# How do you factor 6x^4y^3 + 21x^3y^2 − 9x^2y?

Apr 24, 2016

$6 {x}^{4} {y}^{3} + 21 {x}^{3} {y}^{2} - 9 {x}^{2} y$

$= 3 {x}^{2} y \left(2 {x}^{2} {y}^{2} + 7 x y - 3\right)$

$= 6 {x}^{2} y \left(x y + \frac{7}{4} - \frac{\sqrt{73}}{4}\right) \left(x y + \frac{7}{4} + \frac{\sqrt{73}}{4}\right)$

#### Explanation:

First notice that all of the terms are divisible by $3 {x}^{2} y$, so separate that out as a factor:

$6 {x}^{4} {y}^{3} + 21 {x}^{3} {y}^{2} - 9 {x}^{2} y = 3 {x}^{2} y \left(2 {x}^{2} {y}^{2} + 7 x y - 3\right)$

The remaining factor is a quadratic in $x y$, whose zeros we can find using the quadratic formula:

$x y = \frac{- 7 \pm \sqrt{{7}^{2} - \left(4 \cdot 2 \cdot - 3\right)}}{2 \cdot 2}$

$= \frac{- 7 \pm \sqrt{49 + 24}}{4}$

$= \frac{- 7 \pm \sqrt{73}}{4}$

Hence:

$2 {x}^{2} {y}^{2} + 7 x y - 3 = 2 \left(x y + \frac{7}{4} - \frac{\sqrt{73}}{4}\right) \left(x y + \frac{7}{4} + \frac{\sqrt{73}}{4}\right)$

Putting it all together:

$6 {x}^{4} {y}^{3} + 21 {x}^{3} {y}^{2} - 9 {x}^{2} y$

$= 3 {x}^{2} y \left(2 {x}^{2} {y}^{2} + 7 x y - 3\right)$

$= 6 {x}^{2} y \left(x y + \frac{7}{4} - \frac{\sqrt{73}}{4}\right) \left(x y + \frac{7}{4} + \frac{\sqrt{73}}{4}\right)$