# How do you factor completely: 16x^4 - 1?

Apr 25, 2018

16x^4-1=color(blue)((4x^2+1)(2x+1)(2x-1)

#### Explanation:

Remember
$\textcolor{w h i t e}{\text{XXX}} \left({a}^{2} - {b}^{2}\right) = \left(a + b\right) \left(a - b\right)$

$\left(16 {x}^{4} - 1\right)$ is of this form (with $a = \left(4 {x}^{2}\right)$ and $b = 1$).

So
$\textcolor{w h i t e}{\text{XXX}} \left(16 {x}^{4} - 1\right) = \left(4 {x}^{2} + 1\right) \left(4 {x}^{2} - 1\right)$

...but we note that $\left(4 {x}^{2} - 1\right)$ is also of this form (with $a = 2 x$ and $b = 1$)
So $\left(4 {x}^{2} - 1\right) = \left(2 x + 1\right) \left(2 x - 1\right)$

and, completely factored we have
$\textcolor{w h i t e}{\text{XXX}} \left(16 {x}^{4} - 1\right) = \left(4 {x}^{2} + 1\right) \left(2 x + 1\right) \left(2 x - 1\right)$