How do you factor #36p^4-48p^3+16p^2#?

1 Answer
Aug 25, 2016

Answer:

#4p^2*(3p-2)^2#
OR
#(6p^2-4p)^2#

Explanation:

Fist we factor out #p^2# leaving us with:
#p^2(36p^2-48p+16)#
We can also factor out #4# from all 3 terms so we get
#4p^2(9p^2-12p+4)#
Now, looking at #9p^2-12p+4# I see that #3p*3p=9p^2# and #(-2)*(-2)=4# and #(-2*3)+(-2*3)=-12#. That means that:
#9p^2-12p+4=(3p-2)(3p-2)=(3p-2)^2 #

So final equation is #4p^2*(3p-2)^2#

Another answer is if we recognize that #4p^2=(2p)^2# and distribute #2p# into each of the #(3p-2)# leaving each term as #(6p^2-4p)# so the final answer would be:
#(6p^2-4p)^2#