# How do you factor completely 2x^3 -32x?

Nov 15, 2015

$2 x \left(x + 4\right) \left(x - 4\right)$

#### Explanation:

Look for a common factor between the two terms. If you focus on just the constants, $2$ and $32$, it is clear that their greatest common factor is $2$.
So, we can "take a $2$" out of both terms in $2 {x}^{3} - 32 x$.
We can rewrite it as $2 \left({x}^{3} - 16 x\right)$.
We can also factor out an $x$ from both terms: $2 x \left({x}^{2} - 16\right)$
We are not done. The term ${x}^{2} - 16$ is a "difference of squares".
Differences of squares, like ${a}^{2} - {b}^{2}$, can be factored into $\left(a + b\right) \left(a - b\right)$.
Therefore, we can factor ${x}^{2} - 16$ into $\left(x + 4\right) \left(x - 4\right)$.

So, we can factor the entire term into $2 x \left(x + 4\right) \left(x - 4\right)$.