# How do you factor x^ { 3} y - x ^ { 2} y ^ { 2} - 20x y ^ { 3} completely?

Mar 4, 2017

${x}^{3} y - {x}^{2} {y}^{2} - 20 x {y}^{3} = x y \left(4 y + x\right) \left(5 y - x\right)$

#### Explanation:

This is a homogeneous formula so making $y = \lambda x$ and substituting we have

${x}^{4} \left(\lambda - {\lambda}^{2} - 20 {\lambda}^{3}\right)$. The roots of

$\lambda - {\lambda}^{2} - 20 {\lambda}^{3} = 0$ are

$\lambda = \left\{0 , - \frac{1}{4} , \frac{1}{5}\right\}$ or

$\lambda - {\lambda}^{2} - 20 {\lambda}^{3} = 20 \lambda \left(\lambda + \frac{1}{4}\right) \left(\lambda - \frac{1}{5}\right)$

then

${x}^{3} y - {x}^{2} {y}^{2} - 20 x {y}^{3} = 20 {x}^{4} \lambda \left(\lambda + \frac{1}{4}\right) \left(\lambda - \frac{1}{5}\right)$ or

${x}^{3} y - {x}^{2} {y}^{2} - 20 x {y}^{3} = 20 x y \left(y + \frac{x}{4}\right) \left(y - \frac{x}{5}\right)$ or

${x}^{3} y - {x}^{2} {y}^{2} - 20 x {y}^{3} = x y \left(4 y + x\right) \left(5 y - x\right)$