Question

factor the following expression.
$x^3 + 2x^2 - 16x - 32$
a. $(x - 4)(x + 2)(x - 2)$
b. $(x + 4)(x - 4)(x - 2)$
c. $(x + 4)(x - 4)(x + 2)$
d. $(x + 4)(x - 4)(x + 2)^2$

Explanation:

Step1: Group the terms

We have the expression \(x^{3}+2x^{2}-16x - 32\). Group the first two terms and the last two terms:
\((x^{3}+2x^{2})+(-16x - 32)\)

Step2: Factor out the GCF from each group

From the first group \(x^{3}+2x^{2}\), the greatest common factor (GCF) is \(x^{2}\), so we get \(x^{2}(x + 2)\).
From the second group \(-16x - 32\), the GCF is \(-16\), so we get \(-16(x + 2)\).
Now the expression becomes \(x^{2}(x + 2)-16(x + 2)\).

Step3: Factor out the common binomial factor

We can see that \((x + 2)\) is a common factor, so we factor it out:
\((x + 2)(x^{2}-16)\)

Step4: Factor the difference of squares

The term \(x^{2}-16\) is a difference of squares, which can be factored as \((x + 4)(x - 4)\) (since \(a^{2}-b^{2}=(a + b)(a - b)\) with \(a=x\) and \(b = 4\)).
So the fully factored form is \((x + 2)(x + 4)(x - 4)\), which is the same as \((x + 4)(x - 4)(x + 2)\).

Answer:

C. \((x + 4)(x - 4)(x + 2)\)