# How do you factor completely: 10x^4y^3 − 5x^3y^2 + 20x^2y?

Jul 23, 2015

$10 {x}^{4} {y}^{3} - 5 {x}^{3} {y}^{2} + 20 {x}^{2} y$

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

$= 10 {x}^{2} y \left(x y - \frac{1}{4} - i \frac{\sqrt{31}}{4}\right) \left(x y - \frac{1}{4} + i \frac{\sqrt{31}}{4}\right)$

#### Explanation:

$10 {x}^{4} {y}^{3} - 5 {x}^{3} {y}^{2} + 20 {x}^{2} y$

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

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

This is as far as we can go with real coefficients.

The roots of $2 {t}^{2} - t + 4 = 0$ are given by the quadratic formula as:

$t = \frac{1 \pm \sqrt{{1}^{2} - \left(4 \times 2 \times 4\right)}}{2 \cdot 2}$

$= \frac{1 \pm \sqrt{- 31}}{4}$

$= \frac{1}{4} \pm i \frac{\sqrt{31}}{4}$

So if we allow complex coefficients we get:

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

$= 2 \left(x y - \frac{1}{4} - i \frac{\sqrt{31}}{4}\right) \left(x y - \frac{1}{4} + i \frac{\sqrt{31}}{4}\right)$

Hence:

$10 {x}^{4} {y}^{3} - 5 {x}^{3} {y}^{2} + 20 {x}^{2} y$

$= 10 {x}^{2} y \left(x y - \frac{1}{4} - i \frac{\sqrt{31}}{4}\right) \left(x y - \frac{1}{4} + i \frac{\sqrt{31}}{4}\right)$