomial or the sum of monomials, so 3x2 is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of 3x2 is 2. The

polynomial has one term, so it is a monomial.

4.

SOLUTION: A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer exponents. It has

only one term. is a division of two monomials, so it is not a monomial.

5. 5m2p

3 + 6

SOLUTION:

A polynomial is a monomial or the sum of monomials. 5m2p

3 + 6

is the sum of 2 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 5 and 0,

so the degree of 5m2p

3 + 6 is 5. The polynomial has two terms, so it is a binomial.

6. 5q–4

+ 6q

SOLUTION: A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer exponents. It has

only one term. 5q-4 is equal to , which is a division of two monomials, so it is not a monomial.

Write each polynomial in standard form. Identify the leading coefficient.

7. –4d4 + 1 – d

2

SOLUTION: Find the degree of each term.

–4d4→ 4

1 → 0

–d2 → 2

The greatest degree is 4, from the term –4d4, so the leading coefficient of –4d

4 + 1 – d

2 is –4.

Rewrite the polynomial with each monomial in descending order according to degree.

–4d4 – d

2 + 1

8. 2x5 – 12 + 3x

SOLUTION: Find the degree of each term.

2x5 → 5

–12 → 0 3x → 1

The greatest degree is 5, from the term 2x5, so the leading coefficient of 2x

5 – 12 + 3x is 2.

Rewrite the polynomial with each monomial in descending order according to degree.

2x5 + 3x – 12

9. 4z – 2z2 – 5z

4

SOLUTION: Find the degree of each term. 4z → 1

– 2z2 → 2

– 5z4→ 4

The greatest degree is 4, from the term – 5z4, so the leading coefficient of 4z – 2z

2 – 5z

4 is –5.

Rewrite the polynomial with each monomial in descending order according to degree.

–5z4 – 2z

2 + 4z

10. 2a + 4a3 – 5a

2 – 1

SOLUTION: Find the degree of each term. 2a → 1

4a3→ 3

– 5a2 → 2

– 1 → 0

The greatest degree is 3, from the term 4a3, so the leading coefficient of 2a + 4a

3 – 5a

2 – 1 is 4.

Rewrite the polynomial with each monomial in descending order according to degree.

4a3– 5a

2 + 2a – 1

Find each sum or difference.

11. (6x3 − 4) + (−2x

3 + 9)

SOLUTION:

12. (g3 − 2g2 + 5g + 6) − (g2 + 2g)

SOLUTION:

13. (4 + 2a2 − 2a) − (3a

2 − 8a + 7)

SOLUTION:

14. (8y − 4y2) + (3y − 9y

2)

SOLUTION:

15. (−4z3 − 2z + 8) − (4z

3 + 3z2 − 5)

SOLUTION:

16. (−3d2 − 8 + 2d) + (4d − 12 + d2

)

SOLUTION:

17. (y + 5) + (2y + 4y2 – 2)

SOLUTION:

18. (3n3 − 5n + n2

) − (−8n2 + 3n

3)

SOLUTION:

19. CCSS SENSE-MAKING The total number of students T who traveled for spring break consists of two groups: students who flew to their destinations F and students who drove to their destination D. The number (in thousands) of students who flew and the total number of students who flew or drove can be modeled by the following equations,where n is the number of years since 1995. T = 14n + 21

F = 8n + 7 a. Write an equation that models the number of students who drove to their destination for this time period. b. Predict the number of students who will drive to their destination in 2012. c. How many students will drive or fly to their destination in 2015?

SOLUTION: a.

D = 6n + 14 b. n = 2012 – 1995 = 17

The number of students who will drive to their destination in 2012 is 116,000 students. c. n = 2015 – 1995 = 20

The number of students who will drive or fly to their destination in 2015 is 301,000 students.

Determine whether each expression is a polynomial. If it is a polynomial, find the degree and determine whether it is a monomial, binomial, or trinomial.

20.

SOLUTION: A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer exponents. It has

only one term. is a division of two monomials, so it is not a monomial.

21.

SOLUTION: A polynomial is a monomial or the sum of monomials. 21 is a monomial, so it is also a polynomial. The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of 21 is 0. The polynomial has only one term, so it is a monomial.

22. c4 – 2c

2 + 1

SOLUTION:

A polynomial is a monomial or the sum of monomials. c4 – 2c

2 + 1

is the sum of 3 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 4, 2, and

0, so the degree of c4 – 2c2 + 1 is 4. The polynomial has three terms, so it is a trinomial.

23. d + 3dc

SOLUTION: A polynomial is a monomial or the sum of monomials. A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer

exponents. 3dc

has a variable in the exponent, so it is not a monomial.

24. a – a2

SOLUTION:

A polynomial is a monomial or the sum of monomials. a – a2 is the sum of 2 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 1 and 2,

so the degree of a – a2 is 2. The polynomial has two terms, so it is a binomial.

25. 5n3 + nq

3

SOLUTION:

A polynomial is a monomial or the sum of monomials. 5n3 + nq3 is the sum of 2 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 3 and 4,

so the degree of 5n3 + nq3 is 4. The polynomial has two terms, so it is a binomial.

Write each polynomial in standard form. Identify the leading coefficient.

26. 5x2 – 2 + 3x

SOLUTION: Find the degree of each term.

5x2 → 2

– 2 → 0

3x → 1

The greatest degree is 2, from the term 5x2, so the leading coefficient of 5x

2 – 2 + 3x is 5.

Rewrite the polynomial with each monomial in descending order according to degree.

5x2 + 3x – 2

27. 8y + 7y3

SOLUTION: Find the degree of each term.

7y3 → 3

8y → 1

The greatest degree is 3, from the term 7y3, so the leading coefficient of 8y + 7y

3 is 7.

Rewrite the polynomial with each monomial in descending order according to degree.

7y3 + 8y

28. 4 – 3c – 5c2

SOLUTION: Find the degree of each term. 4 → 0 3c → 1

– 5c2 → 2

The greatest degree is 2, from the term – 5c2, so the leading coefficient of 4 – 3c – 5c

2 is –5.

Rewrite the polynomial with each monomial in descending order according to degree.

–5c2 – 3c + 4

29. –y3 + 3y – 3y

2 + 2

SOLUTION: Find the degree of each term.

–y3 → 3

3y → 1

3y2 → 2

2 → 0

The greatest degree is 3, from the term –y3, so the leading coefficient of –y

3 + 3y – 3y

2 + 2 is –1.

Rewrite the polynomial with each monomial in descending order according to degree. –y 3 – 3y 2 + 3y + 2

30. 11t + 2t2 – 3 + t

5

SOLUTION: Find the degree of each term. 11t → 1

2t2 → 2

–3 → 0

t5 → 5

The greatest degree is 5, from the term t5, so the leading coefficient of 11t + 2t2 – 3 + t

5 is 1.

Rewrite the polynomial with each monomial in descending order according to degree.

t5 + 2t

2 + 11t – 3

31. 2 + r – r3

SOLUTION: Find the degree of each term. 2 → 0 r → 1

– r3→ 3

The greatest degree is 3, from the term – r3, so the leading coefficient of 2 + r – r3

is –1. Rewrite the polynomial with each monomial in descending order according to degree.

–r3 + r + 2

32.

SOLUTION: Find the degree of each term.

→ 0

–3x4 → 4

7 → 0

The greatest degree is 4, from the term –3x4, so the leading coefficient of is –3.

Rewrite the polynomial with each monomial in descending order according to degree.

33. –9b2 + 10b – b

6

SOLUTION: Find the degree of each term.

–9b2 → 2

10b → 1

–b6→ 6

The greatest degree is 6, from the term –b6, so the leading coefficient of –9b2 + 10b – b6 is –1. Rewrite the polynomial with each monomial in descending order according to degree.

–b6 – 9b

2 + 10b

Find each sum or difference.

34. (2c2 + 6c + 4) + (5c – 7)

SOLUTION:

35. (2x + 3x2) − (7 − 8x

2)

SOLUTION:

36. (3c3 − c + 11) − (c2 + 2c + 8)

SOLUTION:

37. (z2 + z) + (z

2 − 11)

SOLUTION:

38. (2x − 2y + 1) − (3y + 4x)

SOLUTION:

39. (4a − 5b2 + 3) + (6 − 2a + 3b

2)

SOLUTION:

40. (x2y − 3x

2 + y) + (3y − 2x2y)

SOLUTION:

41. (−8xy + 3x2 − 5y) + (4x

2 − 2y + 6xy)

SOLUTION:

42. (5n − 2p2 + 2np) − (4p

2 + 4n)

SOLUTION:

43. (4rxt − 8r2x + x2

) − (6rx2 + 5rxt − 2x

2)

SOLUTION:

44. PETS From 1999 through 2009, the number of dogs D and the number of cats C (in hundreds) adopted from animal

shelters in the United States are modeled by the equations D = 2n + 3 and C = n + 4, where n is the number of yearssince 1999. a. Write an equation that models the total number T of dogs and cats adopted in hundreds for this time period. b. If this trend continues, how many dogs and cats will be adopted in 2013?

SOLUTION: a.

So, an equation that models the total number of dogs and cats adopted is T = 3n + 7. b. Evaluate the equation for the total number of dogs and cats for n = 2013 – 1999 = 14.

The number of cats and dogs adopted in 2013 will be 49 × 100 or 4900 cats and dogs.

Classify each polynomial according to its degree and number of terms.

45. 4x – 3x2 + 5

SOLUTION:

Find the degree of each term of 4x – 3x2 + 5.

4x → 1

– 3x2 → 2

5 → 0

The greatest degree is 2 and there are 3 terms, so 4x – 3x2 + 5 is a quadratic trinomial.

46. 11z3

SOLUTION: Find the degree of each term. 11z3

→ 3

The greatest degree is 3 and there is one term, so 11z3 is a cubic monomial.

47. 9 + y4

SOLUTION:

Find the degree of each term of 9 + y 4.

9 → 1

y4→ 4

The greatest degree is 4 and there are 2 terms, so 9 + y 4 is a quartic binomial.

48. 3x3 – 7

SOLUTION:

Find the degree of each term of 3x3 – 7.

3x3 → 3

–7 → 0

The greatest degree is 3 and there are 2 terms, so 3x3 – 7 is a cubic binomial.

49. –2x5 – x

2 + 5x – 8

SOLUTION:

Find the degree of each term of –2x5 – x2 + 5x – 8.

–2x5 → 5

–x2 → 2

5x → 1 – 8 → 0

The greatest degree is 5 and there are 4 terms, so –2x5 – x2 + 5x – 8 is a quintic polynomial.

50. 10t – 4t2 + 6t

3

SOLUTION:

Find the degree of each term of 10t – 4t2 + 6t

3.

10t → 1

4t2 → 2

6t3 → 3

The greatest degree is 3 and there are 3 terms, so 10t – 4t2 + 6t

3 is a cubic trinomial.

51. ENROLLMENT In a rapidly growing school system, the numbers (in hundreds) of total students N and

K-5 students P enrolled from 2000 to 2009 are modeled by the equations N = 1.25t2 – t + 7.5 and P =

0.7t2 – 0.95t + 3.8, where t is the number of years since 2000.

a. Write an equation modeling the number of 6-12 students S enrolled for this time period. b. How many 6-12 students were enrolled in the school system in 2007?

SOLUTION: a. To write an equation that represents the number of 6-12 students enrolled, subtract the equations that represent the total number of students and the number of K-5 students.

b. Replace t with 7 in the equation for S to determine the number of students enrolled in 6-12 in 2007.

In 2007 there were 30.3 hundreds or 3030 students enrolled in 6-12.

52. CCSS REASONING The perimeter of the figure shown is represented by the expression 3x2 − 7x + 2. Write a

polynomial that represents the measure of the third side.

SOLUTION:

53. GEOMETRY Consider the rectangle.

a. What does (4x2 + 2x – 1)(2x

2 – x + 3) represent?

b. What does 2(4x2 + 2x – 1) + 2(2x

2 – x + 3) represent?

SOLUTION:

a. (4x2 + 2x – 1)(2x

2 – x + 3) is a multiplication of the length and the width of the rectangle, which is the formula for

the area of a rectangle.

b. 2(4x2 + 2x – 1) + 2(2x

2 – x + 3) is the sum of twice the length and twice the width of a rectangle, which is the

formula for the perimeter of the rectangle

Find each sum or difference.

54. (4x + 2y − 6z) + (5y − 2z + 7x) + (−9z − 2x − 3y)

SOLUTION:

55. (5a2 − 4) + (a

2 − 2a + 12) + (4a2 − 6a + 8)

SOLUTION:

56. (3c2 − 7) + (4c + 7) − (c

2 + 5c − 8)

SOLUTION:

57. (3n3 + 3n − 10) − (4n

2 − 5n) + (4n3 − 3n

2 − 9n + 4)

SOLUTION:

58. FOOTBALL The National Football League is divided into two conferences, the American A and the National N. From 2002 through 2009, the total attendance T (in thousands) for both conferences and for the American Conference games are modeled by the following equations, where x is the number of years since 2002.

T = –0.69x3 + 55.83x

2 + 643.31x + 10,538

A = –3.78x3 + 58.96x

2 + 265.96x + 5257

Determine how many people attended National Conference football games in 2009.

SOLUTION:

Let x = 7 represent 2009, then find how many people attended a National Conference football game in 2009.

In 2009 the number of people who attended National Conference football games was about 8829 thousand. Multiply by 1000 to find the attendance in standard form. 8829 ×1000 = 8,829,000. So, about 8,829,000 people attended National Conference football games in 2009.

59. CAR RENTAL The cost to rent a car for a day is $15 plus $0.15 for each mile driven. a. Write a polynomial that represents the cost of renting a car for m miles. b. If a car is driven 145 miles, how much would it cost to rent? c. If a car is driven 105 miles each day for four days, how much would it cost to rent a car? d. If a car is driven 220 miles each day for seven days, how much would it cost to rent a car?

SOLUTION: a. The cost to rent a car is the daily rate and the mileage cost or 15 + 0.15m. b. Substitute 145 for m to find the daily cost to drive 145 miles.

The cost to rent the car would be $36.75. c. The expression represents the cost per day. For 4 days, multiply the entire expression by 4 to find the cost for a 4 day trip. Substitute 105 for m.

The cost to rent the car would be $123. d. The expression represents the cost per day. For 7 days, multiply the entire expression by 7 to find the cost for a 7 day trip. Substitute 220 for m.

The cost to rent the car would be $336.

60. MULTIPLE REPRESENTATIONS In this problem, you will explore perimeter and area. a. Geometric Draw three rectangles that each have a perimeter of 400 feet. b. Tabular Record the width and length of each rectangle in a table like the one shown below. Find the area of eachrectangle.

c. Graphical On a coordinate system, graph the area of rectangle 4 in terms of the length, x. Use the graph to determine the largest area possible. d. Analytical Determine the length and width that produce the largest area.

SOLUTION: a. For the perimeters to be 400, create lengths and widths that sum to 200. Note that some lengths are already provided in the table in part b.

b. The area is length multiplied by width. Make sure the units are squared for area. The sum of the length and width must be 200, so if the length is x, the width must be 200 – x.

c. The length will be the x-values on the horizontal and the area will be the y-values on the vertical. The length cannot pass 200 since the sum of the length and width is 200. Set the intervals for the x-axis to 25 feet. Extend the table of values to find more points to plot on the graph. It appears that 10,000 is the greatest area, so set the intervals

for the y-axis to 1000 ft2. After graphing, it appears that the highest point on the graph is at an area of 10,000 ft

2.

d. The associated x-value with the maximum area is x = 100, so the length must be 100 and the width must be 200 – 100, or 100. The length and width of the rectangle must be 100 feet each to have the largest area.

61. CCSS CRITIQUE Cheyenne and Sebastian are finding (2x2 − x) − (3x + 3x

2 − 2). Is either of them correct? Explain your reasoning.

SOLUTION:

Neither is correct. Cheyenne, did not distribute the negative to the 2nd and 3rd terms when she found the additive inverse. Sebastian did not distribute the negate to the 3rd terms when he found the additive inverse. To find the additive inverse, all terms should be multiplied by −1.

62. REASONING Determine whether each of the following statements is true or false . Explain your reasoning. a. A binomial can have a degree of zero. b. The order in which polynomials are subtracted does not matter.

SOLUTION: a. If a binomial has two terms that are each a degree of 0, then those terms can be combined and the binomial becomes a monomial. For example, 18 + 7 = 25. If one of the terms of the binomial does not have a degree of 0, then the binomial cannot have a degree of 0, since the degree of a polynomial is the greatest degree of any term in the polynomial. b. Subtraction is not commutative. While 2 + 5 = 5 + 2, 2 – 5 ≠ 5 – 2. This is also true for polynomials. Sample answer: (2x – 3) – (4x – 3) = –2x, but (4x – 3) – (2x – 3) = 2x

63. CHALLENGE Write a polynomial that represents the sum of an odd integer 2n + 1 and the next two consecutive odd integers.

SOLUTION:

64. WRITING IN MATH Why would you add or subtract equations that represent real-world situations? Explain.

SOLUTION:

65. WRITING IN MATH Describe how to add and subtract polynomials using both the vertical and horizontal formats.

SOLUTION: To add polynomials in a horizontal format, you combine like terms. For the vertical format, you write the polynomials in standard form, align like terms in columns, and combine like terms.

To subtract polynomials in a horizontal format you find the additive inverse of the polynomial you are subtracting, andthen combine like terms. For the vertical format, you write the polynomials in standard form, align like terms in columns, and subtract by adding the additive inverse.

66. Three consecutive integers can be represented by x, x + 1, and x + 2. What is the sum of these three integers? A x(x + 1)(x + 2)

B x3 + 3

C 3x + 3 D x + 3

SOLUTION:

The correct choice is C.

67. SHORT RESPONSE What is the perimeter of a square with sides that measure 2x + 3 units?

SOLUTION:

The perimeter of the square is 8x + 12 units.

68. Jim cuts a board in the shape of a regular hexagon and pounds in a nail at each vertex, as shown. How many rubber bands will he need to stretch a rubber band across every possible pair of nails?

F 15 G 14 H 12 J 9

SOLUTION: The first nail would connect to 5 others, the second to 4 others, the third to 3 others, etc.5 + 4 + 3 + 2 + 1 = 15 The correct choice is F.

69. Which ordered pair is in the solution set of the system of inequalities shown in the graph?

A (−3, 0) B (0, −3) C (5, 0) D (0, 5)

SOLUTION: Choice A is outside the shaded area for both inequalities. Choices B and D are inside the shaded area for only one inequality. Choice C is the only point in the solution for both inequalities. So, the correct choice is C.

70. COMPUTERS A computer technician charges by the hour to fix and repair computer equipment. The total cost of the technician for one hour is $75, for two hours is $125, for three hours is $175, for four hours is $225, and so on. Write a recursive formula for the sequence.

SOLUTION: Write out the terms. $75, $125, $175, $225, ... The first term is 75, and 50 is added to form each following term. Therefore, we have a1 = 75, an = an – 1 + 50, n ≥ 2.

Determine whether each sequence is arithmetic, geometric, or neither. Explain.71. 8, –32, 128, –512, ...

SOLUTION: Check for a common difference. –32 – 8 = –40 128 – (–32) = 160 There is no common difference. Check for a common ratio. –32 ÷ 8 = –4 128 ÷ (–32) = –4 Geometric; the common ratio is –4.

72. 25, 8, –9, –26, ...

SOLUTION: Check for a common difference. 8 – 25 = –17 –9 – 8 = –17 Arithmetic; the common difference is –17.

73.

SOLUTION: Check for a common difference.

There is no common difference. Check for a common ratio.

There is no common ratio, so the sequence is not arithmetic or geometric.

74. 43, 52, 61, 70, ...

SOLUTION: Check for a common difference. 52 – 43 = 9 61 – 52 = 9 Arithmetic; the common difference is 9.

75. –27, –16, –5, 6, ...

SOLUTION: Check for a common difference. –16 – (–27) = 11 –5 – (–16) = 11 Arithmetic; the common difference is 11.

76. 200, 100, 50, 25, …

SOLUTION: Check for a common difference. 100 – 200 = –100 50 – 100 = –50 There is no common difference. Check for a common ratio. 100 ÷ 200 = 0.5 50 ÷ 100 = 0.5

Geometric; the common ratio is 0.5 or .

77. JOBS Kimi received an offer for a new job. She wants to compare the offer with her current job. What is total amount of sales that Kimi must get each month to make the same income at either job?

SOLUTION: Let s be Kimi’s monthly sales.

Kimi must sell $80,000 each month to make the same income at either job.

Determine whether each sequence is an arithmetic sequence. If it is, state the common difference.78. 24, 16, 8, 0, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is –8, because 16 – 24 = –8; 8 –16 = –8; etc.

79. , 13, 26, …

SOLUTION: Find the difference between the terms.

– =

13 – =

26 – 13 = 13 There is not a common different. The sequence is not an arithmetic sequence

80. 7, 6, 5, 4, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is –1, because 6 – 7 = –1; 5 – 6 = –1; 4 – 5 = –1;etc.

81. 10, 12, 15, 18, …

SOLUTION: Find the difference between the terms. 12 – 10 = 2 15 – 12 = 3 18 – 15 = 3 There is no common difference. The sequence is not an arithmetic sequence.

82. −15, −11, −7, −3, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is 4, because –11 – (–15) = 4; –7 – (–11) = 4; –3 – (–7) = 4;etc.

83. −0.3, 0.2, 0.7, 1.2, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is 0.5, because 0.2 – (–0.3) = 0.5; 0.7 – 0.2 = 0.5; 1.2 – 0.7 = 0.5; etc.

Simplify.

84. t(t5)(t

7)

SOLUTION:

85. n3(n

2)(−2n

3)

SOLUTION:

86. (5t5v

2)(10t

3v

4)

SOLUTION:

87. (−8u4z

5)(5uz

4)

SOLUTION:

88. [(3)2]3

SOLUTION:

89. [(2)3]2

SOLUTION:

90. (2m4k

3)2(−3mk

2)3

SOLUTION:

91. (6xy2)2(2x

2y

2z

2)3

SOLUTION:

eSolutions Manual - Powered by Cognero Page 31

8-1 Adding and Subtracting Polynomials

Determine whether each expression is a polynomial. If it is a polynomial, find the degree and determine whether it is a monomial, binomial, or trinomial.

1. 7ab + 6b2 – 2a

3

SOLUTION:

A polynomial is a monomial or the sum of monomials. 7ab + 6b2 – 2a

3 is the sum of 3 monomials, so it is a

polynomial. The degree of a polynomial is the greatest degree of any term in the polynomial.The degree of each term is 2, 2, and

3, so the degree of 7ab + 6b2 – 2a

3 is 3. The polynomial has three terms, so it is a trinomial.

2. 2y – 5 + 3y2

SOLUTION:

2y – 5 + 3y2

is the sum of monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 1, 0, and

2, so the degree of 2y – 5 + 3y2 is 2. The polynomial has three terms, so it is a trinomial.

3. 3x2

SOLUTION:

A polynomial is a monomial or the sum of monomials, so 3x2 is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of 3x2 is 2. The

polynomial has one term, so it is a monomial.

4.

SOLUTION: A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer exponents. It has

only one term. is a division of two monomials, so it is not a monomial.

5. 5m2p

3 + 6

SOLUTION:

A polynomial is a monomial or the sum of monomials. 5m2p

3 + 6

is the sum of 2 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 5 and 0,

so the degree of 5m2p

3 + 6 is 5. The polynomial has two terms, so it is a binomial.

6. 5q–4

+ 6q

SOLUTION: A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer exponents. It has

only one term. 5q-4 is equal to , which is a division of two monomials, so it is not a monomial.

Write each polynomial in standard form. Identify the leading coefficient.

7. –4d4 + 1 – d

2

SOLUTION: Find the degree of each term.

–4d4→ 4

1 → 0

–d2 → 2

The greatest degree is 4, from the term –4d4, so the leading coefficient of –4d

4 + 1 – d

2 is –4.

Rewrite the polynomial with each monomial in descending order according to degree.

–4d4 – d

2 + 1

8. 2x5 – 12 + 3x

SOLUTION: Find the degree of each term.

2x5 → 5

–12 → 0 3x → 1

The greatest degree is 5, from the term 2x5, so the leading coefficient of 2x

5 – 12 + 3x is 2.

Rewrite the polynomial with each monomial in descending order according to degree.

2x5 + 3x – 12

9. 4z – 2z2 – 5z

4

SOLUTION: Find the degree of each term. 4z → 1

– 2z2 → 2

– 5z4→ 4

The greatest degree is 4, from the term – 5z4, so the leading coefficient of 4z – 2z

2 – 5z

4 is –5.

Rewrite the polynomial with each monomial in descending order according to degree.

–5z4 – 2z

2 + 4z

10. 2a + 4a3 – 5a

2 – 1

SOLUTION: Find the degree of each term. 2a → 1

4a3→ 3

– 5a2 → 2

– 1 → 0

The greatest degree is 3, from the term 4a3, so the leading coefficient of 2a + 4a

3 – 5a

2 – 1 is 4.

Rewrite the polynomial with each monomial in descending order according to degree.

4a3– 5a

2 + 2a – 1

Find each sum or difference.

11. (6x3 − 4) + (−2x

3 + 9)

SOLUTION:

12. (g3 − 2g2 + 5g + 6) − (g2 + 2g)

SOLUTION:

13. (4 + 2a2 − 2a) − (3a

2 − 8a + 7)

SOLUTION:

14. (8y − 4y2) + (3y − 9y

2)

SOLUTION:

15. (−4z3 − 2z + 8) − (4z

3 + 3z2 − 5)

SOLUTION:

16. (−3d2 − 8 + 2d) + (4d − 12 + d2

)

SOLUTION:

17. (y + 5) + (2y + 4y2 – 2)

SOLUTION:

18. (3n3 − 5n + n2

) − (−8n2 + 3n

3)

SOLUTION:

19. CCSS SENSE-MAKING The total number of students T who traveled for spring break consists of two groups: students who flew to their destinations F and students who drove to their destination D. The number (in thousands) of students who flew and the total number of students who flew or drove can be modeled by the following equations,where n is the number of years since 1995. T = 14n + 21

F = 8n + 7 a. Write an equation that models the number of students who drove to their destination for this time period. b. Predict the number of students who will drive to their destination in 2012. c. How many students will drive or fly to their destination in 2015?

SOLUTION: a.

D = 6n + 14 b. n = 2012 – 1995 = 17

The number of students who will drive to their destination in 2012 is 116,000 students. c. n = 2015 – 1995 = 20

The number of students who will drive or fly to their destination in 2015 is 301,000 students.

Determine whether each expression is a polynomial. If it is a polynomial, find the degree and determine whether it is a monomial, binomial, or trinomial.

20.

SOLUTION: A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer exponents. It has

only one term. is a division of two monomials, so it is not a monomial.

21.

SOLUTION: A polynomial is a monomial or the sum of monomials. 21 is a monomial, so it is also a polynomial. The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of 21 is 0. The polynomial has only one term, so it is a monomial.

22. c4 – 2c

2 + 1

SOLUTION:

A polynomial is a monomial or the sum of monomials. c4 – 2c

2 + 1

is the sum of 3 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 4, 2, and

0, so the degree of c4 – 2c2 + 1 is 4. The polynomial has three terms, so it is a trinomial.

23. d + 3dc

SOLUTION: A polynomial is a monomial or the sum of monomials. A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer

exponents. 3dc

has a variable in the exponent, so it is not a monomial.

24. a – a2

SOLUTION:

A polynomial is a monomial or the sum of monomials. a – a2 is the sum of 2 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 1 and 2,

so the degree of a – a2 is 2. The polynomial has two terms, so it is a binomial.

25. 5n3 + nq

3

SOLUTION:

A polynomial is a monomial or the sum of monomials. 5n3 + nq3 is the sum of 2 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 3 and 4,

so the degree of 5n3 + nq3 is 4. The polynomial has two terms, so it is a binomial.

Write each polynomial in standard form. Identify the leading coefficient.

26. 5x2 – 2 + 3x

SOLUTION: Find the degree of each term.

5x2 → 2

– 2 → 0

3x → 1

The greatest degree is 2, from the term 5x2, so the leading coefficient of 5x

2 – 2 + 3x is 5.

Rewrite the polynomial with each monomial in descending order according to degree.

5x2 + 3x – 2

27. 8y + 7y3

SOLUTION: Find the degree of each term.

7y3 → 3

8y → 1

The greatest degree is 3, from the term 7y3, so the leading coefficient of 8y + 7y

3 is 7.

Rewrite the polynomial with each monomial in descending order according to degree.

7y3 + 8y

28. 4 – 3c – 5c2

SOLUTION: Find the degree of each term. 4 → 0 3c → 1

– 5c2 → 2

The greatest degree is 2, from the term – 5c2, so the leading coefficient of 4 – 3c – 5c

2 is –5.

Rewrite the polynomial with each monomial in descending order according to degree.

–5c2 – 3c + 4

29. –y3 + 3y – 3y

2 + 2

SOLUTION: Find the degree of each term.

–y3 → 3

3y → 1

3y2 → 2

2 → 0

The greatest degree is 3, from the term –y3, so the leading coefficient of –y

3 + 3y – 3y

2 + 2 is –1.

Rewrite the polynomial with each monomial in descending order according to degree. –y 3 – 3y 2 + 3y + 2

30. 11t + 2t2 – 3 + t

5

SOLUTION: Find the degree of each term. 11t → 1

2t2 → 2

–3 → 0

t5 → 5

The greatest degree is 5, from the term t5, so the leading coefficient of 11t + 2t2 – 3 + t

5 is 1.

Rewrite the polynomial with each monomial in descending order according to degree.

t5 + 2t

2 + 11t – 3

31. 2 + r – r3

SOLUTION: Find the degree of each term. 2 → 0 r → 1

– r3→ 3

The greatest degree is 3, from the term – r3, so the leading coefficient of 2 + r – r3

is –1. Rewrite the polynomial with each monomial in descending order according to degree.

–r3 + r + 2

32.

SOLUTION: Find the degree of each term.

→ 0

–3x4 → 4

7 → 0

The greatest degree is 4, from the term –3x4, so the leading coefficient of is –3.

Rewrite the polynomial with each monomial in descending order according to degree.

33. –9b2 + 10b – b

6

SOLUTION: Find the degree of each term.

–9b2 → 2

10b → 1

–b6→ 6

The greatest degree is 6, from the term –b6, so the leading coefficient of –9b2 + 10b – b6 is –1. Rewrite the polynomial with each monomial in descending order according to degree.

–b6 – 9b

2 + 10b

Find each sum or difference.

34. (2c2 + 6c + 4) + (5c – 7)

SOLUTION:

35. (2x + 3x2) − (7 − 8x

2)

SOLUTION:

36. (3c3 − c + 11) − (c2 + 2c + 8)

SOLUTION:

37. (z2 + z) + (z

2 − 11)

SOLUTION:

38. (2x − 2y + 1) − (3y + 4x)

SOLUTION:

39. (4a − 5b2 + 3) + (6 − 2a + 3b

2)

SOLUTION:

40. (x2y − 3x

2 + y) + (3y − 2x2y)

SOLUTION:

41. (−8xy + 3x2 − 5y) + (4x

2 − 2y + 6xy)

SOLUTION:

42. (5n − 2p2 + 2np) − (4p

2 + 4n)

SOLUTION:

43. (4rxt − 8r2x + x2

) − (6rx2 + 5rxt − 2x

2)

SOLUTION:

44. PETS From 1999 through 2009, the number of dogs D and the number of cats C (in hundreds) adopted from animal

shelters in the United States are modeled by the equations D = 2n + 3 and C = n + 4, where n is the number of yearssince 1999. a. Write an equation that models the total number T of dogs and cats adopted in hundreds for this time period. b. If this trend continues, how many dogs and cats will be adopted in 2013?

SOLUTION: a.

So, an equation that models the total number of dogs and cats adopted is T = 3n + 7. b. Evaluate the equation for the total number of dogs and cats for n = 2013 – 1999 = 14.

The number of cats and dogs adopted in 2013 will be 49 × 100 or 4900 cats and dogs.

Classify each polynomial according to its degree and number of terms.

45. 4x – 3x2 + 5

SOLUTION:

Find the degree of each term of 4x – 3x2 + 5.

4x → 1

– 3x2 → 2

5 → 0

The greatest degree is 2 and there are 3 terms, so 4x – 3x2 + 5 is a quadratic trinomial.

46. 11z3

SOLUTION: Find the degree of each term. 11z3

→ 3

The greatest degree is 3 and there is one term, so 11z3 is a cubic monomial.

47. 9 + y4

SOLUTION:

Find the degree of each term of 9 + y 4.

9 → 1

y4→ 4

The greatest degree is 4 and there are 2 terms, so 9 + y 4 is a quartic binomial.

48. 3x3 – 7

SOLUTION:

Find the degree of each term of 3x3 – 7.

3x3 → 3

–7 → 0

The greatest degree is 3 and there are 2 terms, so 3x3 – 7 is a cubic binomial.

49. –2x5 – x

2 + 5x – 8

SOLUTION:

Find the degree of each term of –2x5 – x2 + 5x – 8.

–2x5 → 5

–x2 → 2

5x → 1 – 8 → 0

The greatest degree is 5 and there are 4 terms, so –2x5 – x2 + 5x – 8 is a quintic polynomial.

50. 10t – 4t2 + 6t

3

SOLUTION:

Find the degree of each term of 10t – 4t2 + 6t

3.

10t → 1

4t2 → 2

6t3 → 3

The greatest degree is 3 and there are 3 terms, so 10t – 4t2 + 6t

3 is a cubic trinomial.

51. ENROLLMENT In a rapidly growing school system, the numbers (in hundreds) of total students N and

K-5 students P enrolled from 2000 to 2009 are modeled by the equations N = 1.25t2 – t + 7.5 and P =

0.7t2 – 0.95t + 3.8, where t is the number of years since 2000.

a. Write an equation modeling the number of 6-12 students S enrolled for this time period. b. How many 6-12 students were enrolled in the school system in 2007?

SOLUTION: a. To write an equation that represents the number of 6-12 students enrolled, subtract the equations that represent the total number of students and the number of K-5 students.

b. Replace t with 7 in the equation for S to determine the number of students enrolled in 6-12 in 2007.

In 2007 there were 30.3 hundreds or 3030 students enrolled in 6-12.

52. CCSS REASONING The perimeter of the figure shown is represented by the expression 3x2 − 7x + 2. Write a

polynomial that represents the measure of the third side.

SOLUTION:

53. GEOMETRY Consider the rectangle.

a. What does (4x2 + 2x – 1)(2x

2 – x + 3) represent?

b. What does 2(4x2 + 2x – 1) + 2(2x

2 – x + 3) represent?

SOLUTION:

a. (4x2 + 2x – 1)(2x

2 – x + 3) is a multiplication of the length and the width of the rectangle, which is the formula for

the area of a rectangle.

b. 2(4x2 + 2x – 1) + 2(2x

2 – x + 3) is the sum of twice the length and twice the width of a rectangle, which is the

formula for the perimeter of the rectangle

Find each sum or difference.

54. (4x + 2y − 6z) + (5y − 2z + 7x) + (−9z − 2x − 3y)

SOLUTION:

55. (5a2 − 4) + (a

2 − 2a + 12) + (4a2 − 6a + 8)

SOLUTION:

56. (3c2 − 7) + (4c + 7) − (c

2 + 5c − 8)

SOLUTION:

57. (3n3 + 3n − 10) − (4n

2 − 5n) + (4n3 − 3n

2 − 9n + 4)

SOLUTION:

58. FOOTBALL The National Football League is divided into two conferences, the American A and the National N. From 2002 through 2009, the total attendance T (in thousands) for both conferences and for the American Conference games are modeled by the following equations, where x is the number of years since 2002.

T = –0.69x3 + 55.83x

2 + 643.31x + 10,538

A = –3.78x3 + 58.96x

2 + 265.96x + 5257

Determine how many people attended National Conference football games in 2009.

SOLUTION:

Let x = 7 represent 2009, then find how many people attended a National Conference football game in 2009.

In 2009 the number of people who attended National Conference football games was about 8829 thousand. Multiply by 1000 to find the attendance in standard form. 8829 ×1000 = 8,829,000. So, about 8,829,000 people attended National Conference football games in 2009.

59. CAR RENTAL The cost to rent a car for a day is $15 plus $0.15 for each mile driven. a. Write a polynomial that represents the cost of renting a car for m miles. b. If a car is driven 145 miles, how much would it cost to rent? c. If a car is driven 105 miles each day for four days, how much would it cost to rent a car? d. If a car is driven 220 miles each day for seven days, how much would it cost to rent a car?

SOLUTION: a. The cost to rent a car is the daily rate and the mileage cost or 15 + 0.15m. b. Substitute 145 for m to find the daily cost to drive 145 miles.

The cost to rent the car would be $36.75. c. The expression represents the cost per day. For 4 days, multiply the entire expression by 4 to find the cost for a 4 day trip. Substitute 105 for m.

The cost to rent the car would be $123. d. The expression represents the cost per day. For 7 days, multiply the entire expression by 7 to find the cost for a 7 day trip. Substitute 220 for m.

The cost to rent the car would be $336.

60. MULTIPLE REPRESENTATIONS In this problem, you will explore perimeter and area. a. Geometric Draw three rectangles that each have a perimeter of 400 feet. b. Tabular Record the width and length of each rectangle in a table like the one shown below. Find the area of eachrectangle.

c. Graphical On a coordinate system, graph the area of rectangle 4 in terms of the length, x. Use the graph to determine the largest area possible. d. Analytical Determine the length and width that produce the largest area.

SOLUTION: a. For the perimeters to be 400, create lengths and widths that sum to 200. Note that some lengths are already provided in the table in part b.

b. The area is length multiplied by width. Make sure the units are squared for area. The sum of the length and width must be 200, so if the length is x, the width must be 200 – x.

c. The length will be the x-values on the horizontal and the area will be the y-values on the vertical. The length cannot pass 200 since the sum of the length and width is 200. Set the intervals for the x-axis to 25 feet. Extend the table of values to find more points to plot on the graph. It appears that 10,000 is the greatest area, so set the intervals

for the y-axis to 1000 ft2. After graphing, it appears that the highest point on the graph is at an area of 10,000 ft

2.

d. The associated x-value with the maximum area is x = 100, so the length must be 100 and the width must be 200 – 100, or 100. The length and width of the rectangle must be 100 feet each to have the largest area.

61. CCSS CRITIQUE Cheyenne and Sebastian are finding (2x2 − x) − (3x + 3x

2 − 2). Is either of them correct? Explain your reasoning.

SOLUTION:

Neither is correct. Cheyenne, did not distribute the negative to the 2nd and 3rd terms when she found the additive inverse. Sebastian did not distribute the negate to the 3rd terms when he found the additive inverse. To find the additive inverse, all terms should be multiplied by −1.

62. REASONING Determine whether each of the following statements is true or false . Explain your reasoning. a. A binomial can have a degree of zero. b. The order in which polynomials are subtracted does not matter.

SOLUTION: a. If a binomial has two terms that are each a degree of 0, then those terms can be combined and the binomial becomes a monomial. For example, 18 + 7 = 25. If one of the terms of the binomial does not have a degree of 0, then the binomial cannot have a degree of 0, since the degree of a polynomial is the greatest degree of any term in the polynomial. b. Subtraction is not commutative. While 2 + 5 = 5 + 2, 2 – 5 ≠ 5 – 2. This is also true for polynomials. Sample answer: (2x – 3) – (4x – 3) = –2x, but (4x – 3) – (2x – 3) = 2x

63. CHALLENGE Write a polynomial that represents the sum of an odd integer 2n + 1 and the next two consecutive odd integers.

SOLUTION:

64. WRITING IN MATH Why would you add or subtract equations that represent real-world situations? Explain.

SOLUTION:

65. WRITING IN MATH Describe how to add and subtract polynomials using both the vertical and horizontal formats.

SOLUTION: To add polynomials in a horizontal format, you combine like terms. For the vertical format, you write the polynomials in standard form, align like terms in columns, and combine like terms.

To subtract polynomials in a horizontal format you find the additive inverse of the polynomial you are subtracting, andthen combine like terms. For the vertical format, you write the polynomials in standard form, align like terms in columns, and subtract by adding the additive inverse.

66. Three consecutive integers can be represented by x, x + 1, and x + 2. What is the sum of these three integers? A x(x + 1)(x + 2)

B x3 + 3

C 3x + 3 D x + 3

SOLUTION:

The correct choice is C.

67. SHORT RESPONSE What is the perimeter of a square with sides that measure 2x + 3 units?

SOLUTION:

The perimeter of the square is 8x + 12 units.

68. Jim cuts a board in the shape of a regular hexagon and pounds in a nail at each vertex, as shown. How many rubber bands will he need to stretch a rubber band across every possible pair of nails?

F 15 G 14 H 12 J 9

SOLUTION: The first nail would connect to 5 others, the second to 4 others, the third to 3 others, etc.5 + 4 + 3 + 2 + 1 = 15 The correct choice is F.

69. Which ordered pair is in the solution set of the system of inequalities shown in the graph?

A (−3, 0) B (0, −3) C (5, 0) D (0, 5)

SOLUTION: Choice A is outside the shaded area for both inequalities. Choices B and D are inside the shaded area for only one inequality. Choice C is the only point in the solution for both inequalities. So, the correct choice is C.

70. COMPUTERS A computer technician charges by the hour to fix and repair computer equipment. The total cost of the technician for one hour is $75, for two hours is $125, for three hours is $175, for four hours is $225, and so on. Write a recursive formula for the sequence.

SOLUTION: Write out the terms. $75, $125, $175, $225, ... The first term is 75, and 50 is added to form each following term. Therefore, we have a1 = 75, an = an – 1 + 50, n ≥ 2.

Determine whether each sequence is arithmetic, geometric, or neither. Explain.71. 8, –32, 128, –512, ...

SOLUTION: Check for a common difference. –32 – 8 = –40 128 – (–32) = 160 There is no common difference. Check for a common ratio. –32 ÷ 8 = –4 128 ÷ (–32) = –4 Geometric; the common ratio is –4.

72. 25, 8, –9, –26, ...

SOLUTION: Check for a common difference. 8 – 25 = –17 –9 – 8 = –17 Arithmetic; the common difference is –17.

73.

SOLUTION: Check for a common difference.

There is no common difference. Check for a common ratio.

There is no common ratio, so the sequence is not arithmetic or geometric.

74. 43, 52, 61, 70, ...

SOLUTION: Check for a common difference. 52 – 43 = 9 61 – 52 = 9 Arithmetic; the common difference is 9.

75. –27, –16, –5, 6, ...

SOLUTION: Check for a common difference. –16 – (–27) = 11 –5 – (–16) = 11 Arithmetic; the common difference is 11.

76. 200, 100, 50, 25, …

SOLUTION: Check for a common difference. 100 – 200 = –100 50 – 100 = –50 There is no common difference. Check for a common ratio. 100 ÷ 200 = 0.5 50 ÷ 100 = 0.5

Geometric; the common ratio is 0.5 or .

77. JOBS Kimi received an offer for a new job. She wants to compare the offer with her current job. What is total amount of sales that Kimi must get each month to make the same income at either job?

SOLUTION: Let s be Kimi’s monthly sales.

Kimi must sell $80,000 each month to make the same income at either job.

Determine whether each sequence is an arithmetic sequence. If it is, state the common difference.78. 24, 16, 8, 0, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is –8, because 16 – 24 = –8; 8 –16 = –8; etc.

79. , 13, 26, …

SOLUTION: Find the difference between the terms.

– =

13 – =

26 – 13 = 13 There is not a common different. The sequence is not an arithmetic sequence

80. 7, 6, 5, 4, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is –1, because 6 – 7 = –1; 5 – 6 = –1; 4 – 5 = –1;etc.

81. 10, 12, 15, 18, …

SOLUTION: Find the difference between the terms. 12 – 10 = 2 15 – 12 = 3 18 – 15 = 3 There is no common difference. The sequence is not an arithmetic sequence.

82. −15, −11, −7, −3, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is 4, because –11 – (–15) = 4; –7 – (–11) = 4; –3 – (–7) = 4;etc.

83. −0.3, 0.2, 0.7, 1.2, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is 0.5, because 0.2 – (–0.3) = 0.5; 0.7 – 0.2 = 0.5; 1.2 – 0.7 = 0.5; etc.

Simplify.

84. t(t5)(t

7)

SOLUTION:

85. n3(n

2)(−2n

3)

SOLUTION:

86. (5t5v

2)(10t

3v

4)

SOLUTION:

87. (−8u4z

5)(5uz

4)

SOLUTION:

88. [(3)2]3

SOLUTION:

89. [(2)3]2

SOLUTION:

90. (2m4k

3)2(−3mk

2)3

SOLUTION:

91. (6xy2)2(2x

2y

2z

2)3

SOLUTION:

eSolutions Manual - Powered by Cognero Page 32

8-1 Adding and Subtracting Polynomials

Determine whether each expression is a polynomial. If it is a polynomial, find the degree and determine whether it is a monomial, binomial, or trinomial.

1. 7ab + 6b2 – 2a

3

SOLUTION:

A polynomial is a monomial or the sum of monomials. 7ab + 6b2 – 2a

3 is the sum of 3 monomials, so it is a

polynomial. The degree of a polynomial is the greatest degree of any term in the polynomial.The degree of each term is 2, 2, and

3, so the degree of 7ab + 6b2 – 2a

3 is 3. The polynomial has three terms, so it is a trinomial.

2. 2y – 5 + 3y2

SOLUTION:

2y – 5 + 3y2

is the sum of monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 1, 0, and

2, so the degree of 2y – 5 + 3y2 is 2. The polynomial has three terms, so it is a trinomial.

3. 3x2

SOLUTION:

A polynomial is a monomial or the sum of monomials, so 3x2 is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of 3x2 is 2. The

polynomial has one term, so it is a monomial.

4.

SOLUTION: A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer exponents. It has

only one term. is a division of two monomials, so it is not a monomial.

5. 5m2p

3 + 6

SOLUTION:

A polynomial is a monomial or the sum of monomials. 5m2p

3 + 6

is the sum of 2 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 5 and 0,

so the degree of 5m2p

3 + 6 is 5. The polynomial has two terms, so it is a binomial.

6. 5q–4

+ 6q

SOLUTION: A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer exponents. It has

only one term. 5q-4 is equal to , which is a division of two monomials, so it is not a monomial.

Write each polynomial in standard form. Identify the leading coefficient.

7. –4d4 + 1 – d

2

SOLUTION: Find the degree of each term.

–4d4→ 4

1 → 0

–d2 → 2

The greatest degree is 4, from the term –4d4, so the leading coefficient of –4d

4 + 1 – d

2 is –4.

Rewrite the polynomial with each monomial in descending order according to degree.

–4d4 – d

2 + 1

8. 2x5 – 12 + 3x

SOLUTION: Find the degree of each term.

2x5 → 5

–12 → 0 3x → 1

The greatest degree is 5, from the term 2x5, so the leading coefficient of 2x

5 – 12 + 3x is 2.

Rewrite the polynomial with each monomial in descending order according to degree.

2x5 + 3x – 12

9. 4z – 2z2 – 5z

4

SOLUTION: Find the degree of each term. 4z → 1

– 2z2 → 2

– 5z4→ 4

The greatest degree is 4, from the term – 5z4, so the leading coefficient of 4z – 2z

2 – 5z

4 is –5.

Rewrite the polynomial with each monomial in descending order according to degree.

–5z4 – 2z

2 + 4z

10. 2a + 4a3 – 5a

2 – 1

SOLUTION: Find the degree of each term. 2a → 1

4a3→ 3

– 5a2 → 2

– 1 → 0

The greatest degree is 3, from the term 4a3, so the leading coefficient of 2a + 4a

3 – 5a

2 – 1 is 4.

Rewrite the polynomial with each monomial in descending order according to degree.

4a3– 5a

2 + 2a – 1

Find each sum or difference.

11. (6x3 − 4) + (−2x

3 + 9)

SOLUTION:

12. (g3 − 2g2 + 5g + 6) − (g2 + 2g)

SOLUTION:

13. (4 + 2a2 − 2a) − (3a

2 − 8a + 7)

SOLUTION:

14. (8y − 4y2) + (3y − 9y

2)

SOLUTION:

15. (−4z3 − 2z + 8) − (4z

3 + 3z2 − 5)

SOLUTION:

16. (−3d2 − 8 + 2d) + (4d − 12 + d2

)

SOLUTION:

17. (y + 5) + (2y + 4y2 – 2)

SOLUTION:

18. (3n3 − 5n + n2

) − (−8n2 + 3n

3)

SOLUTION:

19. CCSS SENSE-MAKING The total number of students T who traveled for spring break consists of two groups: students who flew to their destinations F and students who drove to their destination D. The number (in thousands) of students who flew and the total number of students who flew or drove can be modeled by the following equations,where n is the number of years since 1995. T = 14n + 21

F = 8n + 7 a. Write an equation that models the number of students who drove to their destination for this time period. b. Predict the number of students who will drive to their destination in 2012. c. How many students will drive or fly to their destination in 2015?

SOLUTION: a.

D = 6n + 14 b. n = 2012 – 1995 = 17

The number of students who will drive to their destination in 2012 is 116,000 students. c. n = 2015 – 1995 = 20

The number of students who will drive or fly to their destination in 2015 is 301,000 students.

Determine whether each expression is a polynomial. If it is a polynomial, find the degree and determine whether it is a monomial, binomial, or trinomial.

20.

SOLUTION: A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer exponents. It has

only one term. is a division of two monomials, so it is not a monomial.

21.

SOLUTION: A polynomial is a monomial or the sum of monomials. 21 is a monomial, so it is also a polynomial. The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of 21 is 0. The polynomial has only one term, so it is a monomial.

22. c4 – 2c

2 + 1

SOLUTION:

A polynomial is a monomial or the sum of monomials. c4 – 2c

2 + 1

is the sum of 3 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 4, 2, and

0, so the degree of c4 – 2c2 + 1 is 4. The polynomial has three terms, so it is a trinomial.

23. d + 3dc

SOLUTION: A polynomial is a monomial or the sum of monomials. A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer

exponents. 3dc

has a variable in the exponent, so it is not a monomial.

24. a – a2

SOLUTION:

A polynomial is a monomial or the sum of monomials. a – a2 is the sum of 2 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 1 and 2,

so the degree of a – a2 is 2. The polynomial has two terms, so it is a binomial.

25. 5n3 + nq

3

SOLUTION:

A polynomial is a monomial or the sum of monomials. 5n3 + nq3 is the sum of 2 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 3 and 4,

so the degree of 5n3 + nq3 is 4. The polynomial has two terms, so it is a binomial.

Write each polynomial in standard form. Identify the leading coefficient.

26. 5x2 – 2 + 3x

SOLUTION: Find the degree of each term.

5x2 → 2

– 2 → 0

3x → 1

The greatest degree is 2, from the term 5x2, so the leading coefficient of 5x

2 – 2 + 3x is 5.

Rewrite the polynomial with each monomial in descending order according to degree.

5x2 + 3x – 2

27. 8y + 7y3

SOLUTION: Find the degree of each term.

7y3 → 3

8y → 1

The greatest degree is 3, from the term 7y3, so the leading coefficient of 8y + 7y

3 is 7.

Rewrite the polynomial with each monomial in descending order according to degree.

7y3 + 8y

28. 4 – 3c – 5c2

SOLUTION: Find the degree of each term. 4 → 0 3c → 1

– 5c2 → 2

The greatest degree is 2, from the term – 5c2, so the leading coefficient of 4 – 3c – 5c

2 is –5.

Rewrite the polynomial with each monomial in descending order according to degree.

–5c2 – 3c + 4

29. –y3 + 3y – 3y

2 + 2

SOLUTION: Find the degree of each term.

–y3 → 3

3y → 1

3y2 → 2

2 → 0

The greatest degree is 3, from the term –y3, so the leading coefficient of –y

3 + 3y – 3y

2 + 2 is –1.

Rewrite the polynomial with each monomial in descending order according to degree. –y 3 – 3y 2 + 3y + 2

30. 11t + 2t2 – 3 + t

5

SOLUTION: Find the degree of each term. 11t → 1

2t2 → 2

–3 → 0

t5 → 5

The greatest degree is 5, from the term t5, so the leading coefficient of 11t + 2t2 – 3 + t

5 is 1.

Rewrite the polynomial with each monomial in descending order according to degree.

t5 + 2t

2 + 11t – 3

31. 2 + r – r3

SOLUTION: Find the degree of each term. 2 → 0 r → 1

– r3→ 3

The greatest degree is 3, from the term – r3, so the leading coefficient of 2 + r – r3

is –1. Rewrite the polynomial with each monomial in descending order according to degree.

–r3 + r + 2

32.

SOLUTION: Find the degree of each term.

→ 0

–3x4 → 4

7 → 0

The greatest degree is 4, from the term –3x4, so the leading coefficient of is –3.

Rewrite the polynomial with each monomial in descending order according to degree.

33. –9b2 + 10b – b

6

SOLUTION: Find the degree of each term.

–9b2 → 2

10b → 1

–b6→ 6

The greatest degree is 6, from the term –b6, so the leading coefficient of –9b2 + 10b – b6 is –1. Rewrite the polynomial with each monomial in descending order according to degree.

–b6 – 9b

2 + 10b

Find each sum or difference.

34. (2c2 + 6c + 4) + (5c – 7)

SOLUTION:

35. (2x + 3x2) − (7 − 8x

2)

SOLUTION:

36. (3c3 − c + 11) − (c2 + 2c + 8)

SOLUTION:

37. (z2 + z) + (z

2 − 11)

SOLUTION:

38. (2x − 2y + 1) − (3y + 4x)

SOLUTION:

39. (4a − 5b2 + 3) + (6 − 2a + 3b

2)

SOLUTION:

40. (x2y − 3x

2 + y) + (3y − 2x2y)

SOLUTION:

41. (−8xy + 3x2 − 5y) + (4x

2 − 2y + 6xy)

SOLUTION:

42. (5n − 2p2 + 2np) − (4p

2 + 4n)

SOLUTION:

43. (4rxt − 8r2x + x2

) − (6rx2 + 5rxt − 2x

2)

SOLUTION:

44. PETS From 1999 through 2009, the number of dogs D and the number of cats C (in hundreds) adopted from animal

shelters in the United States are modeled by the equations D = 2n + 3 and C = n + 4, where n is the number of yearssince 1999. a. Write an equation that models the total number T of dogs and cats adopted in hundreds for this time period. b. If this trend continues, how many dogs and cats will be adopted in 2013?

SOLUTION: a.

So, an equation that models the total number of dogs and cats adopted is T = 3n + 7. b. Evaluate the equation for the total number of dogs and cats for n = 2013 – 1999 = 14.

The number of cats and dogs adopted in 2013 will be 49 × 100 or 4900 cats and dogs.

Classify each polynomial according to its degree and number of terms.

45. 4x – 3x2 + 5

SOLUTION:

Find the degree of each term of 4x – 3x2 + 5.

4x → 1

– 3x2 → 2

5 → 0

The greatest degree is 2 and there are 3 terms, so 4x – 3x2 + 5 is a quadratic trinomial.

46. 11z3

SOLUTION: Find the degree of each term. 11z3

→ 3

The greatest degree is 3 and there is one term, so 11z3 is a cubic monomial.

47. 9 + y4

SOLUTION:

Find the degree of each term of 9 + y 4.

9 → 1

y4→ 4

The greatest degree is 4 and there are 2 terms, so 9 + y 4 is a quartic binomial.

48. 3x3 – 7

SOLUTION:

Find the degree of each term of 3x3 – 7.

3x3 → 3

–7 → 0

The greatest degree is 3 and there are 2 terms, so 3x3 – 7 is a cubic binomial.

49. –2x5 – x

2 + 5x – 8

SOLUTION:

Find the degree of each term of –2x5 – x2 + 5x – 8.

–2x5 → 5

–x2 → 2

5x → 1 – 8 → 0

The greatest degree is 5 and there are 4 terms, so –2x5 – x2 + 5x – 8 is a quintic polynomial.

50. 10t – 4t2 + 6t

3

SOLUTION:

Find the degree of each term of 10t – 4t2 + 6t

3.

10t → 1

4t2 → 2

6t3 → 3

The greatest degree is 3 and there are 3 terms, so 10t – 4t2 + 6t

3 is a cubic trinomial.

51. ENROLLMENT In a rapidly growing school system, the numbers (in hundreds) of total students N and

K-5 students P enrolled from 2000 to 2009 are modeled by the equations N = 1.25t2 – t + 7.5 and P =

0.7t2 – 0.95t + 3.8, where t is the number of years since 2000.

a. Write an equation modeling the number of 6-12 students S enrolled for this time period. b. How many 6-12 students were enrolled in the school system in 2007?

SOLUTION: a. To write an equation that represents the number of 6-12 students enrolled, subtract the equations that represent the total number of students and the number of K-5 students.

b. Replace t with 7 in the equation for S to determine the number of students enrolled in 6-12 in 2007.

In 2007 there were 30.3 hundreds or 3030 students enrolled in 6-12.

52. CCSS REASONING The perimeter of the figure shown is represented by the expression 3x2 − 7x + 2. Write a

polynomial that represents the measure of the third side.

SOLUTION:

53. GEOMETRY Consider the rectangle.

a. What does (4x2 + 2x – 1)(2x

2 – x + 3) represent?

b. What does 2(4x2 + 2x – 1) + 2(2x

2 – x + 3) represent?

SOLUTION:

a. (4x2 + 2x – 1)(2x

2 – x + 3) is a multiplication of the length and the width of the rectangle, which is the formula for

the area of a rectangle.

b. 2(4x2 + 2x – 1) + 2(2x

2 – x + 3) is the sum of twice the length and twice the width of a rectangle, which is the

formula for the perimeter of the rectangle

Find each sum or difference.

54. (4x + 2y − 6z) + (5y − 2z + 7x) + (−9z − 2x − 3y)

SOLUTION:

55. (5a2 − 4) + (a

2 − 2a + 12) + (4a2 − 6a + 8)

SOLUTION:

56. (3c2 − 7) + (4c + 7) − (c

2 + 5c − 8)

SOLUTION:

57. (3n3 + 3n − 10) − (4n

2 − 5n) + (4n3 − 3n

2 − 9n + 4)

SOLUTION:

58. FOOTBALL The National Football League is divided into two conferences, the American A and the National N. From 2002 through 2009, the total attendance T (in thousands) for both conferences and for the American Conference games are modeled by the following equations, where x is the number of years since 2002.

T = –0.69x3 + 55.83x

2 + 643.31x + 10,538

A = –3.78x3 + 58.96x

2 + 265.96x + 5257

Determine how many people attended National Conference football games in 2009.

SOLUTION:

Let x = 7 represent 2009, then find how many people attended a National Conference football game in 2009.

In 2009 the number of people who attended National Conference football games was about 8829 thousand. Multiply by 1000 to find the attendance in standard form. 8829 ×1000 = 8,829,000. So, about 8,829,000 people attended National Conference football games in 2009.

59. CAR RENTAL The cost to rent a car for a day is $15 plus $0.15 for each mile driven. a. Write a polynomial that represents the cost of renting a car for m miles. b. If a car is driven 145 miles, how much would it cost to rent? c. If a car is driven 105 miles each day for four days, how much would it cost to rent a car? d. If a car is driven 220 miles each day for seven days, how much would it cost to rent a car?

SOLUTION: a. The cost to rent a car is the daily rate and the mileage cost or 15 + 0.15m. b. Substitute 145 for m to find the daily cost to drive 145 miles.

The cost to rent the car would be $36.75. c. The expression represents the cost per day. For 4 days, multiply the entire expression by 4 to find the cost for a 4 day trip. Substitute 105 for m.

The cost to rent the car would be $123. d. The expression represents the cost per day. For 7 days, multiply the entire expression by 7 to find the cost for a 7 day trip. Substitute 220 for m.

The cost to rent the car would be $336.

60. MULTIPLE REPRESENTATIONS In this problem, you will explore perimeter and area. a. Geometric Draw three rectangles that each have a perimeter of 400 feet. b. Tabular Record the width and length of each rectangle in a table like the one shown below. Find the area of eachrectangle.

c. Graphical On a coordinate system, graph the area of rectangle 4 in terms of the length, x. Use the graph to determine the largest area possible. d. Analytical Determine the length and width that produce the largest area.

SOLUTION: a. For the perimeters to be 400, create lengths and widths that sum to 200. Note that some lengths are already provided in the table in part b.

b. The area is length multiplied by width. Make sure the units are squared for area. The sum of the length and width must be 200, so if the length is x, the width must be 200 – x.

c. The length will be the x-values on the horizontal and the area will be the y-values on the vertical. The length cannot pass 200 since the sum of the length and width is 200. Set the intervals for the x-axis to 25 feet. Extend the table of values to find more points to plot on the graph. It appears that 10,000 is the greatest area, so set the intervals

for the y-axis to 1000 ft2. After graphing, it appears that the highest point on the graph is at an area of 10,000 ft

2.

d. The associated x-value with the maximum area is x = 100, so the length must be 100 and the width must be 200 – 100, or 100. The length and width of the rectangle must be 100 feet each to have the largest area.

61. CCSS CRITIQUE Cheyenne and Sebastian are finding (2x2 − x) − (3x + 3x

2 − 2). Is either of them correct? Explain your reasoning.

SOLUTION:

Neither is correct. Cheyenne, did not distribute the negative to the 2nd and 3rd terms when she found the additive inverse. Sebastian did not distribute the negate to the 3rd terms when he found the additive inverse. To find the additive inverse, all terms should be multiplied by −1.

62. REASONING Determine whether each of the following statements is true or false . Explain your reasoning. a. A binomial can have a degree of zero. b. The order in which polynomials are subtracted does not matter.

SOLUTION: a. If a binomial has two terms that are each a degree of 0, then those terms can be combined and the binomial becomes a monomial. For example, 18 + 7 = 25. If one of the terms of the binomial does not have a degree of 0, then the binomial cannot have a degree of 0, since the degree of a polynomial is the greatest degree of any term in the polynomial. b. Subtraction is not commutative. While 2 + 5 = 5 + 2, 2 – 5 ≠ 5 – 2. This is also true for polynomials. Sample answer: (2x – 3) – (4x – 3) = –2x, but (4x – 3) – (2x – 3) = 2x

63. CHALLENGE Write a polynomial that represents the sum of an odd integer 2n + 1 and the next two consecutive odd integers.

SOLUTION:

64. WRITING IN MATH Why would you add or subtract equations that represent real-world situations? Explain.

SOLUTION:

65. WRITING IN MATH Describe how to add and subtract polynomials using both the vertical and horizontal formats.

SOLUTION: To add polynomials in a horizontal format, you combine like terms. For the vertical format, you write the polynomials in standard form, align like terms in columns, and combine like terms.

To subtract polynomials in a horizontal format you find the additive inverse of the polynomial you are subtracting, andthen combine like terms. For the vertical format, you write the polynomials in standard form, align like terms in columns, and subtract by adding the additive inverse.

66. Three consecutive integers can be represented by x, x + 1, and x + 2. What is the sum of these three integers? A x(x + 1)(x + 2)

B x3 + 3

C 3x + 3 D x + 3

SOLUTION:

The correct choice is C.

67. SHORT RESPONSE What is the perimeter of a square with sides that measure 2x + 3 units?

SOLUTION:

The perimeter of the square is 8x + 12 units.

68. Jim cuts a board in the shape of a regular hexagon and pounds in a nail at each vertex, as shown. How many rubber bands will he need to stretch a rubber band across every possible pair of nails?

F 15 G 14 H 12 J 9

SOLUTION: The first nail would connect to 5 others, the second to 4 others, the third to 3 others, etc.5 + 4 + 3 + 2 + 1 = 15 The correct choice is F.

69. Which ordered pair is in the solution set of the system of inequalities shown in the graph?

A (−3, 0) B (0, −3) C (5, 0) D (0, 5)

SOLUTION: Choice A is outside the shaded area for both inequalities. Choices B and D are inside the shaded area for only one inequality. Choice C is the only point in the solution for both inequalities. So, the correct choice is C.

70. COMPUTERS A computer technician charges by the hour to fix and repair computer equipment. The total cost of the technician for one hour is $75, for two hours is $125, for three hours is $175, for four hours is $225, and so on. Write a recursive formula for the sequence.

SOLUTION: Write out the terms. $75, $125, $175, $225, ... The first term is 75, and 50 is added to form each following term. Therefore, we have a1 = 75, an = an – 1 + 50, n ≥ 2.

Determine whether each sequence is arithmetic, geometric, or neither. Explain.71. 8, –32, 128, –512, ...

SOLUTION: Check for a common difference. –32 – 8 = –40 128 – (–32) = 160 There is no common difference. Check for a common ratio. –32 ÷ 8 = –4 128 ÷ (–32) = –4 Geometric; the common ratio is –4.

72. 25, 8, –9, –26, ...

SOLUTION: Check for a common difference. 8 – 25 = –17 –9 – 8 = –17 Arithmetic; the common difference is –17.

73.

SOLUTION: Check for a common difference.

There is no common difference. Check for a common ratio.

There is no common ratio, so the sequence is not arithmetic or geometric.

74. 43, 52, 61, 70, ...

SOLUTION: Check for a common difference. 52 – 43 = 9 61 – 52 = 9 Arithmetic; the common difference is 9.

75. –27, –16, –5, 6, ...

SOLUTION: Check for a common difference. –16 – (–27) = 11 –5 – (–16) = 11 Arithmetic; the common difference is 11.

76. 200, 100, 50, 25, …

SOLUTION: Check for a common difference. 100 – 200 = –100 50 – 100 = –50 There is no common difference. Check for a common ratio. 100 ÷ 200 = 0.5 50 ÷ 100 = 0.5

Geometric; the common ratio is 0.5 or .

77. JOBS Kimi received an offer for a new job. She wants to compare the offer with her current job. What is total amount of sales that Kimi must get each month to make the same income at either job?

SOLUTION: Let s be Kimi’s monthly sales.

Kimi must sell $80,000 each month to make the same income at either job.

Determine whether each sequence is an arithmetic sequence. If it is, state the common difference.78. 24, 16, 8, 0, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is –8, because 16 – 24 = –8; 8 –16 = –8; etc.

79. , 13, 26, …

SOLUTION: Find the difference between the terms.

– =

13 – =

26 – 13 = 13 There is not a common different. The sequence is not an arithmetic sequence

80. 7, 6, 5, 4, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is –1, because 6 – 7 = –1; 5 – 6 = –1; 4 – 5 = –1;etc.

81. 10, 12, 15, 18, …

SOLUTION: Find the difference between the terms. 12 – 10 = 2 15 – 12 = 3 18 – 15 = 3 There is no common difference. The sequence is not an arithmetic sequence.

82. −15, −11, −7, −3, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is 4, because –11 – (–15) = 4; –7 – (–11) = 4; –3 – (–7) = 4;etc.

83. −0.3, 0.2, 0.7, 1.2, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is 0.5, because 0.2 – (–0.3) = 0.5; 0.7 – 0.2 = 0.5; 1.2 – 0.7 = 0.5; etc.

Simplify.

84. t(t5)(t

7)

SOLUTION:

85. n3(n

2)(−2n

3)

SOLUTION:

86. (5t5v

2)(10t

3v

4)

SOLUTION:

87. (−8u4z

5)(5uz

4)

SOLUTION:

88. [(3)2]3

SOLUTION:

89. [(2)3]2

SOLUTION:

90. (2m4k

3)2(−3mk

2)3

SOLUTION:

91. (6xy2)2(2x

2y

2z

2)3

SOLUTION:

eSolutions Manual - Powered by Cognero Page 33

8-1 Adding and Subtracting Polynomials

Determine whether each expression is a polynomial. If it is a polynomial, find the degree and determine whether it is a monomial, binomial, or trinomial.

1. 7ab + 6b2 – 2a

3

SOLUTION:

A polynomial is a monomial or the sum of monomials. 7ab + 6b2 – 2a

3 is the sum of 3 monomials, so it is a

polynomial. The degree of a polynomial is the greatest degree of any term in the polynomial.The degree of each term is 2, 2, and

3, so the degree of 7ab + 6b2 – 2a

3 is 3. The polynomial has three terms, so it is a trinomial.

2. 2y – 5 + 3y2

SOLUTION:

2y – 5 + 3y2

is the sum of monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 1, 0, and

2, so the degree of 2y – 5 + 3y2 is 2. The polynomial has three terms, so it is a trinomial.

3. 3x2

SOLUTION:

A polynomial is a monomial or the sum of monomials, so 3x2 is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of 3x2 is 2. The

polynomial has one term, so it is a monomial.

4.

SOLUTION: A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer exponents. It has

only one term. is a division of two monomials, so it is not a monomial.

5. 5m2p

3 + 6

SOLUTION:

A polynomial is a monomial or the sum of monomials. 5m2p

3 + 6

is the sum of 2 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 5 and 0,

so the degree of 5m2p

3 + 6 is 5. The polynomial has two terms, so it is a binomial.

6. 5q–4

+ 6q

SOLUTION: A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer exponents. It has

only one term. 5q-4 is equal to , which is a division of two monomials, so it is not a monomial.

Write each polynomial in standard form. Identify the leading coefficient.

7. –4d4 + 1 – d

2

SOLUTION: Find the degree of each term.

–4d4→ 4

1 → 0

–d2 → 2

The greatest degree is 4, from the term –4d4, so the leading coefficient of –4d

4 + 1 – d

2 is –4.

Rewrite the polynomial with each monomial in descending order according to degree.

–4d4 – d

2 + 1

8. 2x5 – 12 + 3x

SOLUTION: Find the degree of each term.

2x5 → 5

–12 → 0 3x → 1

The greatest degree is 5, from the term 2x5, so the leading coefficient of 2x

5 – 12 + 3x is 2.

Rewrite the polynomial with each monomial in descending order according to degree.

2x5 + 3x – 12

9. 4z – 2z2 – 5z

4

SOLUTION: Find the degree of each term. 4z → 1

– 2z2 → 2

– 5z4→ 4

The greatest degree is 4, from the term – 5z4, so the leading coefficient of 4z – 2z

2 – 5z

4 is –5.

Rewrite the polynomial with each monomial in descending order according to degree.

–5z4 – 2z

2 + 4z

10. 2a + 4a3 – 5a

2 – 1

SOLUTION: Find the degree of each term. 2a → 1

4a3→ 3

– 5a2 → 2

– 1 → 0

The greatest degree is 3, from the term 4a3, so the leading coefficient of 2a + 4a

3 – 5a

2 – 1 is 4.

Rewrite the polynomial with each monomial in descending order according to degree.

4a3– 5a

2 + 2a – 1

Find each sum or difference.

11. (6x3 − 4) + (−2x

3 + 9)

SOLUTION:

12. (g3 − 2g2 + 5g + 6) − (g2 + 2g)

SOLUTION:

13. (4 + 2a2 − 2a) − (3a

2 − 8a + 7)

SOLUTION:

14. (8y − 4y2) + (3y − 9y

2)

SOLUTION:

15. (−4z3 − 2z + 8) − (4z

3 + 3z2 − 5)

SOLUTION:

16. (−3d2 − 8 + 2d) + (4d − 12 + d2

)

SOLUTION:

17. (y + 5) + (2y + 4y2 – 2)

SOLUTION:

18. (3n3 − 5n + n2

) − (−8n2 + 3n

3)

SOLUTION:

19. CCSS SENSE-MAKING The total number of students T who traveled for spring break consists of two groups: students who flew to their destinations F and students who drove to their destination D. The number (in thousands) of students who flew and the total number of students who flew or drove can be modeled by the following equations,where n is the number of years since 1995. T = 14n + 21

F = 8n + 7 a. Write an equation that models the number of students who drove to their destination for this time period. b. Predict the number of students who will drive to their destination in 2012. c. How many students will drive or fly to their destination in 2015?

SOLUTION: a.

D = 6n + 14 b. n = 2012 – 1995 = 17

The number of students who will drive to their destination in 2012 is 116,000 students. c. n = 2015 – 1995 = 20

The number of students who will drive or fly to their destination in 2015 is 301,000 students.

Determine whether each expression is a polynomial. If it is a polynomial, find the degree and determine whether it is a monomial, binomial, or trinomial.

20.

SOLUTION: A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer exponents. It has

only one term. is a division of two monomials, so it is not a monomial.

21.

SOLUTION: A polynomial is a monomial or the sum of monomials. 21 is a monomial, so it is also a polynomial. The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of 21 is 0. The polynomial has only one term, so it is a monomial.

22. c4 – 2c

2 + 1

SOLUTION:

A polynomial is a monomial or the sum of monomials. c4 – 2c

2 + 1

is the sum of 3 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 4, 2, and

0, so the degree of c4 – 2c2 + 1 is 4. The polynomial has three terms, so it is a trinomial.

23. d + 3dc

SOLUTION: A polynomial is a monomial or the sum of monomials. A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer

exponents. 3dc

has a variable in the exponent, so it is not a monomial.

24. a – a2

SOLUTION:

A polynomial is a monomial or the sum of monomials. a – a2 is the sum of 2 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 1 and 2,

so the degree of a – a2 is 2. The polynomial has two terms, so it is a binomial.

25. 5n3 + nq

3

SOLUTION:

A polynomial is a monomial or the sum of monomials. 5n3 + nq3 is the sum of 2 monomials, so it is a polynomial.

The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of each term is 3 and 4,

so the degree of 5n3 + nq3 is 4. The polynomial has two terms, so it is a binomial.

Write each polynomial in standard form. Identify the leading coefficient.

26. 5x2 – 2 + 3x

SOLUTION: Find the degree of each term.

5x2 → 2

– 2 → 0

3x → 1

The greatest degree is 2, from the term 5x2, so the leading coefficient of 5x

2 – 2 + 3x is 5.

Rewrite the polynomial with each monomial in descending order according to degree.

5x2 + 3x – 2

27. 8y + 7y3

SOLUTION: Find the degree of each term.

7y3 → 3

8y → 1

The greatest degree is 3, from the term 7y3, so the leading coefficient of 8y + 7y

3 is 7.

Rewrite the polynomial with each monomial in descending order according to degree.

7y3 + 8y

28. 4 – 3c – 5c2

SOLUTION: Find the degree of each term. 4 → 0 3c → 1

– 5c2 → 2

The greatest degree is 2, from the term – 5c2, so the leading coefficient of 4 – 3c – 5c

2 is –5.

Rewrite the polynomial with each monomial in descending order according to degree.

–5c2 – 3c + 4

29. –y3 + 3y – 3y

2 + 2

SOLUTION: Find the degree of each term.

–y3 → 3

3y → 1

3y2 → 2

2 → 0

The greatest degree is 3, from the term –y3, so the leading coefficient of –y

3 + 3y – 3y

2 + 2 is –1.

Rewrite the polynomial with each monomial in descending order according to degree. –y 3 – 3y 2 + 3y + 2

30. 11t + 2t2 – 3 + t

5

SOLUTION: Find the degree of each term. 11t → 1

2t2 → 2

–3 → 0

t5 → 5

The greatest degree is 5, from the term t5, so the leading coefficient of 11t + 2t2 – 3 + t

5 is 1.

Rewrite the polynomial with each monomial in descending order according to degree.

t5 + 2t

2 + 11t – 3

31. 2 + r – r3

SOLUTION: Find the degree of each term. 2 → 0 r → 1

– r3→ 3

The greatest degree is 3, from the term – r3, so the leading coefficient of 2 + r – r3

is –1. Rewrite the polynomial with each monomial in descending order according to degree.

–r3 + r + 2

32.

SOLUTION: Find the degree of each term.

→ 0

–3x4 → 4

7 → 0

The greatest degree is 4, from the term –3x4, so the leading coefficient of is –3.

Rewrite the polynomial with each monomial in descending order according to degree.

33. –9b2 + 10b – b

6

SOLUTION: Find the degree of each term.

–9b2 → 2

10b → 1

–b6→ 6

The greatest degree is 6, from the term –b6, so the leading coefficient of –9b2 + 10b – b6 is –1. Rewrite the polynomial with each monomial in descending order according to degree.

–b6 – 9b

2 + 10b

Find each sum or difference.

34. (2c2 + 6c + 4) + (5c – 7)

SOLUTION:

35. (2x + 3x2) − (7 − 8x

2)

SOLUTION:

36. (3c3 − c + 11) − (c2 + 2c + 8)

SOLUTION:

37. (z2 + z) + (z

2 − 11)

SOLUTION:

38. (2x − 2y + 1) − (3y + 4x)

SOLUTION:

39. (4a − 5b2 + 3) + (6 − 2a + 3b

2)

SOLUTION:

40. (x2y − 3x

2 + y) + (3y − 2x2y)

SOLUTION:

41. (−8xy + 3x2 − 5y) + (4x

2 − 2y + 6xy)

SOLUTION:

42. (5n − 2p2 + 2np) − (4p

2 + 4n)

SOLUTION:

43. (4rxt − 8r2x + x2

) − (6rx2 + 5rxt − 2x

2)

SOLUTION:

44. PETS From 1999 through 2009, the number of dogs D and the number of cats C (in hundreds) adopted from animal

shelters in the United States are modeled by the equations D = 2n + 3 and C = n + 4, where n is the number of yearssince 1999. a. Write an equation that models the total number T of dogs and cats adopted in hundreds for this time period. b. If this trend continues, how many dogs and cats will be adopted in 2013?

SOLUTION: a.

So, an equation that models the total number of dogs and cats adopted is T = 3n + 7. b. Evaluate the equation for the total number of dogs and cats for n = 2013 – 1999 = 14.

The number of cats and dogs adopted in 2013 will be 49 × 100 or 4900 cats and dogs.

Classify each polynomial according to its degree and number of terms.

45. 4x – 3x2 + 5

SOLUTION:

Find the degree of each term of 4x – 3x2 + 5.

4x → 1

– 3x2 → 2

5 → 0

The greatest degree is 2 and there are 3 terms, so 4x – 3x2 + 5 is a quadratic trinomial.

46. 11z3

SOLUTION: Find the degree of each term. 11z3

→ 3

The greatest degree is 3 and there is one term, so 11z3 is a cubic monomial.

47. 9 + y4

SOLUTION:

Find the degree of each term of 9 + y 4.

9 → 1

y4→ 4

The greatest degree is 4 and there are 2 terms, so 9 + y 4 is a quartic binomial.

48. 3x3 – 7

SOLUTION:

Find the degree of each term of 3x3 – 7.

3x3 → 3

–7 → 0

The greatest degree is 3 and there are 2 terms, so 3x3 – 7 is a cubic binomial.

49. –2x5 – x

2 + 5x – 8

SOLUTION:

Find the degree of each term of –2x5 – x2 + 5x – 8.

–2x5 → 5

–x2 → 2

5x → 1 – 8 → 0

The greatest degree is 5 and there are 4 terms, so –2x5 – x2 + 5x – 8 is a quintic polynomial.

50. 10t – 4t2 + 6t

3

SOLUTION:

Find the degree of each term of 10t – 4t2 + 6t

3.

10t → 1

4t2 → 2

6t3 → 3

The greatest degree is 3 and there are 3 terms, so 10t – 4t2 + 6t

3 is a cubic trinomial.

51. ENROLLMENT In a rapidly growing school system, the numbers (in hundreds) of total students N and

K-5 students P enrolled from 2000 to 2009 are modeled by the equations N = 1.25t2 – t + 7.5 and P =

0.7t2 – 0.95t + 3.8, where t is the number of years since 2000.

a. Write an equation modeling the number of 6-12 students S enrolled for this time period. b. How many 6-12 students were enrolled in the school system in 2007?

SOLUTION: a. To write an equation that represents the number of 6-12 students enrolled, subtract the equations that represent the total number of students and the number of K-5 students.

b. Replace t with 7 in the equation for S to determine the number of students enrolled in 6-12 in 2007.

In 2007 there were 30.3 hundreds or 3030 students enrolled in 6-12.

52. CCSS REASONING The perimeter of the figure shown is represented by the expression 3x2 − 7x + 2. Write a

polynomial that represents the measure of the third side.

SOLUTION:

53. GEOMETRY Consider the rectangle.

a. What does (4x2 + 2x – 1)(2x

2 – x + 3) represent?

b. What does 2(4x2 + 2x – 1) + 2(2x

2 – x + 3) represent?

SOLUTION:

a. (4x2 + 2x – 1)(2x

2 – x + 3) is a multiplication of the length and the width of the rectangle, which is the formula for

the area of a rectangle.

b. 2(4x2 + 2x – 1) + 2(2x

2 – x + 3) is the sum of twice the length and twice the width of a rectangle, which is the

formula for the perimeter of the rectangle

Find each sum or difference.

54. (4x + 2y − 6z) + (5y − 2z + 7x) + (−9z − 2x − 3y)

SOLUTION:

55. (5a2 − 4) + (a

2 − 2a + 12) + (4a2 − 6a + 8)

SOLUTION:

56. (3c2 − 7) + (4c + 7) − (c

2 + 5c − 8)

SOLUTION:

57. (3n3 + 3n − 10) − (4n

2 − 5n) + (4n3 − 3n

2 − 9n + 4)

SOLUTION:

58. FOOTBALL The National Football League is divided into two conferences, the American A and the National N. From 2002 through 2009, the total attendance T (in thousands) for both conferences and for the American Conference games are modeled by the following equations, where x is the number of years since 2002.

T = –0.69x3 + 55.83x

2 + 643.31x + 10,538

A = –3.78x3 + 58.96x

2 + 265.96x + 5257

Determine how many people attended National Conference football games in 2009.

SOLUTION:

Let x = 7 represent 2009, then find how many people attended a National Conference football game in 2009.

In 2009 the number of people who attended National Conference football games was about 8829 thousand. Multiply by 1000 to find the attendance in standard form. 8829 ×1000 = 8,829,000. So, about 8,829,000 people attended National Conference football games in 2009.

59. CAR RENTAL The cost to rent a car for a day is $15 plus $0.15 for each mile driven. a. Write a polynomial that represents the cost of renting a car for m miles. b. If a car is driven 145 miles, how much would it cost to rent? c. If a car is driven 105 miles each day for four days, how much would it cost to rent a car? d. If a car is driven 220 miles each day for seven days, how much would it cost to rent a car?

SOLUTION: a. The cost to rent a car is the daily rate and the mileage cost or 15 + 0.15m. b. Substitute 145 for m to find the daily cost to drive 145 miles.

The cost to rent the car would be $36.75. c. The expression represents the cost per day. For 4 days, multiply the entire expression by 4 to find the cost for a 4 day trip. Substitute 105 for m.

The cost to rent the car would be $123. d. The expression represents the cost per day. For 7 days, multiply the entire expression by 7 to find the cost for a 7 day trip. Substitute 220 for m.

The cost to rent the car would be $336.

60. MULTIPLE REPRESENTATIONS In this problem, you will explore perimeter and area. a. Geometric Draw three rectangles that each have a perimeter of 400 feet. b. Tabular Record the width and length of each rectangle in a table like the one shown below. Find the area of eachrectangle.

c. Graphical On a coordinate system, graph the area of rectangle 4 in terms of the length, x. Use the graph to determine the largest area possible. d. Analytical Determine the length and width that produce the largest area.

SOLUTION: a. For the perimeters to be 400, create lengths and widths that sum to 200. Note that some lengths are already provided in the table in part b.

b. The area is length multiplied by width. Make sure the units are squared for area. The sum of the length and width must be 200, so if the length is x, the width must be 200 – x.

c. The length will be the x-values on the horizontal and the area will be the y-values on the vertical. The length cannot pass 200 since the sum of the length and width is 200. Set the intervals for the x-axis to 25 feet. Extend the table of values to find more points to plot on the graph. It appears that 10,000 is the greatest area, so set the intervals

for the y-axis to 1000 ft2. After graphing, it appears that the highest point on the graph is at an area of 10,000 ft

2.

d. The associated x-value with the maximum area is x = 100, so the length must be 100 and the width must be 200 – 100, or 100. The length and width of the rectangle must be 100 feet each to have the largest area.

61. CCSS CRITIQUE Cheyenne and Sebastian are finding (2x2 − x) − (3x + 3x

2 − 2). Is either of them correct? Explain your reasoning.

SOLUTION:

Neither is correct. Cheyenne, did not distribute the negative to the 2nd and 3rd terms when she found the additive inverse. Sebastian did not distribute the negate to the 3rd terms when he found the additive inverse. To find the additive inverse, all terms should be multiplied by −1.

62. REASONING Determine whether each of the following statements is true or false . Explain your reasoning. a. A binomial can have a degree of zero. b. The order in which polynomials are subtracted does not matter.

SOLUTION: a. If a binomial has two terms that are each a degree of 0, then those terms can be combined and the binomial becomes a monomial. For example, 18 + 7 = 25. If one of the terms of the binomial does not have a degree of 0, then the binomial cannot have a degree of 0, since the degree of a polynomial is the greatest degree of any term in the polynomial. b. Subtraction is not commutative. While 2 + 5 = 5 + 2, 2 – 5 ≠ 5 – 2. This is also true for polynomials. Sample answer: (2x – 3) – (4x – 3) = –2x, but (4x – 3) – (2x – 3) = 2x

63. CHALLENGE Write a polynomial that represents the sum of an odd integer 2n + 1 and the next two consecutive odd integers.

SOLUTION:

64. WRITING IN MATH Why would you add or subtract equations that represent real-world situations? Explain.

SOLUTION:

65. WRITING IN MATH Describe how to add and subtract polynomials using both the vertical and horizontal formats.

SOLUTION: To add polynomials in a horizontal format, you combine like terms. For the vertical format, you write the polynomials in standard form, align like terms in columns, and combine like terms.

To subtract polynomials in a horizontal format you find the additive inverse of the polynomial you are subtracting, andthen combine like terms. For the vertical format, you write the polynomials in standard form, align like terms in columns, and subtract by adding the additive inverse.

66. Three consecutive integers can be represented by x, x + 1, and x + 2. What is the sum of these three integers? A x(x + 1)(x + 2)

B x3 + 3

C 3x + 3 D x + 3

SOLUTION:

The correct choice is C.

67. SHORT RESPONSE What is the perimeter of a square with sides that measure 2x + 3 units?

SOLUTION:

The perimeter of the square is 8x + 12 units.

68. Jim cuts a board in the shape of a regular hexagon and pounds in a nail at each vertex, as shown. How many rubber bands will he need to stretch a rubber band across every possible pair of nails?

F 15 G 14 H 12 J 9

SOLUTION: The first nail would connect to 5 others, the second to 4 others, the third to 3 others, etc.5 + 4 + 3 + 2 + 1 = 15 The correct choice is F.

69. Which ordered pair is in the solution set of the system of inequalities shown in the graph?

A (−3, 0) B (0, −3) C (5, 0) D (0, 5)

SOLUTION: Choice A is outside the shaded area for both inequalities. Choices B and D are inside the shaded area for only one inequality. Choice C is the only point in the solution for both inequalities. So, the correct choice is C.

70. COMPUTERS A computer technician charges by the hour to fix and repair computer equipment. The total cost of the technician for one hour is $75, for two hours is $125, for three hours is $175, for four hours is $225, and so on. Write a recursive formula for the sequence.

SOLUTION: Write out the terms. $75, $125, $175, $225, ... The first term is 75, and 50 is added to form each following term. Therefore, we have a1 = 75, an = an – 1 + 50, n ≥ 2.

Determine whether each sequence is arithmetic, geometric, or neither. Explain.71. 8, –32, 128, –512, ...

SOLUTION: Check for a common difference. –32 – 8 = –40 128 – (–32) = 160 There is no common difference. Check for a common ratio. –32 ÷ 8 = –4 128 ÷ (–32) = –4 Geometric; the common ratio is –4.

72. 25, 8, –9, –26, ...

SOLUTION: Check for a common difference. 8 – 25 = –17 –9 – 8 = –17 Arithmetic; the common difference is –17.

73.

SOLUTION: Check for a common difference.

There is no common difference. Check for a common ratio.

There is no common ratio, so the sequence is not arithmetic or geometric.

74. 43, 52, 61, 70, ...

SOLUTION: Check for a common difference. 52 – 43 = 9 61 – 52 = 9 Arithmetic; the common difference is 9.

75. –27, –16, –5, 6, ...

SOLUTION: Check for a common difference. –16 – (–27) = 11 –5 – (–16) = 11 Arithmetic; the common difference is 11.

76. 200, 100, 50, 25, …

SOLUTION: Check for a common difference. 100 – 200 = –100 50 – 100 = –50 There is no common difference. Check for a common ratio. 100 ÷ 200 = 0.5 50 ÷ 100 = 0.5

Geometric; the common ratio is 0.5 or .

77. JOBS Kimi received an offer for a new job. She wants to compare the offer with her current job. What is total amount of sales that Kimi must get each month to make the same income at either job?

SOLUTION: Let s be Kimi’s monthly sales.

Kimi must sell $80,000 each month to make the same income at either job.

Determine whether each sequence is an arithmetic sequence. If it is, state the common difference.78. 24, 16, 8, 0, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is –8, because 16 – 24 = –8; 8 –16 = –8; etc.

79. , 13, 26, …

SOLUTION: Find the difference between the terms.

– =

13 – =

26 – 13 = 13 There is not a common different. The sequence is not an arithmetic sequence

80. 7, 6, 5, 4, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is –1, because 6 – 7 = –1; 5 – 6 = –1; 4 – 5 = –1;etc.

81. 10, 12, 15, 18, …

SOLUTION: Find the difference between the terms. 12 – 10 = 2 15 – 12 = 3 18 – 15 = 3 There is no common difference. The sequence is not an arithmetic sequence.

82. −15, −11, −7, −3, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is 4, because –11 – (–15) = 4; –7 – (–11) = 4; –3 – (–7) = 4;etc.

83. −0.3, 0.2, 0.7, 1.2, …

SOLUTION: The sequence is an arithmetic sequence because the items differ by a constant. The common difference is 0.5, because 0.2 – (–0.3) = 0.5; 0.7 – 0.2 = 0.5; 1.2 – 0.7 = 0.5; etc.

Simplify.

84. t(t5)(t

7)

SOLUTION:

85. n3(n

2)(−2n

3)

SOLUTION:

86. (5t5v

2)(10t

3v

4)

SOLUTION:

87. (−8u4z

5)(5uz

4)

SOLUTION:

88. [(3)2]3

SOLUTION:

89. [(2)3]2

SOLUTION:

90. (2m4k

3)2(−3mk

2)3

SOLUTION:

91. (6xy2)2(2x

2y

2z

2)3

SOLUTION:

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8-1 Adding and Subtracting Polynomials