# How do you factor 6x^2 + 3x – 63?

May 15, 2015

$6 {x}^{2} + 3 x - 63 = \left(3 \cdot 2 {x}^{2}\right) + \left(3 \cdot x\right) - \left(3 \cdot 21\right)$

$= 3 \left(2 {x}^{2} + x - 21\right)$

To factor $2 {x}^{2} + x - 21$ I will look for roots of the equation:

$2 {x}^{2} + x - 21 = 0$

This is of the form $a {x}^{2} + b x + c = 0$, which has solutions:

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

In our case $a = 2$, $b = 1$ and $c = - 21$, so

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$= \frac{- 1 \pm \sqrt{{1}^{2} - 4 \cdot 2 \cdot \left(- 21\right)}}{2 \cdot 2}$

$= \frac{- 1 \pm \sqrt{1 + 168}}{4}$

$= \frac{- 1 \pm \sqrt{169}}{4}$

$= \frac{- 1 \pm 13}{4}$

So one root is $\frac{12}{4} = 3$ and the other is $- \frac{14}{4} = - \frac{7}{2}$.

From this we can deduce that $\left(x - 3\right)$ is a factor and $\left(2 x + 7\right)$ is the other factor.

In summary, we have $6 {x}^{2} + 3 x - 63 = 3 \left(x - 3\right) \left(2 x + 7\right)$.