How do you factor #x^4 + 64#?
1 Answer
Apr 7, 2018
Explanation:
Given:
#x^4+64#
Note that this is positive and therefore non-zero for any real value of
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)#