How do you factor #u^4-81#?

1 Answer
Nov 23, 2016

Answer:

#u^4-81 = (u-3)(u+3)(u^2+9)#

Explanation:

The difference of squares identity can be written:

#a^2-b^2 = (a-b)(a+b)#

We can use this a couple of times to derive the factors with Real coefficients, as follows:

#u^4-81 = (u^2)^2-9^2#

#color(white)(u^4-81) = (u^2-9)(u^2+9)#

#color(white)(u^4-81) = (u^2-3^2)(u^2+9)#

#color(white)(u^4-81) = (u-3)(u+3)(u^2+9)#

The remaining quadratic factor has no simpler linear factors with Real coefficients since #u^2+9 >= 9# for any Real value of #u#, hence no Real zeros or corresponding linear factors.