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

Apr 24, 2016

$12 {y}^{5} - 34 x {y}^{4} + 14 {x}^{2} {y}^{3} = 2 {y}^{3} \left(3 y - 7 x\right) \left(2 y - x\right)$

#### Explanation:

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

$12 {y}^{5} - 34 x {y}^{4} + 14 {x}^{2} {y}^{3} = 2 {y}^{3} \left(6 {y}^{2} - 17 x y + 7 {x}^{2}\right)$

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 $A C = 6 \cdot 7 = 42$ with sum $B = 17$.

The pair $3 , 14$ works, in that $3 \cdot 14 = 42$ and $3 + 14 = 17$.

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

$6 {y}^{2} - 17 x y + 7 {x}^{2}$

$= 6 {y}^{2} - 3 x y - 14 x y + 7 {x}^{2}$

$= \left(6 {y}^{2} - 3 x y\right) - \left(14 x y - 7 {x}^{2}\right)$

$= 3 y \left(2 y - x\right) - 7 x \left(2 y - x\right)$

$= \left(3 y - 7 x\right) \left(2 y - x\right)$

Putting it all together:

$12 {y}^{5} - 34 x {y}^{4} + 14 {x}^{2} {y}^{3} = 2 {y}^{3} \left(3 y - 7 x\right) \left(2 y - x\right)$