Basic Concepts of Algebra · 4 Chapter R: Basic Concepts of Algebra 32. x26 x13 = x26−13 = x13...

0 5!5

0!2 2

0!3 !1

0 4 6

0!2

0!5

0 3.8

-3 -2 -1 0 1 2 33

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0 7

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Chapter R

Basic Concepts of Algebra

Exercise Set R.1

1. Rational numbers:23, 6, −2.45, 18.4, −11, 3

√27, 5

16, −8

7,

0,√

16

2. Natural numbers: 6, 3√

27,√

16

3. Irrational numbers:√

3, 6√

26, 7.151551555 . . . , −√

35, 5√

3(Although there is a pattern in 7.151551555 . . . , there isno repeating block of digits.)

4. Integers: 6, −11, 3√

27, 0,√

16

5. Whole numbers: 6, 3√

27, 0,√

16

6. Real numbers: All of them

7. Integers but not natural numbers: −11, 0

8. Integers but not whole numbers: −11

9. Rational numbers but not integers:23, −2.45, 18.4, 5

16,

−87

10. Real numbers but not integers:23,√

3, −2.45, 6√

26, 18.4,

516, 7.151551555 . . . , −

√35, 5

√3, −8

7

11. This is a closed interval, so we use brackets. Interval no-tation is [−5, 5].

12. (−2, 2)

13. This is a half-open interval. We use a parenthesis onthe left and a bracket on the right. Interval notation is(−3,−1].

14. [4, 6)

15. This interval is of unlimited extent in the negative direc-tion, and the endpoint −2 is included. Interval notation is(−∞,−2].

16. (−5,∞)

17. This interval is of unlimited extent in the positive direc-tion, and the endpoint 3.8 is not included. Interval nota-tion is (3.8,∞).

18. [√

3,∞)

19. {x|7 < x}, or {x|x > 7}.This interval is of unlimited extent in the positive directionand the endpoint 7 is not included. Interval notation is(7,∞).

20. (−∞,−3)

21. The endpoints 0 and 5 are not included in the interval, sowe use parentheses. Interval notation is (0, 5).

22. [−1, 2]

23. The endpoint −9 is included in the interval, so we use abracket before the −9. The endpoint −4 is not included,so we use a parenthesis after the −4. Interval notation is[−9,−4).

24. (−9,−5]

25. Both endpoints are included in the interval, so we usebrackets. Interval notation is [x, x + h].

26. (x, x + h]

27. The endpoint p is not included in the interval, so we use aparenthesis before the p. The interval is of unlimited ex-tent in the positive direction, so we use the infinity symbol∞. Interval notation is (p,∞).

Copyright c⃝ 2012 Pearson Education, Inc. Publishing as Addison-Wesley.

2 Chapter R: Basic Concepts of Algebra

28. (−∞, q]

29. Since 6 is an element of the set of natural numbers, thestatement is true.

30. True

31. Since 3.2 is not an element of the set of integers, the state-ment is false.

32. True

33. Since −115

is an element of the set of rational numbers,the statement is true.

34. False

35. Since√

11 is an element of the set of real numbers, thestatement is false.

36. False

37. Since 24 is an element of the set of whole numbers, thestatement is false.

38. True

39. Since 1.089 is not an element of the set of irrational num-bers, the statement is true.

40. True

41. Since every whole number is an integer, the statement istrue.

42. False

43. Since every rational number is a real number, the state-ment is true.

44. True

45. Since there are real numbers that are not integers, thestatement is false.

46. False

47. The sentence 3 + y = y + 3 illustrates the commutativeproperty of addition.

48. Associative property of multiplication

49. The sentence −3 · 1 = −3 illustrates the multiplicativeidentity property.

50. Distributive property

51. The sentence 5 ·x = x ·5 illustrates the commutative prop-erty of multiplication.

52. Associative property of addition

53. The sentence 2(a+b) = (a+b)2 illustrates the commutativeproperty of multiplication.

54. Additive inverse property

55. The sentence −6(m+n) = −6(n+m) illustrates the com-mutative property of addition.

56. Additive identity property

57. The sentence 8 · 18

= 1 illustrates the multiplicative inverseproperty.

58. Distributive property

59. The distance of −8.15 from 0 is 8.15, so |− 8.15| = 8.15.

60. 14.7

61. The distance 295 from 0 is 295, so |295| = 295.

62. 93

63. The distance of −√

97 from 0 is√

97, so |−√

97| =√

97.

64.1219

65. The distance of 0 from 0 is 0, so |0| = 0.

66. 15

67. The distance of54

from 0 is54, so

∣

∣

∣

54

∣

∣

∣=

54.

68.√

3

69. |14 − (−8)| = |14 + 8| = |22| = 22, or|− 8 − 14| = |− 22| = 22

70. |0 − (−5.2)| = |5.2| = 5.2, or|− 5.2 − 0| = |− 5.2| = 5.2

71. |− 3 − (−9)| = |− 3 + 9| = |6| = 6, or|− 9 − (−3)| = |− 9 + 3| = |− 6| = 6

72.∣

∣

∣

158

− 2312

∣

∣

∣=∣

∣

∣

4524

− 4624

∣

∣

∣=∣

∣

∣− 1

24

∣

∣

∣=

124

, or∣

∣

∣

2312

− 158

∣

∣

∣=∣

∣

∣

4624

− 4524

∣

∣

∣=∣

∣

∣

124

∣

∣

∣=

124

73. |12.1 − 6.7| = |5.4| = 5.4, or|6.7 − 12.1| = |− 5.4| = 5.4

74. |− 6 − (−15)| = |− 6 + 15| = |9| = 9, or|− 15 − (−6)| = |− 15 + 6| = |− 9| = 9

75.∣

∣

∣

∣

− 34− 15

8

∣

∣

∣

∣

=∣

∣

∣

∣

− 68− 15

8

∣

∣

∣

∣

=∣

∣

∣

∣

− 218

∣

∣

∣

∣

=218, or

∣

∣

∣

∣

158

−(

− 34

)∣

∣

∣

∣

=∣

∣

∣

∣

158

+34

∣

∣

∣

∣

=∣

∣

∣

∣

158

+68

∣

∣

∣

∣

=∣

∣

∣

∣

218

∣

∣

∣

∣

=218

76. |− 3.4 − 10.2| = |− 13.6| = 13.6, or|10.2 − (−3.4)| = |10.2 + 3.4| = |13.6| = 13.6


Exercise Set R.2 3

77. |− 7 − 0| = |− 7| = 7, or|0 − (−7)| = |0 + 7| = |7| = 7

78. |3 − 19| = |− 16| = 16, or|19 − 3| = |16| = 16

79. Answers may vary. One such number is0.124124412444 . . . .

80. Answers may vary. Since −√

2.01 ≈ −1.418 and −√

2 ≈−1.414, one such number is −1.415.

81. Answers may vary. Since − 1101

= 0.0099 and

− 1100

= −0.01, one such number is −0.00999.

82. Answers may vary. One such number is√

5.995.

83. Since 12 +32 = 10, the hypotenuse of a right triangle withlegs of lengths 1 unit and 3 units has a length of

√10 units.

✏✏✏✏✏✏✏✏

3

1c c2 = 12 + 32

c2 = 10

c =√

10

Exercise Set R.2

1. 3−7 =137

(

a−m =1am

, a ̸= 0)

2.1

(5.9)−4= (5.9)4

3. Observe that each exponent is negative. We move eachfactor to the other side of the fraction bar and change thesign of each exponent.

x−5

y−4=

y4

x5


a−2

b−8=

b8

a2


m−1n−12

t−6=

t6

m1n12, or

t6

mn12


x−9y−17

z−11=

z11

x9y17

7. 230 = 1 (For any nonzero real number, a0 = 1.)

8.(

− 25

)0

= 1

9. z0 · z7 = z0+7 = z7, orz0 · z7 = 1 · z7 = z7

10. x10 · x0 = x10+0 = x10, orx10 · x0 = x10 · 1 = x10

11. 58 · 5−6 = 58+(−6) = 52, or 25

12. 62 · 6−7 = 62+(−7) = 6−5, or165

13. m−5 ·m5 = m−5+5 = m0 = 1

14. n9 · n−9 = n9+(−9) = n0 = 1

15. y3 · y−7 = y3+(−7) = y−4, or1y4

16. b−4 · b12 = b−4+12 = b8

17. (x + 3)4(x + 3)−2 = (x + 3)4+(−2) = (x + 3)2

18. (y − 1)−1(y − 1)5 = (y − 1)−1+5 = (y − 1)4

19. 3−3 · 38 · 3 = 3−3+8+1 = 36, or 729

20. 67 · 6−10 · 62 = 67+(−10)+2 = 6−1, or16

21. 2x3 · 3x2 = 2 · 3 · x3+2 = 6x5

22. 3y4 · 4y3 = 3 · 4 · y4+3 = 12y7

23. (−3a−5)(5a−7) = −3 · 5 · a−5+(−7) = −15a−12, or

− 15a12

24. (−6b−4)(2b−7) = −6 · 2 · b−4+(−7) = −12b−11, or

− 12b11

25. (6x−3y5)(−7x2y−9) = 6(−7)x−3+2y5+(−9) =

−42x−1y−4, or − 42xy4

26. (8ab7)(−7a−5b2) = 8(−7)a1+(−5)b7+2 =

−56a−4b9, or −56b9

a4

27. (2x)4(3x)3 = 24 · x4 · 33 · x3 = 16 · 27 · x4+3 = 432x7

28. (4y)2(3y)3 = 16y2 · 27y3 = 432y5

29. (−2n)3(5n)2 = (−2)3n3 · 52n2 = −8 · 25 · n3+2 =−200n5

30. (2x)5(3x)2 = 25x5 · 32x2 = 32 · 9 · x5+2 = 288x7

31.y35

y31= y35−31 = y4



32.x26

x13= x26−13 = x13

33.b−7

b12= b−7−12 = b−19, or

1b19

34.a−18

a−13= a−18−(−13) = a−18+13 = a−5, or

1a5

35.x2y−2

x−1y= x2−(−1)y−2−1 = x3y−3, or

x3

y3

36.x3y−3

x−1y2= x3−(−1)y−3−2 = x4y−5, or

x4

y5

37.32x−4y3

4x−5y8=

324x−4−(−5)y3−8 = 8xy−5, or

8xy5

38.20a5b−2

5a7b−3=

205a5−7b−2−(−3) = 4a−2b, or

4ba2

39. (2x2y)4 = 24(x2)4y4 = 16x2·4y4 = 16x8y4

40. (3ab5)3 = 33a3(b5)3 = 27a3b15

41. (−2x3)5 = (−2)5(x3)5 = (−2)5x3·5 = −32x15

42. (−3x2)4 = (−3)4(x2)4 = 81x8

43. (−5c−1d−2)−2 = (−5)−2c−1(−2)d−2(−2) =c2d4

(−5)2=

c2d4

25

44. (−4x−5z−2)−3 = (−4)−3(x−5)−3(z−2)−3 =x15z6

(−4)3=

x15z6

−64

45. (3m4)3(2m−5)4 = 33m12 · 24m−20 =

27 · 16m12+(−20) = 432m−8, or432m8

46. (4n−1)2(2n3)3 = 42n−2 · 23n9 = 16 · 8 · n−2+9 =128n7

47.(

2x−3y7

z−1

)3

=(2x−3y7)3

(z−1)3=

23x−9y21

z−3=

8x−9y21

z−3, or

8y21z3

x9

48.(

3x5y−8

z−2

)4

=81x20y−32

z−8, or

81x20z8

y32

49.(

24a10b−8c7

12a6b−3c5

)−5

= (2a4b−5c2)−5 = 2−5a−20b25c−10,

orb25

32a20c10

50.(125p12q−14r22

25p8q6r−15

)−4= (5p4q−20r37)−4 =

5−4p−16q80r−148, orq80

625p16r148

51. Convert 16,500,000 to scientific notation.We want the decimal point to be positioned between the1 and the 6, so we move it 7 places to the left. Since16,500,000 is greater than 10, the exponent must be posi-tive.

16, 500, 000 = 1.65 × 107

52. Position the decimal point 5 places to the left, between the3 and the 5. Since 359,000 is greater than 10, the exponentmust be positive.

359, 000 = 3.59 × 105

53. Convert 0.000000437 to scientific notation.We want the decimal point to be positioned between the4 and the 3, so we move it 7 places to the right. Since0.000000437 is a number between 0 and 1, the exponentmust be negative.

0.000000437 = 4.37 × 10−7

54. Position the decimal point 3 places to the right, betweenthe 5 and the 6. Since 0.0056 is a number between 0 and1, the exponent must be negative.

0.0056 = 5.6 × 10−3

55. Convert 234,600,000,000 to scientific notation. We wantthe decimal point to be positioned between the 2 and the3, so we move it 11 places to the left. Since 234,600,000,000is greater than 10, the exponent must be positive.

234, 600, 000, 000 = 2.346 × 1011

56. Position the decimal point 9 places to the left, between the8 and the 9. Since 8,904,000,000 is greater than 10, theexponent must be positive.

8, 904, 000, 000 = 8.904 × 109

57. Convert 0.00104 to scientific notation. We want the deci-mal point to be positioned between the 1 and the last 0, sowe move it 3 places to the right. Since 0.00104 is a numberbetween 0 and 1, the exponent must be negative.

0.00104 = 1.04 × 10−3

58. Position the decimal point 9 places to the right, betweenthe 5 and the 1. Since 0.00000000514 is a number between0 and 1, the exponent must be negative.

0.00000000514 = 5.14 × 10−9

59. Convert 0.00000000000000000000000000167 to scientificnotation.We want the decimal point to be positioned between the1 and the 6, so we move it 27 places to the right. Since0.00000000000000000000000000167 is a number between 0and 1, the exponent must be negative.

0.00000000000000000000000000167 = 1.67 × 10−27

60. Position the decimal point 10 places to the left, betweenthe 3 and the 7. Since 37,800,000,000 is greater than 10,the exponent must be positive.

37, 800, 000, 000 = 3.78 × 1010


Exercise Set R.2 5

61. Convert 7.6 × 105 to decimal notation.The exponent is positive, so the number is greater than10. We move the decimal point 5 places to the right.

7.6 × 105 = 760, 000

62. The exponent is negative, so the number is between 0 and1. We move the decimal point 6 places to the left.

3.4 × 10−6 = 0.0000034

63. Convert 1.09 × 10−7 to decimal notation.The exponent is negative, so the number is between 0 and1. We move the decimal point 7 places to the left.

1.09 × 10−7 = 0.000000109

64. The exponent is positive, so the number is greater than10. We move the decimal point 8 places to the right.

5.87 × 108 = 587, 000, 000


3.496 × 1010 = 34, 960, 000, 000

66. The exponent is positive, so the number is greater than10. We move the decimal point 11 places to the right.

8.409 × 1011 = 840, 900, 000, 000


5.41 × 10−8 = 0.0000000541


6.27 × 10−10 = 0.000000000627


2.319 × 108 = 231, 900, 000

70. The exponent is negative, so the number is between 0 and1. We move the decimal point 24 places to the left.1.67 × 10−24 g = 0.00000000000000000000000167 g

71. (4.2 × 107)(3.2 × 10−2)= (4.2 × 3.2) × (107 × 10−2)= 13.44 × 105 This is not scientific notation.= (1.344 × 10) × 105

= 1.344 × 106 Writing scientific notation

72. (8.3 × 10−15)(7.7 × 104) = 63.91 × 10−11

= (6.391 × 10) × 10−11

= 6.391 × 10−10

73. (2.6 × 10−18)(8.5 × 107)= (2.6 × 8.5) × (10−18 × 107)= 22.1 × 10−11 This is not scientific notation.= (2.21 × 10) × 10−11

= 2.21 × 10−10

74. (6.4 × 1012)(3.7 × 10−5) = 23.68 × 107

= (2.368 × 10) × 107

= 2.368 × 108

75. 6.4 × 10−7

8.0 × 106=

6.48.0

× 10−7

106

= 0.8 × 10−13 This is not scientificnotation.

= (8 × 10−1) × 10−13

= 8 × 10−14 Writing scientificnotation

76. 1.1 × 10−40

2.0 × 10−71= 0.55 × 1031

= (5.5 × 10−1) × 1031

= 5.5 × 1030

77. 1.8 × 10−3

7.2 × 10−9

=1.87.2

× 10−3

10−9

= 0.25 × 106 This is not scientific notation.= (2.5 × 10−1) × 106

= 2.5 × 105

78. 1.3 × 104

5.2 × 1010= 0.25 × 10−6

= (2.5 × 10−1) × 10−6

= 2.5 × 10−7

79. The average number of pieces of trash per mile is the totalnumber of pieces of trash divided by the number of miles.

51.2 billion76 million

=5.12 × 1010

7.6 × 107

≈ 0.6737 × 103

≈ (6.737 × 10−1) × 103

≈ 6.737 × 102

On average, there are about 6.737× 102 pieces of trash oneach mile of roadway.

80. We multiply the diameter, in nanometers, by the numberof meters in 1 nanometer.

360 × 0.000000001= (3.6 × 102) × 10−9

= 3.6 × (102 × 10−9)= 3.6 × 10−7

The diameter of the wire is 3.6 × 10−7 m.



81. The number of people per square mile is the total numberof people divided by the number of square miles.

38, 0000.75

=3.8 × 104

7.5 × 10−1

≈ 0.50667 × 105

≈ (5.0667 × 10−1) × 105

≈ 5.0667 × 104

There are about 5.0667 × 104 people per square mile.

82. The average cost per mile is the total cost divided by thenumber of miles.

$210 × 106

17.6

=$210 × 106

1.76 × 10≈ $119 × 105

≈ ($1.19 × 102) × 105

≈ $1.19 × 107

The average cost per mile was about $1.19 × 107.

83. We multiply the number of light years by the number ofmiles in a light year.

4.22 × 5.88 × 1012 = 24.8136 × 1012

= (2.48136 × 10) × 1012

= 2.48136 × 1013

The distance from Earth to Alpha Centauri C is2.48136 × 1013 mi.

84. 3.26 × 5.88 × 1012

= 19.1688 × 1012

= (1.91688 × 10) × 1012

= 1.91688 × 1013 mi

85. First find the number of seconds in 1 hour:

1 hour = 1 hr✧ × 60 min✦1 hr✧

× 60 sec1 min✦ = 3600 sec

The number of disintegrations produced in 1 hour is thenumber of disintegrations per second times the number ofseconds in 1 hour.

37 billion × 3600= 37, 000, 000, 000 × 3600= 3.7 × 1010 × 3.6 × 103 Writing scientific

notation= (3.7 × 3.6) × (1010 × 103)= 13.32 × 1013 Multiplying= (1.332 × 10) × 1013

= 1.332 × 1014

One gram of radium produces 1.332× 1014 disintegrationsin 1 hour.

86. 2π × 93, 000, 000= 2π × 9.3 × 107

≈ 58 × 107

≈ (5.8 × 10) × 107

≈ 5.8 × 108 mi

87. = 5 · 3 + 8 · 32 + 4(6 − 2)= 5 · 3 + 8 · 32 + 4 · 4 Working inside parentheses= 5 · 3 + 8 · 9 + 4 · 4 Evaluating 32

= 15 + 72 + 16 Multiplying= 87 + 16 Adding in order= 103 from left to right

88. 5[3 − 8 · 32 + 4 · 6 − 2]= 5[3 − 8 · 9 + 4 · 6 − 2]= 5[3 − 72 + 24 − 2]= 5[−69 + 24 − 2]= 5[−45 − 2]= 5[−47]= −235

89. 16 ÷ 4 · 4 ÷ 2 · 256= 4 · 4 ÷ 2 · 256 Multiplying and dividing

in order from left to right= 16 ÷ 2 · 256= 8 · 256= 2048

90. 26 · 2−3 ÷ 210 ÷ 2−8

= 23 ÷ 210 ÷ 2−8

= 2−7 ÷ 2−8

= 2

91. 4(8 − 6)2 − 4 · 3 + 2 · 831 + 190

=4 · 22 − 4 · 3 + 2 · 8

3 + 1Calculating in thenumerator and inthe denominator

=4 · 4 − 4 · 3 + 2 · 8

4

=16 − 12 + 16

4

=4 + 16

4

=204

= 5


Exercise Set R.2 7

92. [4(8 − 6)2 + 4](3 − 2 · 8)22(23 + 5)

=[4 · 22 + 4](3 − 16)

22(8 + 5)

=[4 · 4 + 4](−13)

22 · 13

=[16 + 4](−13)

4 · 13

=20(−13)

52

=−26052

= −5

93. Since interest is compounded semiannually, n = 2. Substi-tute $3225 for P , 3.1% or 0.031 for i, 2 for n, and 4 for tin the compound interest formula.

A = P(

1 +i

n

)nt

= $3225(

1 +0.031

2

)2·4Substituting

= $3225(1 + 0.0155)2·4 Dividing= $3225(1.0155)2·4 Adding= $3225(1.0155)8 Multiplying 2 and 4≈ $3225(1.130939628) Evaluating the

exponential expression≈ $3647.2803 Multiplying≈ $3647.28 Rounding to the nearest cent

94. A = $7550(

1 +0.028

2

)2·5≈ $8, 676.14

95. Since interest is compounded quarterly, n = 4. Substitute$4100 for P , 2.3% or 0.023 for i, 4 for n, and 6 for t in thecompound interest formula.

A = P(

1 +i

n

)nt

= $4100(

1 +0.023

4

)4·6Substituting

= $4100(1 + 0.00575)4·6 Dividing= $4100(1.00575)4·6 Adding= $4100(1.00575)24 Multiplying 4 and 6≈ $4100(1.147521919) Evaluating the

exponential expression≈ $4704.839868 Multiplying≈ $4704.84 Rounding to the nearest cent

96. A = $4875(

1 +0.018

4

)4·9≈ $5730.24

97. Substitute $250 for P , 0.05 for r and 27 for t and performthe resulting computation.

S = P

⎡

⎣

(

1 +r

12

)12·t− 1

r

12

⎤

⎦

= $250

⎡

⎣

(

1 +0.0512

)12·27− 1

0.0512

⎤

⎦

≈ $170, 797.30

98. t =65 − 25 = 40

S =$100

⎡

⎣

(

1 +0.0412

)12·40− 1

0.0412

⎤

⎦ ≈ $118, 196.13

99. Substitute $120,000 for S, 0.03 for r, and 18 for t and solvefor P .

S = P

⎡

⎣

(

1 +r

12

)12·t− 1

r

12

⎤

⎦

$120, 000 = P

⎡

⎣

(

1 +0.0312

)12·18− 1

0.0312

⎤

⎦

$120, 000 = P

[

(1.0025)216 − 10.0025

]

$120, 000 ≈ P (285.94035)$419.67 ≈ P

100. t =70 − 30 = 40

$200, 000 =P

⎡

⎣

(

1 +0.04512

)12·40− 1

0.04512

⎤

⎦

P ≈$149.13

101. (xt · x3t)2 = (x4t)2 = x4t·2 = x8t

102. (xy · x−y)3 = (x0)3 = 13 = 1

103. (ta+x · tx−a)4 = (t2x)4 = t2x·4 = t8x

104. (mx−b · nx+b)x(mbn−b)x

= (mx2−bxnx2+bx)(mbxn−bx)

= mx2nx2

105.[ (3xayb)3

(−3xayb)2]2

=[27x3ay3b

9x2ay2b

]2

=[

3xayb]2

= 9x2ay2b



106.[(xr

yt

)2(x2r

y4t

)−2]−3

=[(x2r

y2t

)(x−4r

y−8t

)]−3

=(x−2r

y−6t

)−3

=x6r

y18t, or x6ry−18t

Exercise Set R.3

1. 7x3 − 4x2 + 8x + 5 = 7x3 + (−4x2) + 8x + 5Terms: 7x3, −4x2, 8x, 5The degree of the term of highest degree, 7x3, is 3. Thus,the degree of the polynomial is 3.

2. −3n4 − 6n3 + n2 + 2n− 1 =−3n4 + (−6n3) + n2 + 2n + (−1)Terms: −3n4, −6n3, n2, 2n, −1The degree of the term of highest degree, −3n4, is 4. Thus,the degree of the polynomial is 4.

3. 3a4b− 7a3b3 + 5ab− 2 = 3a4b + (−7a3b3) + 5ab + (−2)Terms: 3a4b, −7a3b3, 5ab, −2The degrees of the terms are 5, 6, 2, and, 0, respectively,so the degree of the polynomial is 6.

4. 6p3q2 − p2q4 − 3pq2 + 5 = 6p3q2 + (−p2q4) + (−3pq2) + 5Terms: 6p3q2, −p2q4, −3pq2, 5The degrees of the terms are 5, 6, 3, and 0, respectively,so the degree of the polynomial is 6.

5. (3ab2 − 4a2b− 2ab + 6)+(−ab2 − 5a2b + 8ab + 4)

= (3 − 1)ab2 + (−4 − 5)a2b + (−2 + 8)ab + (6 + 4)= 2ab2 − 9a2b + 6ab + 10

6. −2m2 + 5mn2 − 11mn− 7

7. (2x + 3y + z − 7) + (4x− 2y − z + 8)+(−3x + y − 2z − 4)

= (2 + 4 − 3)x + (3 − 2 + 1)y + (1 − 1 − 2)z+(−7 + 8 − 4)

= 3x + 2y − 2z − 3

8. 7x2 + 12xy − 2x− y − 9

9. (3x2 − 2x− x3 + 2) − (5x2 − 8x− x3 + 4)= (3x2 − 2x− x3 + 2) + (−5x2 + 8x + x3 − 4)= (3 − 5)x2 + (−2 + 8)x + (−1 + 1)x3 + (2 − 4)= −2x2 + 6x− 2

10. −4x2 + 8xy − 5y2 + 3

11. (x4 − 3x2 + 4x) − (3x3 + x2 − 5x + 3)= (x4 − 3x2 + 4x) + (−3x3 − x2 + 5x− 3)= x4 − 3x3 + (−3 − 1)x2 + (4 + 5)x− 3= x4 − 3x3 − 4x2 + 9x− 3

12. 2x4 − 5x3 − 5x2 + 10x− 5

13. (3a2)(−7a4) = [3(−7)](a2 · a4)= −21a6

14. 72y6

15. (6xy3)(9x4y2) = (6 · 9)(x · x4)(y3 · y2)= 54x5y5

16. −30m6n5

17. (a− b)(2a3 − ab + 3b2)= (a− b)(2a3) + (a− b)(−ab) + (a− b)(3b2)

Using the distributive property= 2a4 − 2a3b− a2b + ab2 + 3ab2 − 3b3

Using the distributive propertythree more times

= 2a4 − 2a3b− a2b + 4ab2 − 3b3 Collecting liketerms

18. (n + 1)(n2 − 6n− 4)= (n + 1)(n2) + (n + 1)(−6n) + (n + 1)(−4)= n3 + n2 − 6n2 − 6n− 4n− 4= n3 − 5n2 − 10n− 4

19. (y − 3)(y + 5)= y2 + 5y − 3y − 15 Using FOIL= y2 + 2y − 15 Collecting like terms

20. (z + 4)(z − 2) = z2 − 2z + 4z − 8 = z2 + 2z − 8

21. (x + 6)(x + 3)= x2 + 3x + 6x + 18 Using FOIL= x2 + 9x + 18 Collecting like terms

22. (a− 8)(a− 1) = a2 − a− 8a + 8 = a2 − 9a + 8

23. (2a + 3)(a + 5)= 2a2 + 10a + 3a + 15 Using FOIL= 2a2 + 13a + 15 Collecting like terms

24. (3b + 1)(b− 2) = 3b2 − 6b + b− 2 = 3b2 − 5b− 2

25. (2x + 3y)(2x + y)= 4x2 + 2xy + 6xy + 3y2 Using FOIL= 4x2 + 8xy + 3y2

26. (2a− 3b)(2a− b) = 4a2 − 2ab− 6ab + 3b2 =4a2 − 8ab + 3b2


Exercise Set R.3 9

27. (x + 3)2

= x2 + 2 · x · 3 + 32

[(A + B)2 = A2 + 2AB + B2]= x2 + 6x + 9

28. (z + 6)2 = z2 + 2 · z · 6 + 62 = z2 + 12z + 36

29. (y − 5)2

= y2 − 2 · y · 5 + 52

[(A−B)2 = A2 − 2AB + B2]= y2 − 10y + 25

30. (x− 4)2 = x2 − 2 · x · 4 + 42 = x2 − 8x + 16

31. (5x− 3)2

= (5x)2 − 2 · 5x · 3 + 32

[(A−B)2 = A2 − 2AB +B2]= 25x2 − 30x + 9

32. (3x− 2)2 = (3x)2 − 2 · 3x · 2 + 22 = 9x2 − 12x + 4

33. (2x + 3y)2

= (2x)2 + 2(2x)(3y) + (3y)2

[(A+B)2 = A2+2AB+B2]

= 4x2 + 12xy + 9y2

34. (5x + 2y)2 = (5x)2 + 2 · 5x · 2y + (2y)2 =25x2 + 20xy + 4y2

35. (2x2 − 3y)2

= (2x2)2 − 2(2x2)(3y) + (3y)2

[(A−B)2 = A2 − 2AB + B2]= 4x4 − 12x2y + 9y2

36. (4x2 − 5y)2 = (4x2)2 − 2 · 4x2 · 5y + (5y)2 =16x4 − 40x2y + 25y2

37. (n + 6)(n− 6)= n2 − 62 [(A + B)(A−B) = A2 −B2]= n2 − 36

38. (m + 1)(m− 1) = m2 − 12 = m2 − 1

39. (3y + 4)(3y − 4)= (3y)2 − 42 [(A + B)(A−B) = A2 −B2]= 9y2 − 16

40. (2x− 7)(2x + 7) = (2x)2 − 72 = 4x2 − 49

41. (3x− 2y)(3x + 2y)= (3x)2 − (2y)2 [(A−B)(A+B) = A2−B2]

= 9x2 − 4y2

42. (3x + 5y)(3x− 5y) = (3x)2 − (5y)2 = 9x2 − 25y2

43. (2x + 3y + 4)(2x + 3y − 4)= [(2x + 3y) + 4][(2x + 3y) − 4]= (2x + 3y)2 − 42

= 4x2 + 12xy + 9y2 − 16

44. (5x + 2y + 3)(5x + 2y − 3) = (5x + 2y)2 − 32 =25x2 + 20xy + 4y2 − 9

45. (x + 1)(x− 1)(x2 + 1)= (x2 − 1)(x2 + 1)= x4 − 1

46. (y − 2)(y + 2)(y2 + 4)= (y2 − 4)(y2 + 4)= y4 − 16

47. (an + bn)(an − bn) = (an)2 − (bn)2

= a2n − b2n

48. (ta + 4)(ta − 7) = (ta)2 − 7ta + 4ta − 28 =t2a − 3ta − 28

49. (an + bn)2 = (an)2 + 2 · an · bn + (bn)2

= a2n + 2anbn + b2n

50. (x3m − t5n)2 = (x3m)2 − 2 · x3m · t5n + (t5n)2 =x6m − 2x3mt5n + t10n

51. (x− 1)(x2 + x + 1)(x3 + 1)= [(x− 1)x2 + (x− 1)x + (x− 1) · 1](x3 + 1)= (x3 − x2 + x2 − x + x− 1)(x3 + 1)= (x3 − 1)(x3 + 1)= (x3)2 − 12

= x6 − 1

52. [(2x− 1)2 − 1]2

= [4x2 − 4x + 1 − 1]2

= [4x2 − 4x]2

= (4x2)2 − 2(4x2)(4x) + (4x)2

= 16x4 − 32x3 + 16x2

53. (xa−b)a+b

= x(a−b)(a+b)

= xa2−b2

54. (tm+n)m+n · (tm−n)m−n

= tm2+2mn+n2 · tm2−2mn+n2

= t2m2+2n2

55. (a + b + c)2

= (a + b + c)(a + b + c)= (a + b + c)(a) + (a + b + c)(b) + (a + b + c)(c)= a2 + ab + ac + ab + b2 + bc + ac + bc + c2

= a2 + b2 + c2 + 2ab + 2ac + 2bc



Exercise Set R.4

1. 3x + 18 = 3 · x + 3 · 6 = 3(x + 6)

2. 5y − 20 = 5 · y − 5 · 4 = 5(y − 4)

3. 2z3 − 8z2 = 2z2 · z − 2z2 · 4 = 2z2(z − 4)

4. 12m2 + 3m6 = 3m2 · 4 + 3m2 ·m4 = 3m2(4 + m4)

5. 4a2 − 12a + 16 = 4 · a2 − 4 · 3a + 4 · 4 = 4(a2 − 3a + 4)

6. 6n2 + 24n− 18 = 6 · n2 + 6 · 4n− 6 · 3 = 6(n2 + 4n− 3)

7. a(b− 2) + c(b− 2) = (b− 2)(a + c)

8. a(x2 − 3) − 2(x2 − 3) = (x2 − 3)(a− 2)

9. 3x3 − x2 + 18x− 6= x2(3x− 1) + 6(3x− 1)= (3x− 1)(x2 + 6)

10 x3 + 3x2 + 6x + 18= x2(x + 3) + 6(x + 3)= (x + 3)(x2 + 6)

11. y3 − y2 + 2y − 2= y2(y − 1) + 2(y − 1)= (y − 1)(y2 + 2)

12. y3 − y2 + 3y − 3= y2(y − 1) + 3(y − 1)= (y − 1)(y2 + 3)

13. 24x3 − 36x2 + 72x− 108= 12(2x3 − 3x2 + 6x− 9)= 12[x2(2x− 3) + 3(2x− 3)]= 12(2x− 3)(x2 + 3)

14. 5a3 − 10a2 + 25a− 50= 5(a3 − 2a2 + 5a− 10)= 5[a2(a− 2) + 5(a− 2)]= 5(a− 2)(a2 + 5)

15. x3 − x2 − 5x + 5= x2(x− 1) − 5(x− 1)= (x− 1)(x2 − 5)

16. t3 + 6t2 − 2t− 12= t2(t + 6) − 2(t + 6)= (t + 6)(t2 − 2)

17. a3 − 3a2 − 2a + 6= a2(a− 3) − 2(a− 3)= (a− 3)(a2 − 2)

18. x3 − x2 − 6x + 6= x2(x− 1) − 6(x− 1)= (x− 1)(x2 − 6)

19. w2 − 7w + 10We look for two numbers with a product of 10 and a sumof −7. By trial, we determine that they are −5 and −2.

w2 − 7w + 10 = (w − 5)(w − 2)

20. Note that 2 · 4 = 8 and 2 + 4 = 6. Thenp2 + 6p + 8 = (p + 2)(p + 4).

21. x2 + 6x + 5We look for two numbers with a product of 5 and a sumof 6. By trial, we determine that they are 1 and 5.

x2 + 6x + 5 = (x + 1)(x + 5)

22. Note that (−2)(−6) = 12 and −2 + (−6) = −8. Thenx2 − 8x + 12 = (x− 2)(x− 6).

23. t2 + 8t + 15We look for two numbers with a product of 15 and a sumof 8. By trial, we determine that they are 3 and 5.

t2 + 8t + 15 = (t + 3)(t + 5)

24. Note that 3 · 9 = 27 and 3 + 9 = 12. Theny2 + 12y + 27 = (y + 3)(y + 9).

25. x2 − 6xy − 27y2

We look for two numbers with a product of −27 and a sumof −6. By trial, we determine that they are 3 and −9.

x2 − 6xy − 27y2 = (x + 3y)(x− 9y)

26. Note that 3(−5) = −15 and 3 + (−5) = −2. Thent2 − 2t− 15 = (t + 3)(t− 5).

27. 2n2 − 20n− 48 = 2(n2 − 10n− 24)Now factor n2−10n−24. We look for two numbers with aproduct of −24 and a sum of −10. By trial, we determinethat they are 2 and −12. Then n2 − 10n− 24 =(n + 2)(n − 12). We must include the common factor, 2,to have a factorization of the original trinomial.

2n2 − 20n− 48 = 2(n + 2)(n− 12)

28. 2a2 − 2ab− 24b2 = 2(a2 − ab− 12b2)Note that −4 · 3 = −12 and −4 + 3 = −1. Then

2a2 − 2ab− 24b2 = 2(a− 4b)(a + 3b).

29. y2 − 4y − 21We look for two numbers with a product of −21 and a sumof −4. By trial, we determine that they are 3 and −7.

y2 − 4y − 21 = (y + 3)(y − 7)

30. Note that 9(−10) = −90 and 9 + (−10) = −1. Thenm2 −m− 90 = (m + 9)(m− 10).


Exercise Set R.4 11

31. y4 − 9y3 + 14y2 = y2(y2 − 9y + 14)Now factor y2 − 9y + 14. Look for two numbers with aproduct of 14 and a sum of −9. The numbers are −2 and−7. Then y2 − 9y + 14 = (y − 2)(y − 7). We must includethe common factor, y2, in order to have a factorization ofthe original trinomial.

y4 − 9y3 + 14y2 = y2(y − 2)(y − 7)

32. 3z3 − 21z2 + 18z = 3z(z2 − 7z + 6) = 3z(z − 1)(z − 6)

33. 2x3 − 2x2y − 24xy2 = 2x(x2 − xy − 12y2)Now factor x2 − xy − 12y2. Look for two numbers with aproduct of −12 and a sum of −1. The numbers are −4 and3. Then x2−xy−12y2 = (x−4y)(x+3y). We must includethe common factor, 2x, in order to have a factorization ofthe original trinomial.

2x3 − 2x2y − 24xy2 = 2x(x− 4y)(x + 3y)

34. a3b−9a2b2+20ab3 = ab(a2−9ab+20b2) = ab(a−4b)(a−5b)

35. 2n2 + 9n− 56We use the FOIL method.1. There is no common factor other than 1 or −1.2. The factorization must be of the form

(2n+ )(n+ ).3. Factor the constant term, −56. The possibilities

are −1 ·56, 1(−56), −2 ·28, 2(−28), −4 ·16, 4(−16),−7 · 8, and 7(−8). The factors can be written inthe opposite order as well: 56(−1), −56 · 1, 28(−2),−28 · 2, 16(−4), −16 · 4, 8(−7), and −8 · 7.

4. Find a pair of factors for which the sum of the out-side and the inside products is the middle term,9n. By trial, we determine that the factorizationis (2n− 7)(n + 8).

36. 3y2 + 7y − 20 = (3y − 5)(y + 4)

37. 12x2 + 11x + 2We use the grouping method.1. There is no common factor other than 1 or −1.2. Multiply the leading coefficient and the constant:

12 · 2 = 24.3. Try to factor 24 so that the sum of the factors is the

coefficient of the middle term, 11. The factors wewant are 3 and 8.

4. Split the middle term using the numbers found instep (3):

11x = 3x + 8x5. Factor by grouping.

12x2 + 11x + 2 = 12x2 + 3x + 8x + 2= 3x(4x + 1) + 2(4x + 1)= (4x + 1)(3x + 2)

38. 6x2 − 7x− 20 = (3x + 4)(2x− 5)

39. 4x2 + 15x + 9We use the FOIL method.1. There is no common factor other than 1 or −1.2. The factorization must be of the form

(4x+ )(x+ ) or (2x+ )(2x+ ).3. Factor the constant term, 9. The possibilities are

1 · 9, −1(−9), 3 · 3, and −3(−3). The first two pairsof factors can be written in the opposite order aswell: 9 · 1, −9(−1).

4. Find a pair of factors for which the sum of the out-side and the inside products is the middle term,15x. By trial, we determine that the factorizationis (4x + 3)(x + 3).

40. 2y2 + 7y + 6 = (2y + 3)(y + 2)

41. 2y2 + y − 6We use the grouping method.1. There is no common factor other than 1 or −1.2. Multiply the leading coefficient and the constant:

2(−6) = −12.3. Try to factor −12 so that the sum of the factors is

the coefficient of the middle term, 1. The factors wewant are 4 and −3.


y = 4y − 3y5. Factor by grouping.

2y2 + y − 6 = 2y2 + 4y − 3y − 6= 2y(y + 2) − 3(y + 2)= (y + 2)(2y − 3)

42. 20p2 − 23p + 6 = (4p− 3)(5p− 2)

43. 6a2 − 29ab + 28b2

We use the FOIL method.1. There is no common factor other than 1 or −1.2. The factorization must be of the form

(6x+ )(x+ ) or (3x+ )(2x+ ).3. Factor the coefficient of the last term, 28. The pos-

sibilities are 1 · 28, −1(−28), 2 · 14, −2(−14), 4 · 7,and −4(−7). The factors can be written in the op-posite order as well: 28·1, −28(−1), 14·2, −14(−2),7 · 4, and −7(−4).

4. Find a pair of factors for which the sum of the out-side and the inside products is the middle term,−29. Observe that the second term of each bino-mial factor will contain a factor of b. By trial, wedetermine that the factorization is (3a−4b)(2a−7b).

44. 10m2 + 7mn− 12n2 = (5m− 4n)(2m + 3n)



45. 12a2 − 4a− 16We will use the grouping method.1. Factor out the common factor, 4.

12a2 − 4a− 16 = 4(3a2 − a− 4)2. Now consider 3a2 − a − 4. Multiply the leading

coefficient and the constant: 3(−4) = −12.3. Try to factor −12 so that the sum of the factors is

the coefficient of the middle term, −1. The factorswe want are −4 and 3.


−a = −4a + 3a5. Factor by grouping.

3a2 − a− 4 = 3a2 − 4a + 3a− 4= a(3a− 4) + (3a− 4)= (3a− 4)(a + 1)

We must include the common factor to get a factor-ization of the original trinomial.

12a2 − 4a− 16 = 4(3a− 4)(a + 1)

46. 12a2 − 14a− 20 = 2(6a2 − 7a− 10) = 2(6a + 5)(a− 2)

47. z2 − 81 = z2 − 92 = (z + 9)(z − 9)

48. m2 − 4 = (m + 2)(m− 2)

49. 16x2 − 9 = (4x)2 − 32 = (4x + 3)(4x− 3)

50. 4z2 − 81 = (2z + 9)(2z − 9)

51. 6x2 − 6y2 = 6(x2 − y2) = 6(x + y)(x− y)

52. 8a2 − 8b2 = 8(a2 − b2) = 8(a + b)(a− b)

53. 4xy4 − 4xz2 = 4x(y4 − z2)= 4x[(y2)2 − z2]= 4x(y2 + z)(y2 − z)

54. 5x2y − 5yz4 = 5y(x2 − z4) = 5y(x + z2)(x− z2)

55. 7pq4 − 7py4 = 7p(q4 − y4)= 7p[(q2)2 − (y2)2]= 7p(q2 + y2)(q2 − y2)= 7p(q2 + y2)(q + y)(q − y)

56. 25ab4 − 25az4 = 25a(b4 − z4)= 25a(b2 + z2)(b2 − z2)= 25a(b2 + z2)(b + z)(b− z)

57. x2 + 12x + 36 = x2 + 2 · x · 6 + 62

= (x + 6)2

58. y2 − 6y + 9 = (y − 3)2

59. 9z2 − 12z + 4 = (3z)2 − 2 · 3z · 2 + 22 = (3z − 2)2

60. 4z2 + 12z + 9 = (2z + 3)2

61. 1 − 8x + 16x2 = 12 − 2 · 1 · 4x + (4x)2

= (1 − 4x)2

62. 1 + 10x + 25x2 = (1 + 5x)2

63. a3 + 24a2 + 144a= a(a2 + 24a + 144)= a(a2 + 2 · a · 12 + 122)= a(a + 12)2

64. y3 − 18y2 + 81y = y(y2 − 18y + 81) = y(y − 9)2

65. 4p2 − 8pq + 4q2

= 4(p2 − 2pq + q2)= 4(p− q)2

66. 5a2 − 10ab + 5b2 = 5(a2 − 2ab + b2) = 5(a− b)2

67. x3 + 64 = x3 + 43

= (x + 4)(x2 − 4x + 16)

68. y3 − 8 = (y − 2)(y2 + 2y + 4)

69. m3 − 216 = m3 − 63

= (m− 6)(m2 + 6m + 36)

70. n3 + 1 = (n + 1)(n2 − n + 1)

71. 8t3 + 8 = 8(t3 + 1)= 8(t3 + 13)= 8(t + 1)(t2 − t + 1)

72. 2y3 − 128 = 2(y3 − 64) = 2(y − 4)(y2 + 4y + 16)

73. 3a5 − 24a2 = 3a2(a3 − 8)= 3a2(a3 − 23)= 3a2(a− 2)(a2 + 2a + 4)

74. 250z4 − 2z = 2z(125z3 − 1)= 2z(5z − 1)(25z2 + 5z + 1)

75. t6 + 1 = (t2)3 + 13

= (t2 + 1)(t4 − t2 + 1)

76. 27x6 − 8 = (3x2 − 2)(9x4 + 6x2 + 4)

77. 18a2b− 15ab2 = 3ab · 6a− 3ab · 5b= 3ab(6a− 5b)

78. 4x2y + 12xy2 = 4xy(x + 3y)

79. x3 − 4x2 + 5x− 20 = x2(x− 4) + 5(x− 4)= (x− 4)(x2 + 5)

80. z3 + 3z2 − 3z − 9 = z2(z + 3) − 3(z + 3)= (z + 3)(z2 − 3)

81. 8x2 − 32 = 8(x2 − 4)= 8(x + 2)(x− 2)


Exercise Set R.4 13

82. 6y2 − 6 = 6(y2 − 1) = 6(y + 1)(y − 1)

83. 4y2 − 5There are no common factors. We might try to factorthis polynomial as a difference of squares, but there is nointeger which yields 5 when squared. Thus, the polynomialis prime.

84. There are no common factors and there is no integer whichyields 7 when squared, so 16x2 − 7 is prime.

85. m2 − 9n2 = m2 − (3n)2

= (m + 3n)(m− 3n)

86. 25t2 − 16 = (5t + 4)(5t− 4)

87. x2 + 9x + 20We look for two numbers with a product of 20 and a sumof 9. They are 4 and 5.

x2 + 9x + 20 = (x + 4)(x + 5)

88. Note that 3(−2) = −6 and 3 + (−2) = 1. Theny2 + y − 6 = (y + 3)(y − 2).

89. y2 − 6y + 5We look for two numbers with a product of 5 and a sumof −6. They are −5 and −1.

y2 − 6y + 5 = (y − 5)(y − 1)

90. Note that −7(3) = −21 and −7 + 3 = −4.x2 − 4x− 21 = (x− 7)(x + 3)

91. 2a2 + 9a + 4We use the FOIL method.1. There is no common factor other than 1 or −1.2. The factorization must be of the form

(2a+ )(a+ ).3. Factor the constant term, 4. The possibilities are

1 ·4, −1(−4), and 2 ·2. The first two pairs of factorscan be written in the opposite order as well: 4 · 1,−4(−1).

4. Find a pair of factors for which the sum of the out-side and the inside products is the middle term,9a. By trial, we determine that the factorizationis (2a + 1)(a + 4).

92. 3b2 − b− 2 = (3b + 2)(b− 1)

93. 6x2 + 7x− 3We use the grouping method.1. There is no common factor other than 1 or −1.2. Multiply the leading coefficient and the constant:

6(−3) = −18.3. Try to factor −18 so that the sum of the factors is

the coefficient of the middle term, 7. The factors wewant are 9 and −2.


7x = 9x− 2x5. Factor by grouping.

6x2 + 7x− 3 = 6x2 + 9x− 2x− 3= 3x(2x + 3) − (2x + 3)= (2x + 3)(3x− 1)

94. 8x2 + 2x− 15 = (4x− 5)(2x + 3)

95. y2 − 18y + 81 = y2 − 2 · y · 9 + 92

= (y − 9)2

96. n2 + 2n + 1 = (n + 1)2

97. 9z2 − 24z + 16 = (3z)2 − 2 · 3z · 4 + 42

= (3z − 4)2

98. 4z2 + 20z + 25 = (2z + 5)2

99. x2y2 − 14xy + 49 = (xy)2 − 2 · xy · 7 + 72

= (xy − 7)2

100. x2y2 − 16xy + 64 = (xy − 8)2

101. 4ax2 + 20ax− 56a = 4a(x2 + 5x− 14)= 4a(x + 7)(x− 2)

102. 21x2y + 2xy − 8y = y(21x2 + 2x− 8)= y(7x− 4)(3x + 2)

103. 3z3 − 24 = 3(z3 − 8)= 3(z3 − 23)= 3(z − 2)(z2 + 2z + 4)

104. 4t3 + 108 = 4(t3 + 27)= 4(t + 3)(t2 − 3t + 9)

105. 16a7b + 54ab7

= 2ab(8a6 + 27b6)= 2ab[(2a2)3 + (3b2)3]= 2ab(2a2 + 3b2)(4a4 − 6a2b2 + 9b4)

106. 24a2x4 − 375a8x

= 3a2x(8x3 − 125a6)= 3a2x(2x− 5a2)(4x2 + 10a2x + 25a4)

107. y3 − 3y2 − 4y + 12= y2(y − 3) − 4(y − 3)= (y − 3)(y2 − 4)= (y − 3)(y + 2)(y − 2)

108. p3 − 2p2 − 9p + 18= p2(p− 2) − 9(p− 2)= (p− 2)(p2 − 9)= (p− 2)(p + 3)(p− 3)



109. x3 − x2 + x− 1= x2(x− 1) + (x− 1)= (x− 1)(x2 + 1)

110. x3 − x2 − x + 1= x2(x− 1) − (x− 1)= (x− 1)(x2 − 1)= (x− 1)(x + 1)(x− 1), or

(x− 1)2(x + 1)

111. 5m4 − 20 = 5(m4 − 4)= 5(m2 + 2)(m2 − 2)

112. 2x2 − 288 = 2(x2 − 144) = 2(x + 12)(x− 12)

113. 2x3 + 6x2 − 8x− 24= 2(x3 + 3x2 − 4x− 12)= 2[x2(x + 3) − 4(x + 3)]= 2(x + 3)(x2 − 4)= 2(x + 3)(x + 2)(x− 2)

114. 3x3 + 6x2 − 27x− 54= 3(x3 + 2x2 − 9x− 18)= 3[x2(x + 2) − 9(x + 2)]= 3(x + 2)(x2 − 9)= 3(x + 2)(x + 3)(x− 3)

115. 4c2 − 4cd− d2 = (2c)2 − 2 · 2c · d− d2

= (2c− d)2

116. 9a2 − 6ab + b2 = (3a− b)2

117. m6 + 8m3 − 20 = (m3)2 + 8m3 − 20We look for two numbers with a product of −20 and a sumof 8. They are 10 and −2.

m6 + 8m3 − 20 = (m3 + 10)(m3 − 2)

118. x4 − 37x2 + 36 = (x2 − 1)(x2 − 36)= (x + 1)(x− 1)(x + 6)(x− 6)

119. p− 64p4 = p(1 − 64p3)= p[13 − (4p)3]= p(1 − 4p)(1 + 4p + 16p2)

120. 125a− 8a4 = a(125 − 8a3) = a(5 − 2a)(25 + 10a + 4a2)

121. y4 − 84 + 5y2

= y4 + 5y2 − 84= u2 + 5u− 84 Substituting u for y2

= (u + 12)(u− 7)= (y2 + 12)(y2 − 7) Substituting y2 for u

122. 11x2 + x4 − 80= x4 + 11x2 − 80= u2 + 11u− 80 Substituting u for x2

= (u + 16)(u− 5)= (x2 + 16)(x2 − 5) Substituting x2 for u

123. y2 − 849

+27y = y2 +

27y − 8

49

=(

y +47

)(

y − 27

)

124. t2 − 27100

+35t = t2 +

35t− 27

100=(

t +910

)(

t− 310

)

125. x2 + 3x +94

= x2 + 2 · x · 32

+(

32

)2

=(

x +32

)2

126. x2 − 5x +254

=(

x− 52

)2

127. x2 − x +14

= x2 − 2 · x · 12

+(

12

)2

=(

x− 12

)2

128. x2 − 23x +

19

=(

x− 13

)2

129. (x + h)3 − x3

= [(x + h) − x][(x + h)2 + x(x + h) + x2]= (x + h− x)(x2 + 2xh + h2 + x2 + xh + x2)= h(3x2 + 3xh + h2)

130. (x + 0.01)2 − x2

= (x + 0.01 + x)(x + 0.01 − x)= 0.01(2x + 0.01), or 0.02(x + 0.005)

131. (y − 4)2 + 5(y − 4) − 24= u2 + 5u− 24 Substituting u for y − 4= (u + 8)(u− 3)= (y − 4 + 8)(y − 4 − 3) Substituting y − 4

for u= (y + 4)(y − 7)

132. 6(2p + q)2 − 5(2p + q) − 25= 6u2 − 5u− 25 Substituting u for 2p + q

= (3u + 5)(2u− 5)= [3(2p + q) + 5][2(2p + q) − 5] Substituting

2p + q for u= (6p + 3q + 5)(4p + 2q − 5)

133. x2n + 5xn − 24 = (xn)2 + 5xn − 24= (xn + 8)(xn − 3)

134. 4x2n − 4xn − 3 = (2xn − 3)(2xn + 1)


Exercise Set R.5 15

135. x2 + ax + bx + ab = x(x + a) + b(x + a)= (x + a)(x + b)

136. bdy2 + ady + bcy + ac

= dy(by + a) + c(by + a)= (by + a)(dy + c)

137. 25y2m − (x2n − 2xn + 1)= (5ym)2 − (xn − 1)2

= [5ym + (xn − 1)][5ym − (xn − 1)]= (5ym + xn − 1)(5ym − xn + 1)

138. x6a − t3b = (x2a)3 − (tb)3 =(x2a − tb)(x4a + x2atb + t2b)

139. (y − 1)4 − (y − 1)2

= (y − 1)2[(y − 1)2 − 1]= (y − 1)2[y2 − 2y + 1 − 1]= (y − 1)2(y2 − 2y)= y(y − 1)2(y − 2)

140. x6 − 2x5 + x4 − x2 + 2x− 1= x4(x2 − 2x + 1) − (x2 − 2x + 1)= (x2 − 2x + 1)(x4 − 1)= (x− 1)2(x2 + 1)(x2 − 1)= (x− 1)2(x2 + 1)(x + 1)(x− 1)= (x2 + 1)(x + 1)(x− 1)3

Exercise Set R.5

1. x− 5 = 7x = 12 Adding 5

The solution is 12.

2. y + 3 = 4y = 1

3. 3x + 4 = −83x = −12 Subtracting 4x = −4 Dividing by 3

The solution is −4.

4. 5x− 7 = 235x = 30x = 6

5. 5y − 12 = 35y = 15 Adding 12y = 3 Dividing by 5

The solution is 3.

6. 6x + 23 = 56x = −18x = −3

7. 6x− 15 = 456x = 60 Adding 15x = 10 Dividing by 6

The solution is 10.

8. 4x− 7 = 814x = 88x = 22

9. 5x− 10 = 455x = 55 Adding 10x = 11 Dividing by 5

The solution is 11.

10. 6x− 7 = 116x = 18x = 3

11. 9t + 4 = −59t = −9 Subtracting 4t = −1 Dividing by 9


12. 5x + 7 = −135x = −20x = −4

13. 8x + 48 = 3x− 125x + 48 = −12 Subtracting 3x

5x = −60 Subtracting 48x = −12 Dividing by 5


14. 15x + 40 = 8x− 97x = −49x = −7

15. 7y − 1 = 23 − 5y12y − 1 = 23 Adding 5y

12y = 24 Adding 1y = 2 Dividing by 12

The solution is 2.

16. 3x− 15 = 15 − 3x6x = 30x = 5

17. 3x− 4 = 5 + 12x−9x− 4 = 5 Subtracting 12x

−9x = 9 Adding 4x = −1 Dividing by −9




18. 9t− 4 = 14 + 15t−6t = 18

t = −3

19. 5 − 4a = a− 135 − 5a = −13 Subtracting a

−5a = −18 Subtracting 5

a =185

Dividing by −5

The solution is185

.

20. 6 − 7x = x− 14−8x = −20

x =52

21. 3m− 7 = −13 + m

2m− 7 = −13 Subtracting m

2m = −6 Adding 7m = −3 Dividing by 2


22. 5x− 8 = 2x− 83x = 0x = 0

23. 11 − 3x = 5x + 311 − 8x = 3 Subtracting 5x

−8x = −8 Subtracting 11x = 1

The solution is 1.

24. 20 − 4y = 10 − 6y2y = −10y = −5

25. 2(x + 7) = 5x + 142x + 14 = 5x + 14

−3x + 14 = 14 Subtracting 5x−3x = 0 Subtracting 14

x = 0The solution is 0.

26. 3(y + 4) = 8y3y + 12 = 8y

12 = 5y125

= y

27. 24 = 5(2t + 5)24 = 10t + 25−1 = 10t Subtracting 25

− 110

= t Dividing by 10

The solution is − 110

.

28. 9 = 4(3y − 2)9 = 12y − 8

17 = 12y1712

= y

29. 5y − (2y − 10) = 255y − 2y + 10 = 25

3y + 10 = 25 Collecting like terms3y = 15 Subtracting 10y = 5 Dividing by 3

The solution is 5.

30. 8x− (3x− 5) = 408x− 3x + 5 = 40

5x + 5 = 405x = 35x = 7

31. 7(3x + 6) = 11 − (x + 2)21x + 42 = 11 − x− 221x + 42 = 9 − x Collecting like terms22x + 42 = 9 Adding x

22x = −33 Subtracting 42

x = −32

Dividing by 22


32. 9(2x + 8) = 20 − (x + 5)18x + 72 = 20 − x− 518x + 72 = 15 − x

19x = −57x = −3

33. 4(3y − 1) − 6 = 5(y + 2)12y − 4 − 6 = 5y + 10

12y − 10 = 5y + 10 Collecting like terms7y − 10 = 10 Subtracting 5y

7y = 20 Adding 10

y =207

Dividing by 7

The solution is207

.

34. 3(2n− 5) − 7 = 4(n− 9)6n− 15 − 7 = 4n− 36

6n− 22 = 4n− 362n = −14n = −7


Exercise Set R.5 17

35. x2 + 3x− 28 = 0(x + 7)(x− 4) = 0 Factoring

x + 7 = 0 or x− 4 = 0 Principle of zero productsx = −7 or x = 4

The solutions are −7 and 4.

36. y2 − 4y − 45 = 0(y − 9)(y + 5) = 0

y − 9 = 0 or y + 5 = 0y = 9 or y = −5

37. x2 + 5x = 0x(x + 5) = 0 Factoring

x = 0 or x + 5 = 0 Principle of zero productsx = 0 or x = −5

The solutions are 0 and −5.

38. t2 + 6t = 0t(t + 6) = 0

t = 0 or t + 6 = 0t = 0 or t = −6

39. y2 + 6y + 9 = 0(y + 3)(y + 3) = 0

y + 3 = 0 or y + 3 = 0y = −3 or y = −3


40. n2 + 4n + 4 = 0(n + 2)(n + 2) = 0

n + 2 = 0 or n + 2 = 0n = −2 or n = −2

41. x2 + 100 = 20xx2 − 20x + 100 = 0 Subtracting 20x

(x− 10)(x− 10) = 0

x− 10 = 0 or x− 10 = 0x = 10 or x = 10

The solution is 10.

42. y2 + 25 = 10yy2 − 10y + 25 = 0(y − 5)(y − 5) = 0

y − 5 = 0 or y − 5 = 0y = 5 or y = 5

43. x2 − 4x− 32 = 0(x− 8)(x + 4) = 0

x− 8 = 0 or x + 4 = 0x = 8 or x = −4


44. t2 + 12t + 27 = 0(t + 9)(t + 3) = 0

t + 9 = 0 or t + 3 = 0t = −9 or t = −3

45. 3y2 + 8y + 4 = 0(3y + 2)(y + 2) = 0

3y + 2 = 0 or y + 2 = 03y = −2 or y = −2

y = −23or y = −2

The solutions are −23

and −2.

46. 9y2 + 15y + 4 = 0(3y + 4)(3y + 1) = 0

3y + 4 = 0 or 3y + 1 = 03y = −4 or 3y = −1

y = −43or y = −1

3

47. 12z2 + z = 612z2 + z − 6 = 0

(4z + 3)(3z − 2) = 0

4z + 3 = 0 or 3z − 2 = 04z = −3 or 3z = 2

z = −34or z =

23


and23.

48. 6x2 − 7x = 106x2 − 7x− 10 = 0

(6x + 5)(x− 2) = 0

6x + 5 = 0 or x− 2 = 06x = −5 or x = 2

x = −56or x = 2

49. 12a2 − 28 = 5a12a2 − 5a− 28 = 0

(3a + 4)(4a− 7) = 0

3a + 4 = 0 or 4a− 7 = 03a = −4 or 4a = 7

a = −43or a =

74


and74.



50. 21n2 − 10 = n

21n2 − n− 10 = 0(3n + 2)(7n− 5) = 0

3n + 2 = 0 or 7n− 5 = 03n = −2 or 7n = 5

n = −23or n =

57

51. 14 = x(x− 5)14 = x2 − 5x0 = x2 − 5x− 140 = (x− 7)(x + 2)

x− 7 = 0 or x + 2 = 0x = 7 or x = −2


52. 24 = x(x− 2)24 = x2 − 2x0 = x2 − 2x− 240 = (x− 6)(x + 4)

x− 6 = 0 or x + 4 = 0x = 6 or x = −4

53. x2 − 36 = 0(x + 6)(x− 6) = 0

x + 6 = 0 or x− 6 = 0x = −6 or x = 6


54. y2 − 81 = 0(y + 9)(y − 9) = 0

y + 9 = 0 or y − 9 = 0y = −9 or y = 9

55. z2 = 144z2 − 144 = 0

(z + 12)(z − 12) = 0

z + 12 = 0 or z − 12 = 0z = −12 or z = 12


56. t2 = 25t2 − 25 = 0

(t + 5)(t− 5) = 0

t + 5 = 0 or t− 5 = 0t = −5 or t = 5

57. 2x2 − 20 = 02x2 = 20x2 = 10

x =√

10 or x = −√

10 Principle of square rootsThe solutions are

√10 and −

√10, or ±

√10.

58. 3y2 − 15 = 03y2 = 15y2 = 5

y =√

5 or y = −√

5

59. 6z2 − 18 = 06z2 = 18z2 = 3

z =√

3 or z = −√

3The solutions are

√3 and −

√3, or ±

√3.

60. 5x2 − 75 = 05x2 = 75x2 = 15

x =√

15 or x = −√

15

61. A =12bh

2A = bh Multiplying by 2 on both sides2Ah

= b Dividing by h on both sides

62. A = πr2

A

r2= π

63. P = 2l + 2wP − 2l = 2w Subtracting 2l on both sides

P − 2l2

= w Dividing by 2 on both sides

64. A = P + Prt

A− P = Prt

A− P

Pt= r

65. A =12h(b1 + b2)

2A = h(b1 + b2) Multiplying by 2 onboth sides

2Ab1 + b2

= h Dividing by b1 + b2 on both sides

66. A =12h(b1 + b2)

2Ah

= b1 + b2

2Ah

− b1 = b2, or

2A− b1h

h= b2

67. V =43πr3

3V = 4πr3 Multiplying by 3 on both sides3V4r3

= π Dividing by 4r3 on both sides


Exercise Set R.5 19

68. V =43πr3

3V4π

= r3

69. F =95C + 32

F − 32 =95C Subtracting 32 on both sides

59(F − 32) = C Multiplying by

59

on both sides

70. Ax + By = C

By = C −Ax

y =C −Ax

B

71. Ax + By = C

Ax = C −By Subtracting By on both sides

A =C −By

xDividing by x on both sides

72. 2w + 2h + l = p

2w = p− 2h− l

w =p− 2h− l

2

73. 2w + 2h + l = p

2h = p− 2w − l Subtracting 2w and l

h =p− 2w − l

2Dividing by 2

74. 3x + 4y = 124y = 12 − 3x

y =12 − 3x

4

75. 2x− 3y = 6−3y = 6 − 2x Subtracting 2x

y =6 − 2x−3

, or Dividing by −3

2x− 63

76. T =310

(I − 12, 000)

103T = I − 12, 000

103T + 12, 000 = I, or

10T + 36, 0003

= I

77. a = b + bcd

a = b(1 + cd) Factoringa

1 + cd= b Dividing by 1 + cd

78. q = p− np

q = p(1 − n)q

1 − n= p

79. z = xy − xy2

z = x(y − y2) Factoringz

y − y2= x Dividing by y − y2

80. st = t− 4st− t = −4

t(s− 1) = −4

t =−4s− 1

, or4

1 − s

81. 3[5 − 3(4 − t)] − 2 = 5[3(5t− 4) + 8] − 263[5 − 12 + 3t] − 2 = 5[15t− 12 + 8] − 26

3[−7 + 3t] − 2 = 5[15t− 4] − 26−21 + 9t− 2 = 75t− 20 − 26

9t− 23 = 75t− 46−66t− 23 = −46

−66t = −23

t =2366

The solution is2366

.

82. 6[4(8 − y) − 5(9 + 3y)] − 21 = −7[3(7 + 4y) − 4]6[32 − 4y − 45 − 15y] − 21 = −7[21 + 12y − 4]

6[−13 − 19y] − 21 = −7[17 + 12y]−78 − 114y − 21 = −119 − 84y

−114y − 99 = −119 − 84y−30y = −20

y =23

83. x− {3x− [2x− (5x− (7x− 1))]} = x + 7x− {3x− [2x− (5x− 7x + 1)]} = x + 7

x− {3x− [2x− (−2x + 1)]} = x + 7x− {3x− [2x + 2x− 1]} = x + 7

x− {3x− [4x− 1]} = x + 7x− {3x− 4x + 1} = x + 7

x− {−x + 1} = x + 7x + x− 1 = x + 7

2x− 1 = x + 7x− 1 = 7

x = 8The solution is 8.



84. 23−2[4+3(x−1)]+5[x−2(x+3)]=7{x−2[5−(2x+3)]}23 − 2[4 + 3x− 3] + 5[x− 2x− 6]=7{x− 2[5 − 2x− 3]}

23 − 2[3x + 1] + 5[−x− 6]=7{x− 2[−2x + 2]}23 − 6x− 2 − 5x− 30=7{x + 4x− 4}

−11x− 9=7{5x− 4}−11x− 9=35x− 28

−46x=−19

x=1946

85. (5x2 + 6x)(12x2 − 5x− 2) = 0x(5x + 6)(4x + 1)(3x− 2) = 0

x = 0 or 5x+6 = 0 or 4x+1 = 0 or 3x−2 = 0x = 0 or 5x = −6 or 4x = −1 or 3x = 2

x = 0 or x = −65or x = −1

4or x =

23

The solutions are 0, −65, −1

4, and

23.

86. (3x2 + 7x− 20)(x2 − 4x) = 0(3x− 5)(x + 4)(x)(x− 4) = 0

3x− 5 = 0 or x + 4 = 0 or x = 0 or x− 4 = 0

x =53or x = −4 or x = 0 or x = 4

87. 3x3 + 6x2 − 27x− 54 = 03(x3 + 2x2 − 9x− 18) = 0

3[x2(x + 2) − 9(x + 2)] = 0 Factoring by grouping3(x + 2)(x2 − 9) = 0

3(x + 2)(x + 3)(x− 3) = 0

x + 2 = 0 or x + 3 = 0 or x− 3 = 0x = −2 or x = −3 or x = 3

The solutions are −2, −3, and 3.

88. 2x3 + 6x2 = 8x + 242x3 + 6x2 − 8x− 24 = 0

2(x3 + 3x2 − 4x− 12) = 02[x2(x + 3) − 4(x + 3)] = 0

2(x + 3)(x2 − 4) = 02(x + 3)(x + 2)(x− 2) = 0

x + 3 = 0 or x + 2 = 0 or x− 2 = 0x = −3 or x = −2 or x = 2

Exercise Set R.6

1. Since −34

is defined for all real numbers, the domain is{x|x is a real number}.

2. Solve: 7 − x = 07 = x

Since the denominator is 0 when x = 7, the domain is{x|x is a real number and x ̸= 7}.

3.3x− 3x(x− 1)The denominator is 0 when the factor x = 0 andalso when x − 1 = 0, or x = 1. The domain is{x|x is a real number and x ̸= 0 and x ̸= 1}.

4.15x− 10

2x(3x− 2)Since 2x = 0 when x = 0 and 3x− 2 = 0 whenx =

23, the domain is

{

x

∣

∣

∣

∣

x is a real number and x ̸= 0 and x ̸= 23

}

.

5.x + 5

x2 + 4x− 5=

x + 5(x + 5)(x− 1)

We see that x + 5 = 0 when x = −5 and x− 1 = 0when x = 1. Thus, the domain is{x|x is a real number and x ̸= −5 and x ̸= 1}.

6.(x2 − 4)(x + 1)(x + 2)(x2 − 1)

=(x2 − 4)(x + 1)

(x + 2)(x + 1)(x− 1)x+2 = 0 when x = −2; x+1 = 0 when x = −1; x− 1 = 0when x = 1. The domain is {x|x is a real number andx ̸= −2 and x ̸= −1 and x ̸= 1}.

7. We first factor the denominator completely.7x2 − 28x + 28

(x2−4)(x2+3x−10)=

7x2 − 28x + 28(x+2)(x−2)(x+5)(x−2)

We see that x + 2 = 0 when x = −2, x − 2 = 0 whenx = 2, and x + 5 = 0 when x = −5. Thus, the domain is{x|x is a real number and x ̸=−2 and x ̸=2 and x ̸=−5}.

8.7x2 + 11x− 6x(x2 − x− 6)

=7x2 + 11x− 6x(x− 3)(x + 2)

The denominator is 0 when x = 0 or when x − 3 = 0or when x + 2 = 0. Now x − 3 = 0 when x = 3and x + 2 = 0 when x = −2. Thus, the domain is{x|x is a real number and x ̸= 0 and x ̸= 3 andx ̸= −2}.

9.x2 − 4

x2 − 4x + 4=

(x + 2)(x−2)✏(x− 2)(x−2)✏ =

x + 2x− 2

10.x2 + 2x− 3

x2 − 9=

(x− 1)(x+3)✏(x+3)✏ (x− 3)

=x− 1x− 3

11. x3 − 6x2 + 9xx3 − 3x2

=x(x2 − 6x + 9)

x2(x− 3)

=x/(x−3)✏ (x− 3)

x/ · x(x−3)✏=

x− 3x

12. y5 − 5y4 + 4y3

y3 − 6y2 + 8y=

y3(y2 − 5y + 4)y(y2 − 6y + 8)

=y/ · y · y · (y − 1)(y−4)✏

y/(y − 2)(y−4)✏=

y2(y − 1)y − 2


Exercise Set R.6 21

13. 6y2 + 12y − 483y2 − 9y + 6

=6(y2 + 2y − 8)3(y2 − 3y + 2)

=2 · 3/ · (y + 4)(y−2)✏

3/(y − 1)(y−2)✏=

2(y + 4)y − 1

14. 2x2 − 20x + 5010x2 − 30x− 100

=2(x2 − 10x + 25)10(x2 − 3x− 10)

=2/(x−5)✏ (x− 5)

2/ · 5 · (x−5)✏ (x + 2)

=x− 5

5(x + 2)

15. 4 − x

x2 + 4x− 32=

−1(x−4)✏(x−4)✏ (x + 8)

=−1

x + 8, or − 1

x + 8

16. 6 − x

x2 − 36=

−1(x−6)✏(x + 6)(x−6)✏

=−1

x + 6, or − 1

x + 6

17. r − s

r + s· r2 − s2

(r − s)2=

(r − s)(r2 − s2)(r + s)(r − s)2

=(r−s)✏ (r−s)✏ (r+s)✏ · 1

(r+s)✏ (r−s)✏ (r−s)✏= 1

18. x2 − y2

(x− y)2· 1x + y

=(x2 − y2) · 1

(x− y)2(x + y)

=(x+y)✏ (x−y)✏ · 1

(x−y)✏ (x− y)(x+y)✏=

1x− y

19. x2 + 2x− 353x3 − 2x2

· 9x3 − 4x7x + 49

=(x+7)✏ (x− 5)(x)✧ (3x + 2)(3x−2)✘

x/ · x(3x−2)✘ (7)(x+7)✏=

(x− 5)(3x + 2)7x

20. x2 − 2x− 352x3 − 3x2

· 4x3 − 9x7x− 49

=(x−7)✏ (x + 5)(x)✧ (2x + 3)(2x−3)✘

x/ · x(2x−3)✘ (7)(x−7)✏=

(x + 5)(2x + 3)7x

21. a2 − a− 6a2 − 7a + 12

· a2 − 2a− 8a2 − 3a− 10

=(a−3)✏ (a + 2)(a−4)✏ (a+2)✏(a−4)✏ (a−3)✏ (a− 5)(a+2)✏

=a + 2a− 5

22. a2 − a− 12a2 − 6a + 8

· a2 + a− 6a2 − 2a− 24

=(a−4)✏ (a + 3)(a+3)(a−2)✏(a−2)✏ (a−4)✏ (a− 6)(a+4)

=(a + 3)2

(a− 6)(a + 4)

23. m2 − n2

r + s÷ m− n

r + s

=m2 − n2

r + s· r + s

m− n

=(m + n)(m−n)✏ (r+s)✏

(r+s)✏ (m−n)✏= m + n

24. a2 − b2

x− y÷ a + b

x− y

=a2 − b2

x− y· x− y

a + b

=(a+b)✏ (a− b)(x−y)✏

(x−y)✏ (a+b)✏ · 1= a− b

25. 3x + 122x− 8

÷ (x + 4)2

(x− 4)2

=3x + 122x− 8

· (x− 4)2

(x + 4)2

=3(x+4)✏ (x−4)✏ (x− 4)2(x−4)✏ (x+4)✏ (x + 4)

=3(x− 4)2(x + 4)

26. a2 − a− 2a2 − a− 6

÷ a2 − 2a2a + a2

=a2 − a− 2a2 − a− 6

· 2a + a2

a2 − 2a

=(a−2)✏ (a + 1)(a)✧ (2+a)✏(a− 3)(a+2)✏ (a)✧ (a−2)✏

=a + 1a− 3

27. x2 − y2

x3 − y3· x2 + xy + y2

x2 + 2xy + y2

=(x + y)(x− y)(x2 + xy + y2)

(x− y)(x2 + xy + y2)(x + y)(x + y)

=1

x + y· (x + y)(x− y)(x2 + xy + y2)(x + y)(x− y)(x2 + xy + y2)

=1

x + y· 1 Removing a factor of 1

=1

x + y



28. c3 + 8c2 − 4

÷ c2 − 2c + 4c2 − 4c + 4

=c3 + 8c2 − 4

· c2 − 4c + 4c2 − 2c + 4

=(c + 2)(c2 − 2c + 4)(c− 2)(c− 2)

(c + 2)(c− 2)(c2 − 2c + 4)

=(c + 2)(c2 − 2c + 4)(c− 2)(c + 2)(c2 − 2c + 4)(c− 2)

· c− 21

= c− 2

29. (x− y)2 − z2

(x + y)2 − z2÷ x− y + z

x + y − z

=(x− y)2 − z2

(x + y)2 − z2· x + y − z

x− y + z

=(x− y + z)(x− y − z)(x + y − z)(x + y + z)(x + y − z)(x− y + z)

=(x− y + z)(x + y − z)(x− y + z)(x + y − z)

· x− y − z

x + y + z

= 1 · x− y − z

x + y + zRemoving a factor of 1

=x− y − z

x + y + z

30. (a + b)2 − 9(a− b)2 − 9

· a− b− 3a + b + 3

=(a + b + 3)(a + b− 3)(a− b− 3)(a− b + 3)(a− b− 3)(a + b + 3)

=(a + b + 3)(a− b− 3)(a + b + 3)(a− b− 3)

· a + b− 3a− b + 3

=a + b− 3a− b + 3

31. 75x

+35x

=7 + 35x

=105x

=5/ · 25/ · x

=2x

32. 712y

− 112y

=6

12y

=6/ · 16/ · 2y

=12y

33. 43a + 4

+3a

3a + 4=

4 + 3a3a + 4

= 1 (4 + 3a = 3a + 4)

34. a− 3ba + b

+a + 5ba + b

=2a + 2ba + b

=2(a+b)✏

1 · (a+b)✏= 2

35. 54z

− 38z

, LCD is 8z

=54z

· 22− 3

8z

=108z

− 38z

=78z

36. 12x2y

+5

xy2, LCD is x2y2

=12yx2y2

+5xx2y2

=12y + 5xx2y2

37. 3x + 2

+2

x2 − 4

=3

x + 2+

2(x + 2)(x− 2)

, LCD is (x+2)(x−2)

=3

x + 2· x− 2x− 2

+2

(x + 2)(x− 2)

=3x− 6

(x + 2)(x− 2)+

2(x + 2)(x− 2)

=3x− 4

(x + 2)(x− 2)

38. 5a− 3

− 2a2 − 9

=5

a− 3− 2

(a + 3)(a− 3), LCD is (a + 3)(a− 3)

=5(a + 3) − 2

(a + 3)(a− 3)

=5a + 15 − 2

(a + 3)(a− 3)

=5a + 13

(a + 3)(a− 3)

39. y

y2 − y − 20− 2

y + 4

=y

(y + 4)(y − 5)− 2

y + 4, LCD is (y + 4)(y − 5)

=y

(y + 4)(y − 5)− 2

y + 4· y − 5y − 5

=y

(y + 4)(y − 5)− 2y − 10

(y + 4)(y − 5)

=y − (2y − 10)(y + 4)(y − 5)

=y − 2y + 10

(y + 4)(y − 5)

=−y + 10

(y + 4)(y − 5)


Exercise Set R.6 23

40. 6y2 + 6y + 9

− 5y + 3

=6

(y + 3)2− 5

y + 3, LCD is (y + 3)2

=6 − 5(y + 3)

(y + 3)2

=6 − 5y − 15

(y + 3)2

=−5y − 9(y + 3)2

41. 3x + y

+x− 5yx2 − y2

=3

x + y+

x− 5y(x + y)(x− y)

, LCD is (x + y)(x− y)

=3

x + y· x− y

x− y+

x− 5y(x + y)(x− y)

=3x− 3y

(x + y)(x− y)+

x− 5y(x + y)(x− y)

=4x− 8y

(x + y)(x− y)

42. a2 + 1a2 − 1

− a− 1a + 1

=a2 + 1

(a + 1)(a− 1)− a− 1

a + 1, LCD is (a + 1)(a− 1)

=a2 + 1 − (a− 1)(a− 1)

(a + 1)(a− 1)

=a2 + 1 − a2 + 2a− 1

(a + 1)(a− 1)

=2a

(a + 1)(a− 1)

43. y

y − 1+

21 − y

=y

y − 1+

−1−1

· 21 − y

=y

y − 1+

−2y − 1

=y − 2y − 1

44. a

a− b+

b

b− a=

a

a− b+

−b

a− b

=a− b

a− b= 1

45.x

2x− 3y− y

3y − 2x

=x

2x− 3y− −1

−1· y

3y − 2x

=x

2x− 3y− −y

2x− 3y

=x + y

2x− 3y[x− (−y) = x + y]

46. 3a3a− 2b

− 2a2b− 3a

=3a

3a− 2b− −2a

3a− 2b

=5a

3a− 2b

47. 9x + 23x2 − 2x− 8

+7

3x2 + x− 4

=9x + 2

(3x + 4)(x− 2)+

7(3x + 4)(x− 1)

,

LCD is (3x + 4)(x− 2)(x− 1)

=9x + 2

(3x+4)(x−2)· x− 1x− 1

+7

(3x+4)(x−1)· x− 2x− 2

=9x2 − 7x− 2

(3x+4)(x−2)(x−1)+

7x− 14(3x+4)(x−1)(x−2)

=9x2 − 16

(3x + 4)(x− 2)(x− 1)

=(3x+4)✘ (3x− 4)

(3x+4)✘ (x− 2)(x− 1)

=3x− 4

(x− 2)(x− 1)

48. 3yy2 − 7y + 10

− 2yy2 − 8y + 15

=3y

(y − 2)(y − 5)− 2y

(y − 5)(y − 3),

LCD is (y − 2)(y − 5)(y − 3)

=3y(y − 3) − 2y(y − 2)(y − 2)(y − 5)(y − 3)

=3y2 − 9y − 2y2 + 4y(y − 2)(y − 5)(y − 3)

=y2 − 5y

(y − 2)(y − 5)(y − 3)

=y(y−5)✏

(y − 2)(y−5)✏ (y − 3)

=y

(y − 2)(y − 3)

49. 5aa− b

+ab

a2 − b2+

4ba + b

=5a

a− b+

ab

(a + b)(a− b)+

4ba + b

,

LCD is (a + b)(a− b)

=5a

a− b· a + b

a + b+

ab

(a + b)(a− b)+

4ba + b

· a− b

a− b

=5a2 + 5ab

(a+b)(a−b)+

ab

(a+b)(a−b)+

4ab− 4b2

(a+b)(a−b)

=5a2 + 10ab− 4b2

(a + b)(a− b)



50. 6aa− b

− 3bb− a

+5

a2 − b2

=6a

a− b+

3ba− b

+5

(a + b)(a− b),

LCD is (a + b)(a− b)

=6a(a + b) + 3b(a + b) + 5

(a + b)(a− b)

=6a2 + 6ab + 3ab + 3b2 + 5

(a + b)(a− b)

=6a2 + 9ab + 3b2 + 5

(a + b)(a− b)

51. 7x + 2

− x + 84 − x2

+3x− 2

4 − 4x + x2

=7

x + 2− x + 8

(2 + x)(2 − x)+

3x− 2(2 − x)2

,

LCD is (2 + x)(2 − x)2

=7

2 + x· (2 − x)2

(2 − x)2− x + 8

(2 + x)(2 − x)· 2 − x

2 − x+

3x− 2(2 − x)2

· 2 + x

2 + x

=28−28x+7x2−(16−6x−x2)+3x2+4x−4

(2 + x)(2 − x)2

=28 − 28x + 7x2 − 16 + 6x + x2 + 3x2 + 4x− 4

(2 + x)(2 − x)2

=11x2 − 18x + 8(2 + x)(2 − x)2

, or11x2 − 18x + 8(x + 2)(x− 2)2

52. 6x + 3

− x + 49 − x2

+2x− 3

9 − 6x + x2

=6

x + 3− x + 4

(3 + x)(3 − x)+

2x− 3(3 − x)2

,

LCD is (3 + x)(3 − x)2

=6(3 − x)2 − (x + 4)(3 − x) + (2x− 3)(3 + x)

(3 + x)(3 − x)2

=54 − 36x + 6x2 + x2 + x− 12 + 2x2 + 3x− 9

(3 + x)(3 − x)2

=33 − 32x + 9x2

(3 + x)(3 − x)2, or

9x2 − 32x + 33(x + 3)(x− 3)2

53. 1x + 1

+x

2 − x+

x2 + 2x2 − x− 2

=1

x + 1+

x

2 − x+

x2 + 2(x + 1)(x− 2)

=1

x + 1+

−1−1

· x

2 − x+

x2 + 2(x + 1)(x− 2)

=1

x + 1+

−x

x− 2+

x2 + 2(x + 1)(x− 2)

,

LCD is (x + 1)(x− 2)

=1

x + 1· x− 2x− 2

+−x

x− 2· x + 1x + 1

+x2 + 2

(x + 1)(x− 2)

=x− 2

(x+1)(x−2)+

−x2 − x

(x+1)(x−2)+

x2 + 2(x+1)(x−2)

=x− 2 − x2 − x + x2 + 2

(x + 1)(x− 2)

=0

(x + 1)(x− 2)= 0

54. x− 1x− 2

− x + 1x + 2

− x− 64 − x2

=x− 1x− 2

− x + 1x + 2

− x− 6(2 + x)(2 − x)

=1 − x

2 − x− x + 1

x + 2− x− 6

(2 + x)(2 − x),

LCD is (2 + x)(2 − x)

=(1 − x)(2 + x) − (x + 1)(2 − x) − (x− 6)

(2 + x)(2 − x)

=2 − x− x2 + x2 − x− 2 − x + 6

(2 + x)(2 − x)

=6 − 3x

(2 + x)(2 − x)

=3(2−x)✏

(2 + x)(2−x)✏=

32 + x

55.a− b

ba2 − b2

ab

=a− b

b· ab

a2 − b2

=a− b

b· ab

(a + b)(a− b)

=a b✧(a−b)✏

b/ (a + b)(a−b)✏=

a

a + b


Exercise Set R.6 25

56.x2 − y2

xyx− y

y

=x2 − y2

xy· y

x− y

=(x + y)(x−y)✏ y✧

x y✧ (x−y)✏=

x + y

x

57.

x

y− y

x1y

+1x

=

x

y− y

x1y

+1x

· xyxy

, LCM is xy

=

(x

y− y

x

)

(xy)(1y

+1x

)

(xy)

=x2 − y2

x + y

=(x+y)✏ (x− y)

(x+y)✏ · 1= x− y

58.a

b− b

a1a− 1

b

=a2 − b2

b− aMultiplying by

ab

ab

=(a + b)(a− b)

b− a

=(a + b)(a−b)✏−1 · (a−b)✏

= −a− b

59.c +

8c2

1 +2c

=c · c

2

c2+

8c2

1 · cc

+2c

=

c3 + 8c2

c + 2c

=c3 + 8c2

· c

c + 2

=(c+2)✏ (c2 − 2c + 4)c/

c/ · c(c+2)✏=

c2 − 2c + 4c

60.a− a

b

b− b

a

=

ab− a

bab− b

a

=a(b− 1)

b· a

b(a− 1)

=a2(b− 1)b2(a− 1)

61. x2 + xy + y2

x2

y− y2

x

=x2 + xy + y2

x2

y· xx− y2

x· yy

=x2 + xy + y2

x3 − y3

xy

= (x2 + xy + y2) · xy

x3 − y3

=(x2 + xy + y2)(xy)

(x− y)(x2 + xy + y2)

=x2 + xy + y2

x2 + xy + y2· xy

x− y

= 1 · xy

x− y

=xy

x− y

62.a2

b+

b2

aa2 − ab + b2

=

a3 + b3

aba2 − ab + b2

=(a+b)(a2−ab+b2)

ab· 1a2 − ab + b2

=a + b

ab· a

2 − ab + b2

a2 − ab + b2

=a + b

ab

63. a− a−1

a + a−1=

a− 1a

a +1a

=a · a

a− 1

a

a · aa

+1a

=

a2 − 1a

a2 + 1a

=a2 − 1

a· a

a2 + 1

=a2 − 1a2 + 1

64. x−1 + y−1

x−3 + y−3=

1x

+1y

1x3

+1y3

=

( 1x

+1y

)

(x3y3)( 1x3

+1y3

)

(x3y3)

=x2y3 + x3y2

y3 + x3

=x2y2(y+x)✏

(y+x)✏ (y2 − yx + x2)

=x2y2

y2 − yx + x2



65.

1x− 3

+2

x + 33

x− 1− 4

x + 2

=

1x− 3

· x + 3x + 3

+2

x + 3· x− 3x− 3

3x− 1

· x + 2x + 2

− 4x + 2

· x− 1x− 1

=

x + 3 + 2(x− 3)(x− 3)(x + 3)

3(x + 2) − 4(x− 1)(x− 1)(x + 2)

=

x + 3 + 2x− 6(x− 3)(x + 3)

3x + 6 − 4x + 4(x− 1)(x + 2)

=

3x− 3(x− 3)(x + 3)

−x + 10(x− 1)(x + 2)

=3x− 3

(x− 3)(x + 3)· (x− 1)(x + 2)

−x + 10

=(3x− 3)(x− 1)(x + 2)

(x− 3)(x + 3)(−x + 10), or

3(x− 1)2(x + 2)(x− 3)(x + 3)(−x + 10)

66.5

x + 1− 3

x− 21

x− 5+

2x + 2

=

5(x− 2) − 3(x + 1)(x + 1)(x− 2)x + 2 + 2(x− 5)(x− 5)(x + 2)

=

5x− 10 − 3x− 3(x + 1)(x− 2)x + 2 + 2x− 10(x− 5)(x + 2)

=

2x− 13(x + 1)(x− 2)

3x− 8(x− 5)(x + 2)

=2x− 13

(x + 1)(x− 2)· (x− 5)(x + 2)

3x− 8

=(2x− 13)(x− 5)(x + 2)(x + 1)(x− 2)(3x− 8)

67.

a

1 − a+

1 + a

a1 − a

a+

a

1 + a

=

a

1 − a· aa

+1 + a

a· 1 − a

1 − a1 − a

a· 1 + a

1 + a+

a

1 + a· aa

=

a2 + (1 − a2)a(1 − a)

(1 − a2) + a2

a(1 + a)

=1

a/(1 − a)· a/(1 + a)

1

=1 + a

1 − a

68.1 − x

x+

x

1 + x1 + x

x+

x

1 − x

=

1 − x2 + x2

x(1 + x)1 − x2 + x2

x(1 − x)

=1

x(1 + x)· x(1 − x)

1

=x/(1 − x)x/(1 + x)

=1 − x

1 + x

69.

1a2

+2ab

+1b2

1a2

− 1b2

=

1a2

+2ab

+1b2

1a2

− 1b2

· a2b2

a2b2,

LCM is a2b2

=b2 + 2ab + a2

b2 − a2

=(b+a)✏ (b + a)(b+a)✏ (b− a)

=b + a

b− a

70.

1x2

− 1y2

1x2

− 2xy

+1y2

=y2 − x2

y2 − 2xy + x2

Multiplying byx2y2

x2y2

=(y + x)(y−x)✏(y − x)(y−x)✏

=y + x

y − x

71. (x + h)2 − x2

h=

x2 + 2xh + h2 − x2

h

=2xh + h2

h

=h/(2x + h)

h/ · 1= 2x + h

72.1

x + h− 1

xh

=

x− x− h

x(x + h)h

=−h

x(x + h)· 1h

=−1 · h/

xh/(x + h)

=−1

x(x + h)

73. (x + h)3 − x3

h=

x3 + 3x2h + 3xh2 + h3 − x3

h

=3x2h + 3xh2 + h3

h

=h/(3x2 + 3xh + h2)

h/ · 1= 3x2 + 3xh + h2


Exercise Set R.7 27

74.1

(x + h)2− 1

x2

h=

x2 − x2 − 2xh− h2

x2(x + h)2

h

=−2xh− h2

x2(x + h)2· 1h

=h/(−2x− h)x2h/(x + h)2

=−2x− h

x2(x + h)2

75.

⎡

⎢

⎣

x + 1x− 1

+ 1x + 1x− 1

− 1

⎤

⎥

⎦

5

=

⎡

⎢

⎣

(x + 1) + (x− 1)x− 1

(x + 1) − (x− 1)x− 1

⎤

⎥

⎦

5

=[

2xx− 1

· x− 12

]5

=[

2/x(x−1)✦1 · 2/(x−1)✦

]5

= x5

76. 1 +1

1 +1

1 +1

1 +1x

= 1 +1

1 +1

1 +1

x + 1x

= 1 +1

1 +1

1 +x

x + 1

= 1 +1

1 +1

2x + 1x + 1

= 1 +1

1 +x + 12x + 1

= 1 +1

3x + 22x + 1

= 1 +2x + 13x + 2

=5x + 33x + 2

77. n(n + 1)(n + 2)2 · 3 +

(n + 1)(n + 2)2

=n(n + 1)(n + 2)

2 · 3 +(n + 1)(n + 2)

2· 33,

LCD is 2 · 3

=n(n + 1)(n + 2) + 3(n + 1)(n + 2)

2 · 3

=(n + 1)(n + 2)(n + 3)

2 · 3 Factoring the num-erator by grouping

78. n(n+1)(n+2)(n+3)2 · 3 · 4 +

(n+1)(n+2)(n+3)2 · 3

=n(n+1)(n+2)(n+3) + 4(n+1)(n+2)(n+3)

2 · 3 · 4 ,

LCD is 2 · 3 · 4

=(n + 1)(n + 2)(n + 3)(n + 4)

2 · 3 · 4

79. x2 − 9x3 + 27

· 5x2 − 15x + 45x2 − 2x− 3

+x2 + x

4 + 2x

=(x + 3)(x− 3)(5)(x2 − 3x + 9)

(x + 3)(x2 − 3x + 9)(x− 3)(x + 1)+

x2 + x

4 + 2x

=(x + 3)(x− 3)(x2 − 3x + 9)(x + 3)(x− 3)(x2 − 3x + 9)

· 5x + 1

+x2 + x

4 + 2x

= 1 · 5x + 1

+x2 + x

4 + 2x

=5

x + 1+

x2 + x

2(2 + x)

=5 · 2(2 + x) + (x2 + x)(x + 1)

2(x + 1)(2 + x)

=20 + 10x + x3 + 2x2 + x

2(x + 1)(2 + x)

=x3 + 2x2 + 11x + 20

2(x + 1)(2 + x)

80. x2 + 2x− 3x2 − x− 12

÷ x2 − 1x2 − 16

− 2x + 1x2 + 2x + 1

=x2 + 2x− 3x2 − x− 12

· x2 − 16x2 − 1

− 2x + 1x2 + 2x + 1

=(x+3)✦ (x−1)✦ (x + 4)(x−4)✦(x−4)✦ (x+3)✦ (x + 1)(x−1)✦ − 2x + 1

x2 + 2x + 1

=x + 4x + 1

− 2x + 1(x + 1)(x + 1)

=(x + 4)(x + 1) − (2x + 1)

(x + 1)(x + 1)

=x2 + 5x + 4 − 2x− 1

(x + 1)(x + 1)

=x2 + 3x + 3

(x + 1)2

Exercise Set R.7

1.√

(−21)2 = |− 21| = 21

2.√

(−7)2 = |− 7| = 7

3.√

9y2 =√

(3y)2 = |3y| = 3|y|

4.√

64t2 =√

(8t)2 = |8t| = 8|t|

5.√

(a− 2)2 = |a− 2|

6.√

(2b + 5)2 = |2b + 5|



7. 3√−27x3 = 3

√

(−3x)3 = −3x

8. 3√

−8y3 = −2y

9. 4√

81x8 = 4√

(3x2)4 = |3x2| = 3x2

10. 4√

16z12 = |2z3| = 2|z3| = 2z2|z|

11. 5√

32 = 5√

25 = 2

12. 5√−32 = −2

13.√

180 =√

36 · 5 =√

36 ·√

5 = 6√

5

14.√

48 =√

16 · 3 = 4√

3

15.√

72 =√

36 · 2 =√

36 ·√

2 = 6√

2

16.√

250 =√

25 · 10 = 5√

10

17. 3√

54 = 3√

27 · 2 = 3√

27 · 3√

2 = 3 3√

2

18. 3√

135 = 3√

27 · 5 = 3 3√

5

19.√

128c2d4 =√

64c2d4 · 2 = |8cd2|√

2 = 8√

2 |c|d2

20.√

162c4d6 =√

81c4 · d6 · 2 = 9c2|d3|√

2 =9√

2c2d2|d|

21. 4√

48x6y4 = 4√

16x4y4 · 3x2 = |2xy| 4√

3x2 =

2|x||y| 4√

3x2

22. 4√

243m5n10 = 4√

81m4n8 · 3mn2 = 3|m|n2 4√

3mn2

23.√x2 − 4x + 4 =

√

(x− 2)2 = |x− 2|

24.√x2 + 16x + 64 =

√

(x + 8)2 = |x + 8|

25.√

15√

35 =√

15 · 35 =√

3 · 5 · 5 · 7 =√

52 · 3 · 7 =√

52 ·√

3 · 7 = 5√

21

26.√

21√

6 =√

21 · 6 =√

3 · 7 · 2 · 3 = 3√

14

27.√

8√

10 =√

8 · 10 =√

2 · 4 · 2 · 5 =√

22 · 4 · 5 =2 · 2

√5 = 4

√5

28.√

12√

15 =√

12 · 15 =√

4 · 3 · 3 · 5 = 2 · 3√

5 = 6√

5

29.√

2x3y√

12xy =√

24x4y2 =√

4x4y2 · 6 = 2x2y√

6

30.√

3y4z√

20z =√

60y4z2 =√

4y4z2 · 15 = 2y2z√

15

31. 3√

3x2y 3√

36x = 3√

108x3y = 3√

27x3 · 4y = 3x 3√

4y

32. 5√

8x3y4 5√

4x4y = 5√

32x7y5 = 2xy 5√x2

33. 3√

2(x + 4) 3√

4(x + 4)4 = 3√

8(x + 4)5

= 3√

8(x + 4)3 · (x + 4)2

= 2(x + 4) 3√

(x + 4)2

34. 3√

4(x + 1)2 3√

18(x + 1)2 = 3√

72(x + 1)4

= 3√

8(x + 1)3 · 9(x + 1)

= 2(x + 1) 3√

9(x + 1)

35. 8

√

m16n24

28= 8

√

(

m2n3

2

)8

=m2n3

2

36. 6

√

m12n24

64= 6

√

(m2n4

2

)6=

m2n4

2

37.√

40xy√8x

=√

40xy8x

=√

5y

38.3√

40m3√

5m= 3

√

40m5m

= 3√

8 = 2

39.3√

3x2

3√

24x5= 3

√

3x2

24x5= 3

√

18x3

=12x

40.√

128a2b4√16ab

=

√

128a2b4

16ab=

√8ab3 =

√4 · 2 · a · b2 · b = 2b

√2ab

41. 3

√

64a4

27b3= 3

√

64 · a3 · a27 · b3

=3√

64a3 3√a

3√

27b3

=4a 3

√a

3b

42.

√

9x7

16y8=

√9 · x6 · x√

16 · y8=

3x3√x

4y4

43.

√

7x3

36y6=

√

7 · x2 · x36 · y6

=√x2

√7x

√

36y6

=x√

7x6y3

44. 3

√

2yz250z4

= 3

√

y

125z3=

3√y

3√

125z3=

3√y

5z

45. 5√

2 + 3√

32 = 5√

2 + 3√

16 · 2= 5

√2 + 3 · 4

√2

= 5√

2 + 12√

2

= (5 + 12)√

2

= 17√

2

46. 7√

12 − 2√

3 = 7 · 2√

3 − 2√

3 = 14√

3 − 2√

3 = 12√

3

47. 6√

20 − 4√

45 +√

80 = 6√

4 · 5 − 4√

9 · 5 +√

16 · 5= 6 · 2

√5 − 4 · 3

√5 + 4

√5

= 12√

5 − 12√

5 + 4√

5

= (12 − 12 + 4)√

5

= 4√

5


Exercise Set R.7 29

48. 2√

32 + 3√

8 − 4√

18 = 2 · 4√

2 + 3 · 2√

2 − 4 · 3√

2 =8√

2 + 6√

2 − 12√

2 = 2√

2

49. 8√

2x2 − 6√

20x− 5√

8x2

= 8x√

2 − 6√

4 · 5x− 5√

4x2 · 2= 8x

√2 − 6 · 2

√5x− 5 · 2x

√2

= 8x√

2 − 12√

5x− 10x√

2

= −2x√

2 − 12√

5x

50. 2 3√

8x2 + 5 3√

27x2 − 3√x3

= 4 3√x2 + 15 3

√x2 − 3x

√x

= 19 3√x2 − 3x

√x

51.(√

8 + 2√

5)(√

8 − 2√

5)

=(√

8)2 −

(

2√

5)2

= 8 − 4 · 5= 8 − 20= −12

52.(√

3 −√

2)(√

3 +√

2)

=(√

3)2 −

(√2)2

= 3 − 2= 1

53. (2√

3 +√

5)(√

3 − 3√

5)

= 2√

3 ·√

3 − 2√

3 · 3√

5 +√

5 ·√

3 −√

5 · 3√

5

= 2 · 3 − 6√

15 +√

15 − 3 · 5= 6 − 6

√15 +

√15 − 15

= −9 − 5√

15

54. (√

6 − 4√

7)(3√

6 + 2√

7)

= 3 · 6 + 2√

42 − 12√

42 − 8 · 7= 18 + 2

√42 − 12

√42 − 56

= −38 − 10√

42

55. (√

2 − 5)2 = (√

2)2 − 2 ·√

2 · 5 + 52

= 2 − 10√

2 + 25

= 27 − 10√

2

56. (1 +√

3)2 = 12 + 2 · 1 ·√

3 + (√

3)2

= 1 + 2√

3 + 3

= 4 + 2√

3

57. (√

5 −√

6)2 = (√

5)2 − 2√

5 ·√

6 + (√

6)2

= 5 − 2√

30 + 6

= 11 − 2√

30

58. (√

3 +√

2)2 = 3 + 2√

6 + 2 = 5 + 2√

6

59. We use the Pythagorean theorem. We have a = 47 andb = 25.

c2 = a2 + b2

c2 = 472 + 252

c2 = 2209 + 625c2 = 2834c ≈ 53.2

The distance across the pond is about 53.2 yd.

60. We use the Pythagorean theorem to find b, the airplane’shorizontal distance from the airport. We have a = 3700and c = 14, 200.

c2 = a2 + b2

14, 2002 = 37002 + b2

201, 640, 000 = 13, 690, 000 + b2

187, 950, 000 = b2

13, 709.5 ≈ b

The airplane is about 13, 709.5 ft horizontally from theairport.

61. a) h2 +(a

2

)2= a2 Pythagorean theorem

h2 +a2

4= a2

h2 =3a2

4

h =

√

3a2

4

h =a

2√

3

b) Using the result of part (a) we have

A =12· base · height

A =12a · a

2√

3(a

2+

a

2= a

)

A =a2

4√

3

62. c2 = s2 + s2

c2 = 2s2

c = s√

2 Length of third side

63.

x

x

x x8√

2

❅❅

❅❅

❅❅

❅

x2 + x2 = (8√

2)2 Pythagorean theorem2x2 = 128x2 = 64x = 8



64.

&&

&&

&&❅

❅❅

❅❅

❅&&

&&

&&❅

❅❅

❅❅

❅S x D x

P x B x

R

x

C

x

Q

x

A

x

y y

y y

(2x)2 = 1004x2 = 100x2 = 25x = 5

A = y2 = x2 + x2 = 52 + 52 = 25 + 25 = 50 ft2

65.√

37

=√

37· 77

=√

2149

=√

21√49

=√

217

66.√

23

=√

23· 33

=√

69

=√

6√9

=√

63

67.3√

73√

25=

3√

73√

25·

3√

53√

5=

3√

353√

125=

3√

355

68.3√

53√

4=

3√

53√

4·

3√

23√

2=

3√

103√

8=

3√

102

69. 3

√

169

= 3

√

169

· 33

= 3

√

4827

=3√

483√

27=

3√

8 · 63

=2 3√

63

70. 3

√

35

= 3

√

35· 2525

= 3

√

75125

=3√

755

71.2√

3 − 1=

2√3 − 1

·√

3 + 1√3 + 1

=2(√

3 + 1)3 − 1

=2(√

3 + 1)2

=2/(√

3 + 1)2/ · 1

=√

3 + 1

72.6

3 +√

5=

63 +

√5· 3 −

√5

3 −√

5

=6(

3 −√

5)

9 − 5

=6(

3 −√

5)

4

=3(

3 −√

5)

2=

9 − 3√

52

73.1 −

√2

2√

3 −√

6=

1 −√

22√

3 −√

6· 2

√3 +

√6

2√

3 +√

6

=2√

3 +√

6 − 2√

6 −√

124 · 3 − 6

=2√

3 +√

6 − 2√

6 − 2√

312 − 6

=−√

66

, or −√

66

74.

√5 + 4√

2 + 3√

7=

√5 + 4√

2 + 3√

7·√

2 − 3√

7√2 − 3

√7

=√

10 − 3√

35 + 4√

2 − 12√

72 − 9 · 7

=√

10 − 3√

35 + 4√

2 − 12√

7−61

75.6√

m−√n

=6√

m−√n·√m +

√n√

m +√n

=6(√m +

√n)

(√m)2 − (

√n)2

=6√m + 6

√n

m− n

76.3√

v +√w

=3√

v +√w

·√v −

√w√

v −√w

=3√v − 3

√w

v − w

77.√

503

=√

503

·√

2√2

=√

1003√

2=

103√

2

78.√

125

=√

125

·√

3√3

=√

365√

3=

65√

3

79. 3

√

25

= 3

√

25· 44

= 3

√

820

=3√

83√

20=

23√

20

80. 3

√

72

= 3

√

72· 4949

= 3

√

34398

=3√

3433√

98=

73√

98

81.√

11√3

=√

11√3

·√

11√11

=√

121√33

=11√33

82.√

5√2

=√

5√2·√

5√5

=√

25√10

=5√10

83.9 −

√5

3 −√

3=

9 −√

53 −

√3· 9 +

√5

9 +√

5

=92 − (

√5)2

27 + 3√

5 − 9√

3 −√

15

=81 − 5

27 + 3√

5 − 9√

3 −√

15

=76

27 + 3√

5 − 9√

3 −√

15

84.8 −

√6

5 −√

2=

8 −√

65 −

√2· 8 +

√6

8 +√

6

=64 − 6

40 + 5√

6 − 8√

2 −√

12

=58

40 + 5√

6 − 8√

2 − 2√

3


Exercise Set R.7 31

85.√a +

√b

3a=

√a +

√b

3a·√a−

√b

√a−

√b

=(√a)2 − (

√b)2

3a(√a−

√b)

=a− b

3a√a− 3a

√b

86.√p−√

q

1 + √q

=√p−√

q

1 + √q

·√p + √

q√p + √

q

=p− q

√p + √

q + √pq + q

87. y5/6 = 6√

y5

88. x2/3 = 3√x2

89. 163/4 = (161/4)3 =(

4√

16)3 = 23 = 8

90. 47/2 = (√

4)7 = 27 = 128

91. 125−1/3 =1

1251/3=

13√

125=

15

92. 32−4/5 =( 5√

32)−4 = 2−4 =

116

93. a5/4b−3/4 =a5/4

b3/4=

4√a5

4√b3

=a 4√a

4√b3

, or a 4

√

a

b3

94. x2/5y−1/5 = 5

√

x2

y

95. m5/3n7/3 = 3√m5 3

√n7 = 3

√m5n7 = mn2 3

√m2n

96. p7/6q11/6 = 6√

p7 6√

q11 = 6√

p7q11 = pq 6√

pq5

97. 5√

173 = 173/5

98.(

4√

13)5 = 135/4

99.(

5√

12)4 = 124/5

100. 3√

202 = 202/3

101. 3√√

11 =(√

11)1/3 = (111/2)1/3 = 111/6

102. 3√

4√

7 = (71/4)1/3 = 71/12

103.√

5 3√

5 = 51/2 · 51/3 = 51/2+1/3 = 55/6

104. 3√

2√

2 = 21/3 · 21/2 = 25/6

105. 5√

322 = 322/5 = (321/5)2 = 22 = 4

106. 3√

642 = 642/3 = (641/3)2 = 42 = 16

107. (2a3/2)(4a1/2) = 8a3/2+1/2 = 8a2

108. (3a5/6)(8a2/3) = 24a3/2 = 24a√a

109.( x6

9b−4

)1/2=( x6

32b−4

)1/2=

x3

3b−2, or

x3b2

3

110.( x2/3

4y−2

)1/2=

x1/3

41/2y−1=

3√x

2y−1, or

y 3√x

2

111.x2/3y5/6

x−1/3y1/2= x2/3−(−1/3)y5/6−1/2 = xy1/3 = x 3

√y

112.a1/2b5/8

a1/4b3/8= a1/4b1/4 = 4

√ab

113. (m1/2n5/2)2/3 = m12 ·

23n

52 ·

23 = m1/3n5/3 =

3√m 3√n5 = 3

√mn5 = n 3

√mn2

114. (x5/3y1/3z2/3)3/5 = xy1/5z2/5 = x 5√

yz2

115. a3/4(a2/3 + a4/3) = a3/4+2/3 + a3/4+4/3 =

a17/12 + a25/12 = 12√a17 + 12

√a25 =

a 12√a5 + a2 12

√a

116. m2/3(m7/4 −m5/4) = m29/12 −m23/12 =12√m29 − 12

√m23 = m2 12

√m5 −m 12

√m11

117. 3√

6√

2 = 61/321/2 = 62/623/6

= (6223)1/6

= 6√

36 · 8= 6

√288

118.√

2 4√

8 = 21/2(23)1/4 = 21/223/4 = 25/4 =4√

25 = 2 4√

2

119. 4√xy 3√

x2y = (xy)1/4(x2y)1/3 = (xy)3/12(x2y)4/12

=[

(xy)3(x2y)4]1/12

=[

x3y3x8y4]1/12

= 12√

x11y7

120. 3√ab2

√ab = (ab2)1/3(ab)1/2 = (ab2)2/6(ab)3/6 =

6√

(ab2)2(ab)3 = 6√a5b7 = b 6

√a5b

121. 3√

a4√a3 =

(

a4√a3)1/3 = (a4a3/2)1/3

= (a11/2)1/3

= a11/6

= 6√a11

= a 6√a5

122.√

a3 3√a2 = (a3 · a2/3)1/2 = (a11/3)1/2 = a11/6 =

6√a11 = a 6

√a5

123.√

(a + x)3 3√

(a + x)24√a + x

=(a + x)3/2(a + x)2/3

(a + x)1/4

=(a + x)26/12

(a + x)3/12

= (a + x)23/12

= 12√

(a + x)23

= (a + x) 12√

(a + x)11



124.4√

(x + y)2 3√x + y

√

(x + y)3=

(x + y)2/4(x + y)1/3

(x + y)3/2=

(x + y)−2/3 =1

3√

(x + y)2

125.√

1 + x2 +1√

1 + x2

=√

1 + x2 · 1 + x2

1 + x2+

1√1 + x2

·√

1 + x2

√1 + x2

=(1 + x2)

√1 + x2

1 + x2+

√1 + x2

1 + x2

=(2 + x2)

√1 + x2

1 + x2

126.√

1 − x2 − x2

2√

1 − x2

=√

1 − x2 − x2√

1 − x2

2(1 − x2)Rationalizing thedenominator

=2(1 − x2)

√1 − x2 − x2

√1 − x2

2(1 − x2)

=(2 − 3x2)

√1 − x2

2(1 − x2)

127.

(

√

a√a

)

√a

=(

a√a/2)

√a

= aa/2

128. (2a3b5/4c1/7)4

(54a−2b2/3c6/5)−1/3=

16a12b5c4/7

54−1/3a2/3b−2/9c−2/5

= 16 3√

54a34/3b47/9c34/35

= 24 · 3 · 21/3a34/3b47/9c34/35

= 3 · 213/3a34/3b47/9c34/35, or

48 · 21/3a34/3b47/9c34/35

Chapter R Review Exercises

1. True

2. For any real number a, a ̸= 0, and any integers m and n,am · an = am+n. Thus the given statement is false.

3. True

4. True

5. Rational numbers: −7, 43, −49, 0, 3

√64, 4

34,

127

, 102

6. Whole numbers: 43, 0, 3√

64, 102

7. Integers: −7, 43, 0, 3√

64, 102

8. Real numbers: All of them

9. Natural numbers: 43, 3√

64, 102

10. Irrational numbers:√

17, 2.191191119 . . ., −√

2, 5√

5

11. (−4, 7]

12. The distance of 24 from 0 is 24, so |24| = 24.

13. The distance of −78

from 0 is78, so

∣

∣

∣

∣

− 78

∣

∣

∣

∣

=78.

14. |− 5 − 5| = |− 10| = 10, or|5 − (−5)| = |5 + 5| = |10| = 10

15. 3 · 2 − 4 · 22 + 6(3 − 1)= 3 · 2 − 4 · 22 + 6 · 2 Working inside

parentheses= 3 · 2 − 4 · 4 + 6 · 2 Evaluating 22

= 6 − 16 + 12 Multiplying= −10 + 12 Adding in order= 2 from left to right

16.34 − (6 − 7)4

23 − 24=

34 − (−1)4

23 − 24=

81 − 18 − 16

=80−8

= −10


8.3 × 10−5 = 0.000083


2.07 × 107 = 20, 700, 000

19. Convert 405,000 to scientific notation.We want the decimal point to be positioned between the4 and the first 0, so we move it 5 places to the left. Since405,000 is greater than 10, the exponent must be positive.

405, 000 = 4.05 × 105

20. Convert 0.00000039 to scientific notation.We want the decimal point to be positioned between the3 and the 9, so we move it 7 places to the right. Since0.00000039 is a number between 0 and 1, the exponentmust be negative.

0.00000039 = 3.9 × 10−7

21. (3.1 × 105)(4.5 × 10−3)= (3.1 × 4.5) × (105 × 10−3)= 13.95 × 102 This is not scientific notation.= (1.395 × 10) × 102

= 1.395 × 103 Writing scientific notation

22. 2.5 × 10−8

3.2 × 1013

=2.53.2

× 10−8

1013

= 0.78125 × 10−21

= (7.8125 × 10−1) × 10−21

= 7.8125 × 10−22


Chapter R Review Exercises 33

23. (−3x4y−5)(4x−2y) = −3(4)x4+(−2)y−5+1 =

−12x2y−4, or−12x2

y4, or − 12x2

y4

24. 48a−3b2c5

6a3b−1c4=

486a−3−3b2−(−1)c5−4 =

8a−6b3c, or8b3ca6

25. 4√

81 = 4√

34 = 3

26. 5√−32 = −2

27. b− a−1

a− b−1=

b− 1a

a− 1b

=b · a

a− 1

a

a · bb− 1

b

=

ab

a− 1

aab

b− 1

b

=

ab− 1a

ab− 1b

=ab− 1

a· b

ab− 1

=(ab−1✏ )ba(ab−1✏ )

=b

a

28.x2

y+

y2

xy2 − xy + x2

=

x3 + y3

xyy2 − xy + x2

=x3 + y3

xy· 1y2 − xy + x2

=(x + y)(x2 − xy + y2)

xy(y2 − xy + x2)

=x + y

xy· x

2 − xy + y2

x2 − xy + y2

=x + y

xy

29. (√

3 −√

7)(√

3 +√

7) = (√

3)2 − (√

7)2

= 3 − 7= −4

30. (5 −√

2)2 = 25 − 10√

2 + 2 = 27 − 10√

2

31. 8√

5 +25√

5= 8

√5 +

25√5·√

5√5

= 8√

5 +25

√5

5= 8

√5 + 5

√5

= 13√

5

32. (x + t)(x2 − xt + t2)= (x + t)(x2) + (x + t)(−xt) + (x + t)(t2)= x3 + x2t− x2t− xt2 + xt2 + t3

= x3 + t3

33. (5a + 4b)(2a− 3b)= 10a2 − 15ab + 8ab− 12b2

= 10a2 − 7ab− 12b2

34. (6x2y − 3xy2 + 5xy − 3) − (−4x2y − 4xy2 + 3xy + 8)= 6x2y − 3xy2 + 5xy − 3 + 4x2y + 4xy2 − 3xy − 8= 10x2y + xy2 + 2xy − 11

35. 32x4 − 40xy3 = 8x · 4x3 − 8x · 5y3 = 8x(4x3 − 5y3)

36. y3 + 3y2 − 2y − 6 = y2(y + 3)− 2(y + 3) = (y + 3)(y2 − 2)

37. 24x + 144 + x2

= x2 + 24x + 144= (x + 12)2

38. 9x3 + 35x2 − 4x= x(9x2 + 35x− 4)= x(9x− 1)(x + 4)

39. 9x2 − 30x + 25 = (3x− 5)2

40. 8x3 − 1= (2x)3 − 13

= (2x− 1)(4x2 + 2x + 1)

41. 18x2 − 3x + 6 = 3(6x2 − x + 2)

42. 4x3 − 4x2 − 9x + 9= 4x2(x− 1) − 9(x− 1)= (x− 1)(4x2 − 9)= (x− 1)(2x + 3)(2x− 3)

43. 6x3 + 48= 6(x3 + 8)= 6(x + 2)(x2 − 2x + 4)

44. a2b2 − ab− 6 = (ab− 3)(ab + 2)

45. 2x2 + 5x− 3 = (2x− 1)(x + 3)

46. 2x− 7 = 72x = 14x = 7



47. 5x− 7 = 3x− 92x− 7 = −9

2x = −2x = −1


48. 8 − 3x = −7 + 2x−5x = −15

x = 3

49. 6(2x− 1) = 3 − (x + 10)12x− 6 = 3 − x− 1012x− 6 = −x− 713x− 6 = −7

13x = −1

x = − 113

The solution is − 113

.

50. y2 + 16y + 64 = 0(y + 8)(y + 8) = 0

y + 8 = 0 or y + 8 = 0y = −8 or y = −8

51. x2 − x = 20x2 − x− 20 = 0

(x− 5)(x + 4) = 0

x− 5 = 0 or x + 4 = 0x = 5 or x = −4


52. 2x2 + 11x− 6 = 0(2x− 1)(x + 6) = 0

2x− 1 = 0 or x + 6 = 0

x =12

or x = −6

53. x(x− 2) = 3x2 − 2x = 3

x2 − 2x− 3 = 0(x + 1)(x− 3) = 0

x + 1 = 0 or x− 3 = 0x = −1 or x = 3


54. y2 − 16 = 0(y + 4)(y − 4) = 0

y + 4 = 0 or y − 4 = 0y = −4 or y = 4

55. n2 − 7 = 0n2 = 7

n =√

7 or n = −√

7The solutions are

√7 and −

√7, or ±

√7.

56. 3x2 − 12x2 + 4x + 4

÷ x− 2x + 2

=3x2 − 12

x2 + 4x + 4· x + 2x− 2

=3(x+2)✏ (x−2)✏ (x+2)✏(x+2)✏ (x+2)✏ (x−2)✏

= 3

57. x

x2 + 9x + 20− 4

x2 + 7x + 12

=x

(x + 5)(x + 4)− 4

(x + 4)(x + 3)

LCD is (x + 5)(x + 4)(x + 3)

=x

(x + 5)(x + 4)· x + 3x + 3

− 4(x + 4)(x + 3)

· x + 5x + 5

=x(x + 3) − 4(x + 5)

(x + 5)(x + 4)(x + 3)

=x2 + 3x− 4x− 20

(x + 5)(x + 4)(x + 3)

=x2 − x− 20

(x + 5)(x + 4)(x + 3)

=(x− 5)(x+4)✏

(x + 5)(x+4)✏ (x + 3)

=x− 5

(x + 5)(x + 3)

58.√

y5 · 3√

y2 = (y5)1/2(y2)1/3 = y5/2 · y2/3 = y19/6

= 6√

y19 = y3 6√y

59.√

(a + b)3 3√a + b

6√

(a + b)7

=(a + b)3/2(a + b)1/3

(a + b)7/6

= (a + b)3/2+1/3−7/6

= (a + b)9/6+2/6−7/6

= (a + b)2/3

= 3√

(a + b)2

60. b7/5 = 5√b7 = b 5

√b2

61. 8

√

m32n16

38=(

m32n16

38

)1/8

=m4n2

3

62. 4 −√

35 +

√3

=4 −

√3

5 +√

3· 5 −

√3

5 −√

3

=20 − 4

√3 − 5

√3 + 3

25 − 3

=23 − 9

√3

22


Chapter R Test 35

63. a = 8 and b = 17. Find c.c2 = a2 + b2

c2 = 82 + 172

c2 = 64 + 289c2 = 353c ≈ 18.8

The guy wire is about 18.8 ft long.

64. 128 ÷ (−2)3 ÷ (−2) · 3 = 128 ÷ (−8) ÷ (−2) · 3= −16 ÷ (−2) · 3= 8 · 3= 24

Answer B is correct.

65. 9x2 − 36y2 = 9(x2 − 4y2)= 9[x2 − (2y)2]= 9(x + 2y)(x− 2y)

Answer C is correct.

66. Substitute $98, 000− $16, 000, or $82,000, for P , 0.065 forr, and 12 · 25, or 300, for n and perform the resultingcomputation.

M = P

⎡

⎢

⎢

⎣

r

12

(

1 +r

12

)n

(

1 +r

12

)n

− 1

⎤

⎥

⎥

⎦

= $82, 000

⎡

⎢

⎢

⎣

0.06512

(

1 +0.06512

)300

(

1 +0.06512

)300

− 1

⎤

⎥

⎥

⎦

≈ $553.67

67. Substitute $124, 000 − $20, 000, or $104,000 for P , 0.0575for r, and 12 · 30, or 360, for n and perform the resultingcomputation.

M = P

⎡

⎢

⎢

⎣

r

12

(

1 +r

12

)n

(

1 +r

12

)n

− 1

⎤

⎥

⎥

⎦

= $104, 000

⎡

⎢

⎢

⎣

0.057512

(

1 +0.0575

12

)360

(

1 +0.0575

12

)360

− 1

⎤

⎥

⎥

⎦

≈ $606.92

68. P = $135, 000 − $18, 000 = $117, 000, r = 0.075,n = 12 · 20 = 240.

M = $117, 000

⎡

⎢

⎢

⎣

0.07512

(

1 +0.07512

)240

(

1 +0.07512

)240

− 1

⎤

⎥

⎥

⎦

≈ $942.54

69. P = $151, 000 − $21, 000 = $130, 000, r = 0.0625,n = 12 · 25 = 300.

M = $130, 000

⎡

⎢

⎢

⎣

0.062512

(

1 +0.0625

12

)300

(

1 +0.0625

12

)300

− 1

⎤

⎥

⎥

⎦

≈ $857.57

70. (xn + 10)(xn − 4) = (xn)2 − 4xn + 10xn − 40= x2n + 6xn − 40

71. (ta + t−a)2 = (ta)2 + 2 · ta · t−a + (t−a)2

= t2a + 2 + t−2a

72. (yb − zc)(yb + zc) = (yb)2 − (zc)2

= y2b − z2c

73. (an−bn)3 = (an−bn)(an−bn)2

= (an − bn)(a2n − 2anbn + b2n)= a3n−2a2nbn+anb2n−a2nbn+2anb2n−b3n

= a3n − 3a2nbn + 3anb2n − b3n

74. y2n + 16yn + 64 = (yn)2 + 16yn + 64= (yn + 8)2

75. x2t − 3xt − 28 = (xt)2 − 3xt − 28= (xt − 7)(xt + 4)

76. m6n −m3n = m3n(m3n − 1)= m3n[(mn)3 − 13]= m3n(mn − 1)(m2n + mn + 1)

77. Anya is probably not following the rules for order of op-erations. She is subtracting 6 from 15 first, then dividingthe difference by 3, and finally multiplying the quotient by4. The correct answer is 7.

78. When the least common denominator is used, the multi-plication in the numerators is often simpler and there isusually less simplification required after the addition orsubtraction is performed.

79. A3 −B3 = A3 + (−B)3

= (A + (−B))(A2 −A(−B) + (−B)2)= (A−B)(A2 + AB + B2)

80. Observe that 26 > 25, so√

26 >√

25, or√

26 > 5. Then10

√26 − 50 > 10 · 5 − 50, or 10

√26 − 50 > 0.

Chapter R Test

1. a) Whole numbers: 0, 3√

8, 29b) Irrational numbers:

√12

c) Integers but not natural numbers: 0, −5

d) Rational numbers but not integers: 667, −13

4, −1.2



2. |− 17.6| = 17.6

3.∣

∣

∣

∣

1511

∣

∣

∣

∣

=1511

4. |0| = 0

5. (−3, 6]

6. |− 9 − 6| = |− 15| = 15, or|6 − (−9)| = |6 + 9| = |15| = 15

7. 32 ÷ 23 − 12 ÷ 4 · 3= 32 ÷ 8 − 12 ÷ 4 · 3= 4 − 12 ÷ 4 · 3= 4 − 3 · 3= 4 − 9= −5

8. Position the decimal point 6 places to the left, between the4 and the 5. Since 4,509,000 is a number greater than 10,the exponent must be positive.

4, 509, 000 = 4.509 × 106


8.6 × 10−5 = 0.000086

10. 2.7 × 104

3.6 × 10−3= 0.75 × 107

= (7.5 × 10−1) × 107

= 7.5 × 106

11. x−8 · x5 = x−8+5 = x−3, or1x3

12. (2y2)3(3y4)2 = 23y6 · 32y8 = 8 · 9 · y6+8 = 72y14

13. (−3a5b−4)(5a−1b3)

= −3 · 5 · a5+(−1) · b−4+3

= −15a4b−1, or −15a4

b

14. (5xy4− 7xy2+ 4x2−3)−(−3xy4+ 2xy2−2y+ 4)= (5xy4−7xy2+ 4x2− 3)+(3xy4−2xy2+ 2y − 4)= (5 + 3)xy4+ (−7 − 2)xy2+ 4x2+ 2y + (−3 − 4)= 8xy4 − 9xy2 + 4x2 + 2y − 7

15. (y − 2)(3y + 4) = 3y2 + 4y − 6y − 8 = 3y2 − 2y − 8

16. (4x− 3)2 = (4x)2 − 2 · 4x · 3 + 32 = 16x2 − 24x + 9

17.

x

y− y

xx + y

=

x

y· xx− y

x· yy

x + y

=

x2

xy− y2

xyx + y

=

x2 − y2

xyx + y

=x2 − y2

xy· 1x + y

=(x+y)✏ (x− y)

xy(x+y)✏=

x− y

xy

18.√

45 =√

9 · 5 =√

9√

5 = 3√

5

19. 3√

56 = 3√

8 · 7 = 3√

8 3√

7 = 2 3√

7

20. 3√

75 + 2√

27 = 3√

25 · 3 + 2√

9 · 3= 3 · 5

√3 + 2 · 3

√3

= 15√

3 + 6√

3

= 21√

3

21.√

18√

10 =√

18 · 10 =√

2 · 3 · 3 · 2 · 5 =2 · 3

√5 = 6

√5

22. (2 +√

3)(5 − 2√

3)

= 2 · 5 − 4√

3 + 5√

3 − 2 · 3= 10 − 4

√3 + 5

√3 − 6

= 4 +√

3

23. 8x2 − 18 = 2(4x2 − 9) = 2(2x + 3)(2x− 3)

24. y2 − 3y − 18 = (y + 3)(y − 6)

25. 2n2 + 5n− 12 = (2n− 3)(n + 4)

26. x3 + 10x2 + 25x = x(x2 + 10x + 25) = x(x + 5)2

27. m3 − 8 = (m− 2)(m2 + 2m + 4)

28. 7x− 4 = 247x = 28x = 4

The solution is 4.

29. 3(y − 5) + 6 = 8 − (y + 2)3y − 15 + 6 = 8 − y − 2

3y − 9 = −y + 64y − 9 = 6

4y = 15

y =154

The solution is154

.


Chapter R Test 37

30. 2x2 + 5x + 3 = 0(2x + 3)(x + 1) = 0

2x + 3 = 0 or x + 1 = 02x = −3 or x = −1

x = −32

or x = −1


and −1.

31. z2 − 11 = 0z2 = 11

z =√

11 or z = −√

11The solutions are

√11 and −

√11, or ±

√11.

32. x2 + x− 6x2 + 8x + 15

· x2 − 25x2 − 4x + 4

=(x2 + x− 6)(x2 − 25)

(x2 + 8x + 15)(x2 − 4x + 4)

=(x+3)✏ (x−2)✏ (x+5)✏ (x− 5)(x+3)✏ (x+5)✏ (x−2)✏ (x− 2)

=x− 5x− 2

33. x

x2 − 1− 3

x2 + 4x− 5

=x

(x + 1)(x− 1)− 3

(x− 1)(x + 5)LCD is (x + 1)(x− 1)(x + 5)

=x

(x+ 1)(x−1)· x+ 5x + 5

− 3(x−1)(x + 5)

· x+ 1x+ 1

=x(x + 5) − 3(x + 1)

(x + 1)(x− 1)(x + 5)

=x2 + 5x− 3x− 3

(x + 1)(x− 1)(x + 5)

=x2 + 2x− 3

(x + 1)(x− 1)(x + 5)

=(x + 3)(x−1)✏

(x + 1)(x−1)✏ (x + 5)

=x + 3

(x + 1)(x + 5)

34.5

7 −√

3=

57 −

√3· 7 +

√3

7 +√

3=

35 + 5√

349 − 3

=

35 + 5√

346

35. t5/7 = 7√t5

36. ( 5√

7)3 = (71/5)3 = 73/5

37. a = 5 and b = 12. Find c.c2 = a2 + b2

c2 = 52 + 122

c2 = 25 + 144c2 = 169c = 13

The guy wire is 13 ft long.

38. (x− y − 1)2

= [(x− y) − 1]2

= (x− y)2 − 2(x− y)(1) + 12

= x2 − 2xy + y2 − 2x + 2y + 1


Basic Concepts of Algebra · 4 Chapter R: Basic Concepts of Algebra 32. x26 x13 = x26−13 = x13...

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Transcript of Basic Concepts of Algebra · 4 Chapter R: Basic Concepts of Algebra 32. x26 x13 = x26−13 = x13...