How do you factor completely #x^3-x^2-x+x#?

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Answer 1 · 2018-07-19T23:23:01

#x^2(x-1)#

First, we see that there's a factor of #x# we can easily cancel:
#x^3 - x^2 - x + x = x^3 - x^2 #

From there, we can also factor out #x^2#:
#x^3 - x^2 = x^2(x-1) #

Since each of these is fully factored, we are done.

Answer 2 · 2018-07-19T23:29:12

#x^2(x-1)#

This expression can be simplified to

#x^3-x^2#

From here, we notice that both terms have an #x^2# in common, which we can factor out. At this point, we are essentially dividing.

We get

#x^2(x-1)#

Hope this helps!