How do you factor $x^{8} - 256$ $x^8-256$ ?

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How do you factor $x^{8} - 256$ $x^8-256$ ?

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Answer 1 · 2015-03-27T09:41:40

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

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

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

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

If we put all of this together we get:

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

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How do you factor $x^{8} - 256$ $x^8-256$ ?

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