How do you factor #8u^3-24u^2v+18uv^2#?

2 Answers
Feb 5, 2017

Answer:

The answer is #=2u(2u-3v)^2#

Explanation:

We need

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

Therefore,

#8u^2-24u^2v+18uv^2#

#=2u(4u^2-12uv+9v^2)#

#=2u(2u-3v)(2u-3v)#

#=2u(2u-3v)^2#

Feb 5, 2017

Answer:

#8u^3-24u^2v+18uv^2 = 2u(2u-3v)^2#

Explanation:

Given:

#8u^3-24u^2v+18uv^2#

Note that all of the terms are divisible by #2u#, so we can separate that out as a factor first.

Then the remaining quadratic is a perfect square trinomial, being of the form:

#A^2-2AB+B^2 = (A-B)^2#

with #A=2u# and #B=3v# ...

#8u^3-24u^2v+18uv^2 = 2u(4u^2-12uv+9v^2)#

#color(white)(8u^3-24u^2v+18uv^2) = 2u((2u)^2-2(2u)(3v)+(3v)^2)#

#color(white)(8u^3-24u^2v+18uv^2) = 2u(2u-3v)^2#