# How do you factor x^8-256?

Mar 27, 2015

Remember the special products:
$\left(A - B\right) \left(A + B\right) = {A}^{2} - {B}^{2}$ and vice versa.

${x}^{8} - 256 = {\left({x}^{4}\right)}^{2} - {\left(16\right)}^{2} = \left({x}^{4} - 16\right) \left({x}^{4} + 16\right)$

You can't factor the second term any further, but the first term can be factored according to the same rule:
${x}^{4} - 16 = {\left({x}^{2}\right)}^{2} - {\left(4\right)}^{2} = \left({x}^{2} - 4\right) \left({x}^{2} + 4\right)$

And again:
${x}^{2} - 4 = \left(x - 2\right) \left(x + 2\right)$

If we put all of this together we get:

${x}^{8} - 256 = \left(x - 2\right) \left(x + 2\right) \left({x}^{2} + 4\right) \left({x}^{4} + 16\right)$