# How do you factor x^4 + 64?

Apr 7, 2018

${x}^{4} + 64 = \left({x}^{2} - 4 x + 8\right) \left({x}^{2} + 4 x + 8\right)$

#### Explanation:

Given:

${x}^{4} + 64$

Note that this is positive and therefore non-zero for any real value of $x$. So it has no linear factors with real coefficients. We can find quadratic factors.

We can use the difference of squares identity to help with this:

${A}^{2} - {B}^{2} = \left(A - B\right) \left(A + B\right)$

First note that:

${\left({x}^{2} + 8\right)}^{2} = {x}^{4} + 16 {x}^{2} + 64 = {x}^{4} + 64 + {\left(4 x\right)}^{2}$

So:

${x}^{4} + 64 = {\left({x}^{2} + 8\right)}^{2} - {\left(4 x\right)}^{2}$

$\textcolor{w h i t e}{{x}^{4} + 64} = \left(\left({x}^{2} + 8\right) - 4 x\right) \left(\left({x}^{2} + 8\right) + 4 x\right)$

$\textcolor{w h i t e}{{x}^{4} + 64} = \left({x}^{2} - 4 x + 8\right) \left({x}^{2} + 4 x + 8\right)$