# How do you factor completely 12x^3y + 6x^2y^2 - 9xy^3?

Feb 19, 2017

$12 {x}^{3} y + 6 {x}^{2} {y}^{2} - 9 x {y}^{3} = \frac{3}{4} x y \left(4 x + \left(1 - \sqrt{13}\right) y\right) \left(4 x + \left(1 + \sqrt{13}\right) y\right)$

#### Explanation:

Given:

$12 {x}^{3} y + 6 {x}^{2} {y}^{2} - 9 x {y}^{3}$

Note that all of the terms are of degree $4$ and all are divisible by $3 x y$:

$12 {x}^{3} y + 6 {x}^{2} {y}^{2} - 9 x {y}^{3} = 3 x y \left(4 {x}^{2} + 2 x y - 3 {y}^{2}\right)$

We can factor the remaining quadratic by completing the square as follows:

$4 {x}^{2} + 2 x y - 3 {y}^{2} = \frac{1}{4} \left(16 {x}^{2} + 8 x y - 12 {y}^{2}\right)$

$\textcolor{w h i t e}{4 {x}^{2} + 2 x y - 3 {y}^{2}} = \frac{1}{4} \left({\left(4 x\right)}^{2} + 2 \left(4 x\right) \left(y\right) + {y}^{2} - 13 {y}^{2}\right)$

$\textcolor{w h i t e}{4 {x}^{2} + 2 x y - 3 {y}^{2}} = \frac{1}{4} \left({\left(4 x + y\right)}^{2} - {\left(\sqrt{13} y\right)}^{2}\right)$

$\textcolor{w h i t e}{4 {x}^{2} + 2 x y - 3 {y}^{2}} = \frac{1}{4} \left(\left(4 x + y\right) - \sqrt{13} y\right) \left(\left(4 x + y\right) + \sqrt{13} y\right)$

$\textcolor{w h i t e}{4 {x}^{2} + 2 x y - 3 {y}^{2}} = \frac{1}{4} \left(4 x + \left(1 - \sqrt{13}\right) y\right) \left(4 x + \left(1 + \sqrt{13}\right) y\right)$

Putting it all together:

$12 {x}^{3} y + 6 {x}^{2} {y}^{2} - 9 x {y}^{3} = \frac{3}{4} x y \left(4 x + \left(1 - \sqrt{13}\right) y\right) \left(4 x + \left(1 + \sqrt{13}\right) y\right)$

Feb 19, 2017

${3}^{2} x y \left(y - \frac{x}{3} \left(1 - \sqrt{13}\right)\right) \left(y - \frac{x}{3} \left(1 + \sqrt{13}\right)\right)$

#### Explanation:

$f \left(x , y\right) = 12 {x}^{3} y + 6 {x}^{2} {y}^{2} - 9 x {y}^{3}$ is a homogeneous function

Making $y = \lambda x$ and substituting

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

Solving $\lambda \left(4 + \left(2 - 3 \lambda\right) \lambda\right) = 0$ we have

$\lambda = \left\{0 , \frac{1}{3} \left(1 - \sqrt{13}\right) , \frac{1}{3} \left(1 + \sqrt{13}\right)\right\}$ or

$f \left(x , \lambda\right) = {3}^{2} {x}^{4} \lambda \left(\lambda - \frac{1}{3} \left(1 - \sqrt{13}\right)\right) \left(\lambda - \frac{1}{3} \left(1 + \sqrt{13}\right)\right)$ so

$f \left(x , y\right) = {3}^{2} x y \left(y - \frac{x}{3} \left(1 - \sqrt{13}\right)\right) \left(y - \frac{x}{3} \left(1 + \sqrt{13}\right)\right)$