# How do you factor 16x^4 - 64x^2?

Jul 23, 2015

$16 {x}^{4} - 64 {x}^{2} = 16 {x}^{2} \left({x}^{2} - {2}^{2}\right) = 16 {x}^{2} \left(x - 2\right) \left(x + 2\right)$

using the difference of squares identity:

${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

#### Explanation:

First separate out the common factor $16 {x}^{2}$ to get:

$16 {x}^{4} - 64 {x}^{2} = 16 {x}^{2} \left({x}^{2} - 4\right)$

Next note that $\left({x}^{2} - 4\right) = \left({x}^{2} - {2}^{2}\right)$ is a difference of squares, so we can factor it as $\left(x - 2\right) \left(x + 2\right)$

Putting it all together:

$16 {x}^{4} - 64 {x}^{2} = 16 {x}^{2} \left({x}^{2} - {2}^{2}\right) = 16 {x}^{2} \left(x - 2\right) \left(x + 2\right)$