# How do you factor 8x^2 - 4x - 24?

Aug 21, 2017

Well, you can factor out 4 immediately:

$4 \left(2 {x}^{2} - x - 6\right)$

You can work out that the further factorization will look something like:

$\left(2 x + a\right) \left(x + b\right)$ where $a \cdot b = - 6$

and

$\left(2 x \cdot b\right) + \left(a \cdot x\right) = - 1 x$

so you know that $2 a + b = - 1$
and

$a \cdot b = - 6$

You know that either a or b must be negative, and the other number must be positive.

At this point, I usually resort to trial and error guessing. There are usually a couple of choices, so try one, then the other.

so what if $a = 3$ and $b = - 2$, does that work?

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

...good guess. So your factorization will be:

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

Aug 21, 2017

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

#### Explanation:

First take out the common factor of $4$

$8 {x}^{2} - 4 x - 24$

$= 4 \left(2 {x}^{2} - x - 6\right)$

Find factors of $2 \mathmr{and} 6$ whose products differ by $1$

$\text{ } 2 \mathmr{and} 6$
$\text{ "darr" } \downarrow$

$\text{ "2" "3" } \rightarrow 1 \times 3 = 3$
$\text{ "1" "2" } \rightarrow 2 \times 2 = \underline{4}$
$\textcolor{w h i t e}{\times \times \times \times x . x . \times \times \times} 1$

$8 {x}^{2} - 4 x - 24$

$= 4 \left(2 {x}^{2} - x - 6\right)$

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