Question

1. Find the zeros of the function.

Answer 1

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