How do you factor by grouping #n^3-2n^2+4n-8#?

1 Answer
Mar 27, 2018

Answer:

#(n-2)(n^2+4)#

Explanation:

#n^3-2n^2+4n-8#

Factor #n^2# from the first two terms and factor #4# from the last 2 terms.

#n^2(n-2)+4(n-2)#

You now have two terms both containing a factor of #(n-2)#. Factor#(n-2)# from both of these terms.

#(n-2)(n^2+4)#

Hey, that was easy!