Question

question 9 (5 points)
factor the polynomial: (2x^3 + 16)
a) (2(x + 2)(x^2 - 2x + 4))
b) (2(x + 2)(x - 2)^2)
c) (2(x + 2)(x + 4))
d) (2(x + 2)(x^2 + 2x + 4))

question 10 (5 points)
find the zeros of (f(x) = (x + 2)(x - 1)(x - 3)(x + 1)).
a) (x = -2, -1, 1, 3)
b) (x = -3, -1, 1, 2)
c) (x = -2, 1, 3)
d) (x = 1, 2, 3)

Explanation:

Factor out the greatest common factor

We begin with the polynomial:

$$ 2x^3 + 16 $$

Both terms share a greatest common factor of \(2\). Factoring out \(2\) gives:

$$ 2(x^3 + 8) $$

Apply the sum of cubes formula

The binomial inside the parentheses is a sum of two perfect cubes:

$$ x^3 + 8 = x^3 + 2^3 $$

We use the sum of cubes factoring formula:

$$ a^3 + b^3 = (a + b)(a^2 - ab + b^2) $$

Substituting \(a = x\) and \(b = 2\):

$$ x^3 + 2^3 = (x + 2)(x^2 - 2x + 4) $$

Combining this with the common factor of \(2\), the fully factored form is:

$$ 2(x + 2)(x^2 - 2x + 4) $$

This matches option A.

Find the zeros of the function

Using the Zeros of a Function knowledge point

$$
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$$

The zeros of the function are \(x = -2, -1, 1, 3\). This matches option A.

Answer:

Question 9

• (A) \(2(x + 2)(x^2 - 2x + 4)\) (Correct answer)
• (B) \(2(x + 2)(x - 2)2\)
• (C) \(2(x + 2)(x + 4)\)
• (D) \(2(x + 2)(x^2 + 2x + 4)\)

Question 10

• (A) \(x = -2, -1, 1, 3\) (Correct answer)
• (B) \(x = -3, -1, 1, 2\)
• (C) \(x = -2, 1, 3\)
• (D) \(x = 1, 2, 3\)