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Math 3 VM Part 3 Polynomials A August 6, 2017

Math 3 Variable Manipulation Part 3 Polynomials A

MATH 1 & 2 REVIEW: VOCABULARY Constant: A term that does not have a variable is called a constant. Example: the number 5 is a constant because it does not have a variable attached and will always have the value of 5. Term: A constant or a variable or a product of a constant and a variable is called a term. Example: 2, x, or 3x2 are all terms. Polynomial: An expression formed by adding a finite number of unlike terms is called a polynomial. The variables can only be raised to positive integer exponents. Note that there are no square roots of variables, no fractional powers and no variables in the denominator of any fractions. Example A: 4x3 – 6x2 + 1 is a polynomial Example B: x½ - 2x-1 + 5 is NOT a polynomial Monomial, Binomial, Trinomial: a polynomial with only one term is called a monomial. A polynomial with two terms is called a binomial and a polynomial with three terms is a trinomial. Examples: 6x4 (monomial) 5x + 1 (binomial) 2x2 – x + 3 (trinomial) Standard (General) Form: Polynomials are in standard form when written with exponents in descending order and the constant term is last. Example: 2x4 – 5x3 + 7x2 –x +3 is in standard form Degree: The exponent of a term gives you the degree of the term. For a polynomial, the value of the largest exponent is the degree of the whole polynomial. Example A: -3x2 has degree two Example B: 2x4 – 5x3 + 7x2 –x + 3 has a degree four Coefficient: The number part of a term is called the coefficient when the term contains a variable and a number. Example A: 5x has a coefficient of 5 Example B: -x2 has a coefficient of -1 Leading coefficient: The leading coefficient is the coefficient of the first term when the polynomial is written in standard form. Example: The leading coefficient of 2x4 – 5x3 + 7x2 –x + 3 is 2 General Polynomial: anxn + an-1 xn-1 + . . . a2x2 + a1x + a0

where an is the Leading Coefficient, anxn is the leading term, the degree is n and a0 is a constant.

Sample Questions:

1. (3x2 – 4x + 1) + (-x2 + x – 9)

2. (9x3 – 5x2 + x) + (6x2 + 5x – 10)

3. (x – 4x2 + 7) – (-5x2 + 5x – 3)

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Math 3 VM Part 3 Polynomials A August 6, 2017

MULTIPLYING TWO BINOMIALS Multiply each term in the first binomial by each term in the second binomial. There are several methods to accomplish this. Distributive (also called the Foil Method), Box Method and Vertical Method. Each of these methods all yield the same answer. Example: (5x – 7)(2x + 9) Solution: MULTIPLYING TRINOMIALS Multiplying trinomials is the same process as multiplying binomials except for there are more terms to multiply than just four. Multiply each term in the first polynomial by each term in the second polynomial. This is the distributive method. Box or Vertical Methods could also be used. Example: (2x + 3)(6x2 – 7x – 5) Solution:

Example: (2x2 + 3x – 1)(6x2 – 7x – 5) Solution: Example: (7k − 3)(k2 − 2k + 7) Solution: Multiply 7k by k2, -2k and +7, then multiply -3 by k2, -2k and +7 To get 7k3 - 14k + 49k and -3k2 + 6k – 21 Combine like terms to get: 7k3 − 17k2 + 55k − 21 Sample Questions:

4. (4a + 2)(6a2 − a + 2) 5. (6n2 − 6n − 5)(7n2 + 6n − 5)

6. (7k − 3)(k2 − 2k + 7)

7. (n2 + 6n − 4)(2n − 4)

8. (x2 − 2x − 8)(−x2 + 3x − 5)

9. (4x2 + 6x + 1)(-5x2 – 3x -6)

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Math 3 VM Part 3 Polynomials A August 6, 2017

FACTORING QUADRATIC FORM Some quadratic equations have higher exponent values, but can be factored simply by factoring out a common divisor and then factoring the equation. Or it may be a factored as normal just with the higher exponent value. Example: 3x3 + 24x2 + 21x Solution: 3x is common to each of the terms so first factor out 3x to get 3x(x2 + 8x + 7x) 3x(x + 1)(x + 7) Example: x4 + 9x2 + 20 Solution: The first term is x4 and the second term is x2, so we can just factor as normal except for factoring x2

instead of x and x4 instead of x2. x4 + 9x2 + 20 (x2 + 4)(x2 + 5) Sample Questions:

10. Factor: x4 − 8x2 + 15

11. x4 − 7x2 + 10

12. x4 + 4x2 + 3

13. a4 + 6a2 + 5

14. x4 + x2 − 12

FACTORING TRINOMIALS WHEN a > 1 There are several methods to use when quadratic equations have a > 1. One way is trial and error. First, guess that the first term in one binomial is just the unknown and the other is a times the unknown. Then you can guess the second terms in both binomials by listing the factors that multiply to the last number and then and use trial and error to find the answer. Note: The sign of terms in the polynomial can give clues. If the last number is positive, then you know the signs of both binomials are the same (+ and + or – and -) since when multiplied, you get a positive number. If the middle number is negative, you know both are negative and if the middle number is positive, you know they are both positive. If the last number is negative, then you know that one binomial is positive and the other is negative. Example: 2n2 + 13n + 6 Solution: First guess that the first terms are 2n and n since 2n*n = 2n2 (2n )(n ) Since the last number is positive, we know that both binomials will be the same and since the middle number is positive, we know that both binomials are positive. (2n + )(n + ) Then start to guess numbers that multiply to be 6. Since they are both positive numbers, the only choices are 2 and 3 or 6 and 1. If we try 2 and 3, we get (2n + 2)(n + 3) FOILing this we get 2n2 + 8n + 6. That is not correct and we know that the middle number needs to be a higher number, so try 6 and 1. (2n + 6)(n + 1) FOILing this guess we get 2n2 + 8n + 6. That is not correct and we still know that number needs to be higher. If we put the 6 on the outside spot, it would multiply by the 2, so that would give a higher number so try 1 and 6. (2n + 1)(n + 6) FOILing this guess we get 2n2 + 13n + 6 so (2n + 1)(n + 6) is the correct answer. This may seem like a lot of work, but it gets the answer and it works semi-easily when a is a prime number and there is only one choice of the first terms in the binomials. However, if there are multiply choices to factor a or many choices when factoring c, “factoring by grouping” is another way that is more efficient. Steps to Factor by Grouping when a > 1

1. Multiply a*c = d 2. Find two factors which multiply to d and add to b 3. Split the second term into two parts using the two factors from step 2. Rewrite the entire

polynomial with four terms. Make sure to place the middle terms nearest the first and last terms which correlate the best.

4. Group the polynomial into two separate binomials with parentheses. 5. Factor out a common divisor from both binomials. You know you are correct when a

common binomial is left in each section. 6. Factor out the common binomial to get the final factored binomials.

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Math 3 VM Part 3 Polynomials A August 6, 2017

Example: 2n2 + 13n + 6 Solution: This is the same question from the trial and error method above. This time we use the steps. Step 1: Multiply 2*6 = 12. Step 2: Find two numbers that Multiply to 12 and add to 13. They are 1 and 12. Step 3. Split the second term into + n and + 12n and rewrite the equation replacing the middle term with the two new terms. Notice that the equation is still equivalent just the middle term is split. 2n2 + n + 12n + 6 Step 4. Group the polynomials into two separate binomials with parentheses. (2n2 + n) + (12n + 6) Step 5. Factor out a common divisor from both binomials. n(2n + 1) + 6(2n + 1) We know we are right because we have (2n + 1) in both parts. Step 6. Factor out the common binomial to get the final factored binomials. We will factor out (2n + 1) from both parts leaving n + 6. (2n + 1)(n + 6) For this simple example, both methods are about the same amount of work/time. However, this method works for all polynomials with a > 1. Sample Questions:

15. 2n2 − n – 6

16. 20v2 + 11v – 3

17. 9x2 − 3x – 30

18. −2r2 − 12r + 14

MIX OF ALL FACTORING TECHNIQUES Sample Questions:

19. 15n2 − 27n − 6

20. 6x9n − 30x5n − 300xn

21. −6a2 − 25a − 25

22. 5nu8 − 15nu4 + 40n

23. 16b2 + 60b − 100

24. 2x6 + 13x4 + 6x2

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Math 3 VM Part 3 Polynomials A August 6, 2017

MATH 3 LEVEL: REMAINDER THEOREM For a polynomial p(x) and a number a, the remainder when dividing by x - a is p(a), so p(a) = 0 if and only if x - a is a factor of p(x). Example: Is x + 5 a factor of f(x) = 3x2 + 14x – 5? Solution: First identify a by comparing x – a to the proposed factor x + 5 If x – a = x + 5, a = -5 Next, substitute a for x f(x) = 3x2 + 14x – 5 f(-5) = 3(-5)2 + 14(-5) – 5 = 75 – 70 – 5 = 0 Since f(-5) = 0, the binomial x + 5 is a factor of f(x) = 3x2 + 14x – 5 NOTE: If we factored f(x) = 3x2 + 14x – 5, the result would be f(x) = (3x - 1)(x + 5). Example: Is x − 3 a factor of f(x) = 2x2 -7x – 4? Solution: First identify a by comparing x – a to the proposed factor x - 3 If x – a = x - 3, a = 3 Next, substitute a for x f(x) = 2x2 -7x – 4 f(3) = 3(3)2 - 7(3) – 4 = 18 – 21 – 4 = -7 Since f(3) ≠ 0, the binomial x - 3 is not a factor of f(x) = 2x2 -7x – 4 NOTE: If we factored f(x) = 2x2 -7x – 4, the result would be f(x) = (2x - 1)(x - 4). Example: Is x + 2 a factor of f(x) = x3 - 3x2 -6x + 8? Solution: First identify a by comparing x – a to the proposed factor x + 2 If x – a = x + 2, a = -2 Next, substitute a for x f(x) = x3 - 3x2 -6x + 8 f(-2) = (-2)3 – 3(-2)2 -6(-2) + 8 = -8 – 12 + 12 + 8 = 0 Since f(-2) = 0, the binomial x + 2 is not a factor of f(x) = x3 - 3x2 -6x + 8 NOTE: If we factored f(x) = x3 - 3x2 -6x + 8, the result would be f(x) = (x + 2)(x – 1)(x - 4). Sample Questions: For the given polynomials determine which of the binomials listed are factors.

25. f(x) = -2x3 + 15x2 + 22x - 15

a. x + 3 b. x + 5 c. x – 3

26. k3 − k2 − k − 2 a. k – 2 b. k + 3 c. k + 1

27. p4 − 8p3 + 10p2 + 2p + 4 a. p + 2 b. p – 2 c. P – 3

28. n4 + 10n3 + 21n2 + 6n – 8 a. n – 2 b. n + 1 c. n + 2

29. Is (b - 7) a factor of (b4 − 8b3 − b2 + 62b − 34)

30. Is (p + 5) a factor of (p4 + 6p3 + 11p2 + 29p − 13)

31. Is (k – 2) a factor of (k3 − k2 − k − 2)

USING LONG DIVISION TO DIVIDE POLYNOMIALS Dividing polynomials is similar to long division. First, set up the division problem by putting the dividend (the thing being divided into) inside and the divisor (the thing doing the dividing) outside and to the left. Then divide similar to simple long division. Example: Divide x2 – 9x – 10 by x + 1 Solution:

x2 divided by x is x so put x on top of the division symbol and multiply down. Then subtract.

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Math 3 VM Part 3 Polynomials A August 6, 2017

Then bring down the next term. -10x divided by x is -10 so put -10 on top of the division symbol and multiply down. Then subtract.

Since subtracting gets 0, there is no remainder. If there were something left, it would be written as a remainder. Sample Questions:

32. (x4 − 2x3 − 16x2 + 28x + 9) ÷ (x − 4)

33. (n4 − 6n3 − 10n2 + 20n + 15) ÷ (n + 2)

34. (4n3 − 9n2 + 9n + 3) ÷ (n − 1)

FUNDAMENTAL THEOREM OF ALGEBRA A “zero” is the place where a graph touches or crosses the x-axis and zeros can be repeated. The fundamental theorem of algebra states that a polynomial function of degree n > 0 has n complex zeros. When a zero is repeated, the same factor occurs multiple times. We say the factor has a multiplicity of the number of times it is repeated. For instance, (x −5)4 means that the zero (5, 0) is repeated four times and has a multiplicity of 4. NOTE: When the multiplicity of a zero is even, the graph of the function touches the x-axis. When the multiplicity of a zero is odd, the graph of the function crosses the x-axis.

Example: Given: f(x) = (x – 3)(x + 2)2 a. Identify the zeros of f(x) . b. According to the Fundamental Theorem of Algebra how many zeros should f(x) have? c. Is there a difference between the number of zeros found in part a and the expected number of zeros found in part b? Solution: Example: Identify the zeros, their multiplicity, and determine whether they touch or cross the x-axis at each zero. f(x) = (x – 4)2 (x – 1)3 Example: Identify the zeros, their multiplicity, and determine whether they touch or cross the x-axis at each zero. f(x) = (x – 1)2 (x + 2)

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Math 3 VM Part 3 Polynomials A August 6, 2017

Sample Questions: 35. Identify the zeros, their multiplicity, and determine whether they touch or cross the x-axis at each zero.

f(x) = (x + 1)4 (x - 5)3

36. Identify the zeros, their multiplicity, and determine whether they touch or cross the x-axis at each zero. f(x) = (x – 2)2 (x + 3)2 (x - 4)

37. Identify the zeros, their multiplicity, and determine whether they touch or cross the x-axis at each zero.

f(x) = (x – 1)2 (x - 4)3

38. Identify the zeros, their multiplicity, and determine whether they touch or cross the x-axis at each zero. f(x) = (x – 4)2 (x + 1)3 (x + 3)

39. Identify the zeros, their multiplicity, and determine whether they touch or cross the x-axis at each zero. f(x) = (x – 2)3 (x + 1)2 (x - 5)2

Answer Key

1. 2x2 – 3x – 8 2. 9x3 + x2 + 6x – 10 3. x2 – 4x + 10 4. 24a3 + 8a2 + 6a + 4 5. 42n4 − 6n3 − 101n2 + 25 6. 7k3 − 17k2 + 55k – 21 7. 2n3 + 8n2 − 32n + 16 8. −x4 + 5x3 − 3x2 − 14x + 40 9. -20x4 – 42x3 – 47x2 -39x - 6 10. (x2 − 3)(x2 − 5) 11. (x2 − 2)(x2 − 5) 12. (x2 + 3)(x2 + 1) 13. (a2 + 1)(a2 + 5) 14. (x2 − 3)(x2 + 4) 15. (2n + 3)(n − 2) 16. (5v − 1)(4v + 3) 17. (3x + 5)(3x − 6) 18. (2r − 2)(−r − 7) 19. (15n + 3)(n − 2) 20. 6xn(x4 − 10)(x4 + 5) 21. −(2a + 5)(3a + 5) 22. 5n(u8 − 3u4 + 8) 23. 4(b + 5)(4b − 5) 24. x2(2x2 + 1)(x2 + 6) 25. 26. (a) k – 2 27. (b) p – 2 28. (c) n + 2 29. No 30. No 31. Yes 32. x3 + 2x2 − 8x − 4, R −7 33. n3 − 8n2 + 6n + 8, R −1 34. 4n2 − 5n + 4, R 7

35. 36. .

37. .

38. .

39. .