# How do you factor 2x^2 - x - 3?

May 28, 2015

$2 {x}^{2} - x - 3 = \left(2 x - 3\right) \left(x + 1\right)$

Problem: Factor $2 {x}^{2} - x - 3$.

The generic form of this equation is $a {x}^{2} + b x + c$.

$a = 2$
$b = - 1$
$c = - 3$

Multiply $a$ and $c$.

$2 \cdot \left(- 3\right) = - 6$

Find two factors of $- 6$ that when added equal $- 1$. The numbers $- 3$ and $2$ fit this requirement.

Rewrite the equation so that $- 3 x$ and $2 x$ replace $- 1 x$.

Group the first and second pairs of terms.

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

Factor $x$ out of the first term.

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

$x \left(2 x - 3\right) + 1 \left(2 x - 3\right)$

Factor out the common term $2 x - 3$.

$\left(x + 1\right) \left(2 x - 3\right)$

We can also rewrite the equation as $2 {x}^{2} + 2 x - 3 x - 3$.

Group the two sets of terms.

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

Factor $2 x$ from the first term, and $3$ out of the second term.

$2 x \left(x + 1\right) - 3 \left(x + 1\right)$

Factor out the common term $x + 1$.

$\left(2 x - 3\right) \left(x + 1\right)$