How do you factor completely −12x^3y^5 − 9x^2y^2 + 12xy^3?

Apr 13, 2017

$- 12 {x}^{3} {y}^{5} - 9 {x}^{2} {y}^{2} + 12 x {y}^{3} = - 3 x {y}^{2} \left(4 {x}^{2} {y}^{3} + 3 x - 4 y\right)$

But note that:

$\textcolor{red}{9 {x}^{4} {y}^{4}} - 12 {x}^{3} {y}^{5} - 9 {x}^{2} {y}^{2} + 12 x {y}^{3} = 3 x {y}^{2} \left(x y - 1\right) \left(x y + 1\right) \left(3 x - 4 y\right)$

Explanation:

Given:

$- 12 {x}^{3} {y}^{5} - 9 {x}^{2} {y}^{2} + 12 x {y}^{3}$

First note that all of the terms are divisible by $3$, $x$ and ${y}^{2}$, and hence by $3 x {y}^{2}$. So we can separate that out as a factor. I will use $- 3 x {y}^{2}$ because I prefer positive leading coefficients:

$- 12 {x}^{3} {y}^{5} - 9 {x}^{2} {y}^{2} + 12 x {y}^{3} = - 3 x {y}^{2} \left(4 {x}^{2} {y}^{3} + 3 x - 4 y\right)$

It is not possible to factor this further.

I suspect that a leading term $9 {x}^{4} {y}^{4}$ is missing from the question, since we find:

$9 {x}^{4} {y}^{4} - 12 {x}^{3} {y}^{5} - 9 {x}^{2} {y}^{2} + 12 x {y}^{3} = 3 x {y}^{2} \left(3 {x}^{3} {y}^{2} - 4 {x}^{2} {y}^{3} - 3 x + 4 y\right)$

$\textcolor{w h i t e}{9 {x}^{4} {y}^{4} - 12 {x}^{3} {y}^{5} - 9 {x}^{2} {y}^{2} + 12 x {y}^{3}} = 3 x {y}^{2} \left(\left(3 {x}^{3} {y}^{2} - 4 {x}^{2} {y}^{3}\right) - \left(3 x - 4 y\right)\right)$

$\textcolor{w h i t e}{9 {x}^{4} {y}^{4} - 12 {x}^{3} {y}^{5} - 9 {x}^{2} {y}^{2} + 12 x {y}^{3}} = 3 x {y}^{2} \left({x}^{2} {y}^{2} \left(3 x - 4 y\right) - 1 \left(3 x - 4 y\right)\right)$

$\textcolor{w h i t e}{9 {x}^{4} {y}^{4} - 12 {x}^{3} {y}^{5} - 9 {x}^{2} {y}^{2} + 12 x {y}^{3}} = 3 x {y}^{2} \left({x}^{2} {y}^{2} - 1\right) \left(3 x - 4 y\right)$

$\textcolor{w h i t e}{9 {x}^{4} {y}^{4} - 12 {x}^{3} {y}^{5} - 9 {x}^{2} {y}^{2} + 12 x {y}^{3}} = 3 x {y}^{2} \left({\left(x y\right)}^{2} - {1}^{2}\right) \left(3 x - 4 y\right)$

$\textcolor{w h i t e}{9 {x}^{4} {y}^{4} - 12 {x}^{3} {y}^{5} - 9 {x}^{2} {y}^{2} + 12 x {y}^{3}} = 3 x {y}^{2} \left(x y - 1\right) \left(x y + 1\right) \left(3 x - 4 y\right)$