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

1 Answer
Apr 24, 2016

Answer:

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

Explanation:

First note that all of the terms are divisible by #2y^3#, so separate that out as a factor first:

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

The remaining factor is homogeneous of degree #2#, so we can attempt to factor it using an AC method.

Look for a pair of factors of #AC = 6*7 = 42# with sum #B=17#.

The pair #3, 14# works, in that #3*14 = 42# and #3+14=17#.

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)#