# How do you factor 8x²+20x-48?

Apr 26, 2016

$8 {x}^{2} + 20 x - 48 = 4 \left(2 x - 3\right) \left(x + 4\right)$

#### Explanation:

First notice that all of the terms are divisible by $4$, so separate that out as a factor first:

$8 {x}^{2} + 20 x - 48 = 4 \left(2 {x}^{2} + 5 x - 12\right)$

Next use an AC method to factor $2 {x}^{2} + 5 x - 12$:

Look for a pair of factors of $A C = 2 \cdot 12 = 24$ with difference $B = 5$.

The pair $8 , 3$ works: $8 \times 3 = 24$ and $8 - 3 = 5$.

Use this pair to split the middle term and factor by grouping:

$2 {x}^{2} + 5 x - 12$

$= 2 {x}^{2} + 8 x - 3 x - 12$

$= \left(2 {x}^{2} + 8 x\right) - \left(3 x + 12\right)$

$= 2 x \left(x + 4\right) - 3 \left(x + 4\right)$

$= \left(2 x - 3\right) \left(x + 4\right)$

Hence:

$8 {x}^{2} + 20 x - 48 = 4 \left(2 x - 3\right) \left(x + 4\right)$