# How do you factor completely 40x^2 – 2xy – 24y^2?

May 3, 2016

$40 {x}^{2} - 2 x y - 24 {y}^{2} = 2 \left(4 x + 3 y\right) \left(5 x - 4 y\right)$

#### Explanation:

First separate out the common scalar factor $2$:

$40 {x}^{2} - 2 x y - 24 {y}^{2}$

$= 2 \left(20 {x}^{2} - x y - 12 {y}^{2}\right)$

Since the remaining factor is homogeneous of degree $2$ we can attempt to use an AC method to factor it:

Look for a pair of factors of $A C = 20 \cdot 12 = 240$ with difference $B = 1$

The pair $16 , 15$ works, so we can use that to split the middle term and factor by grouping:

$20 {x}^{2} - x y - 12 {y}^{2}$

$= \left(20 {x}^{2} - 16 x y\right) + \left(15 x y - 12 {y}^{2}\right)$

$= 4 x \left(5 x - 4 y\right) + 3 y \left(5 x - 4 y\right)$

$= \left(4 x + 3 y\right) \left(5 x - 4 y\right)$