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

Question

1372 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-04-24T22:43:25

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

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:

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

How do you factor the expression 12y^5 – 34xy^4 + 14x^2y^312y^5 – 34xy^4 + 14x^2y^3?