How do you factor #x^4 + 64#?

1 Answer
Apr 7, 2018

Answer:

#x^4+64 = (x^2-4x+8)(x^2+4x+8)#

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=(A-B)(A+B)#

First note that:

#(x^2+8)^2 = x^4+16x^2+64 = x^4+64+(4x)^2#

So:

#x^4+64 = (x^2+8)^2-(4x)^2#

#color(white)(x^4+64) = ((x^2+8)-4x)((x^2+8)+4x)#

#color(white)(x^4+64) = (x^2-4x+8)(x^2+4x+8)#