# How do you factor completely 2x^3y^4 − 8x^2y^3 + 6xy^2?

Aug 18, 2016

$2 {x}^{3} {y}^{4} - 8 {x}^{2} {y}^{3} + 6 x {y}^{2} = 2 x {y}^{2} \left(x y - 3\right) \left(x y - 1\right)$

#### Explanation:

Separate out the common factor $2 x {y}^{2}$ of the terms, then factor as a quadratic in $x y$ as follows:

$2 {x}^{3} {y}^{4} - 8 {x}^{2} {y}^{3} + 6 x {y}^{2}$

$= 2 x {y}^{2} \left({x}^{2} {y}^{2} - 4 x y + 3\right)$

$= 2 x {y}^{2} \left({\left(x y\right)}^{2} - 4 \left(x y\right) + 3\right)$

$= 2 x {y}^{2} \left(x y - 3\right) \left(x y - 1\right)$