# How do you factor the expression #12y^5 – 34xy^4 + 14x^2y^3#?

##### 1 Answer

Apr 24, 2016

#### Explanation:

First note that all of the terms are divisible by

#12y^5-34xy^4+14x^2y^3=2y^3(6y^2-17xy+7x^2)#

The remaining factor is homogeneous of degree

Look for a pair of factors of

The pair

Use this pair to split the middle term and factor by grouping:

#6y^2-17xy+7x^2#

#=6y^2-3xy-14xy+7x^2#

#=(6y^2-3xy)-(14xy-7x^2)#

#=3y(2y-x)-7x(2y-x)#

#=(3y-7x)(2y-x)#

Putting it all together:

#12y^5-34xy^4+14x^2y^3=2y^3(3y-7x)(2y-x)#