How do you factor #u^3-2u^2-8u+16#?

2 Answers

Refer to explanation

Explanation:

It is

#u^3-2u^2-8u+16=u^2(u-2)-8(u-2)=(u^2-8)(u-2)=(u-sqrt8)(u+sqrt8)(u-2)=(u-2sqrt2)(u+2sqrt2)(u-2)#

Sep 23, 2015

Try this:

Explanation:

You can collect #u^2# from the first and third term and #-8# from the second and fourth to get:
#u^2(u-2)-8(u-2)=#
Now collect #(u-2)# and get:
#=(u-2)(u^2-8)#