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1. Find the zeros of the function.
f(x) = 9x2 + 6x - 8 (2 points)


a) -2 and 4
b) Negative two divided by three. and Four divided by three.
c) Two divided by three. and Negative four divided by three.
d) 2 and -4

2. Find the zeros of the function.
f(x) = 9x3 - 45x2 + 36x (2 points)


a) 0, 1, and 4
b) -1 and -4
c) 1 and 4
d) 0, -1, and -4

3. Find the zeros of the polynomial function and state the multiplicity of each.
f(x) = 4(x + 7)2(x - 7)3 (2 points)


a) 4, multiplicity 1; -7, multiplicity 3; 7, multiplicity 3
b) -7, multiplicity 3; 7, multiplicity 2
c) 4, multiplicity 1; 7, multiplicity 1; -7, multiplicity 1
d) -7, multiplicity 2; 7, multiplicity 3

4. Find a cubic function with the given zeros. (2 points)
Square root of five. , - Square root of five. , -7


a) f(x) = x3 - 7x2 - 5x - 35
b) f(x) = x3 + 7 x2 - 5x + 35
c) f(x) = x3 + 7x2 - 5x - 35
d) f(x) = x3 + 7x2 + 5x - 35

5. Expand the following using either the Binomial Theorem or Pascal's Triangle. You must show your work for credit.
(2x + 4)3





Respuesta :

1. f(x) = 9x² + 6x - 8
f(x) = (3x - 2)(3x + 4)
When  (3x - 2) = 0, then x = 2/3
When (3x + 4) = 0, then x = -4/3

Answer: The zeros are two divided by three and negative four divided by three.

2. f(x) = 9x³ - 45x² + 36x
  f(x) = 9x(x² - 5x + 4)
        = 9x(x - 1)(x - 4)
  When 9x = 0, then x = 0
  When (x-1) = 0, then x = 1
  When (x-4) = 0, then x = 4

 Answer: 0, 1 and 4

3. f(x) = 4(x+7)²(x-7)³
 When (x+7)² = 0, then x = -7 (twice)
 When (x-7)³ = 0, then x = 7 (thrice)

 Answer: 7, multiplicity 2; -7 multiplicity 3

4. The zeros of f(x) are √5, -√5, -7
 The factors of f(x) are (x-√5)(x+√5)(x+7)
 Note that (x-√5)(x+√5) = x² - (√5)² = x² - 5
 f(x) = (x²-5)(x+7)
       = x³ + 7x² - 5x - 35

Answer: f(x) = x³ + 7x² - 5x - 35

5. Expand (2x + 4)³
 From Pascal's Triangle, the coefficients are 1 3 3 1
 Therefore
 (2x + 4)³ = 1(2x)³(4)⁰ + 3(2x)²(4)¹ + 3(2x)¹(4)² + 1(2x)⁰(4)³
                = 8x³ + 48x² + 96x + 64

 Answer: 8x³ + 48x² + 96x + 64

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