How do you factor $x^{4} + 64$ $x^4 + 64$ ?

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How do you factor $x^{4} + 64$ $x^4 + 64$ ?

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Answer 1 · 2018-04-07T07:54:26

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

Explanation:

Given:

$x^{4} + 64$ $x^4+64$

Note that this is positive and therefore non-zero for any real value of $x$ $x$ . 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:

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

First note that:

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

So:

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

$x^{4} + 64 = ((x^{2} + 8) - 4 x) ((x^{2} + 8) + 4 x)$ $color(white)(x^4+64) = ((x^2+8)-4x)((x^2+8)+4x)$

$x^{4} + 64 = (x^{2} - 4 x + 8) (x^{2} + 4 x + 8)$ $color(white)(x^4+64) = (x^2-4x+8)(x^2+4x+8)$

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How do you factor $x^{4} + 64$ $x^4 + 64$ ?

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