# How do you factor 3v ^ { 2} + 12v - 63?

Mar 20, 2018

$3 {v}^{2} + 12 v - 63 = \textcolor{b l u e}{3 \left(v + 7\right) \left(v - 3\right)}$

#### Explanation:

Given
$\textcolor{w h i t e}{\text{XXX}} 3 {v}^{2} + 12 v - 63$

First factor out the obvious constant
$\textcolor{w h i t e}{\text{XXX}} = \textcolor{red}{3} \left({v}^{2} + 4 v - 21\right)$

Then look for two numbers whose product is $\left(- 21\right)$ and whose sum is $\left(+ 4\right)$.
With a bit of playing around you should come up with $\textcolor{\lim e}{+ 7}$ and $\textcolor{g r e e n}{- 4}$;
enabling the continuation of the factoring:
$\textcolor{w h i t e}{\text{XXX}} \textcolor{red}{3} \left(v \textcolor{\lim e}{+ 7}\right) \left(v \textcolor{g r e e n}{- 4}\right)$

Mar 20, 2018

$3 {x}^{2} + 12 v - 63 = 3 \left(v - 7\right) \left(v + 3\right)$

#### Explanation:

Given:

$3 {x}^{2} + 12 v - 63$

First note that all of the coefficients are divisible by $3$, so we can separate that out as a factor:

$3 {v}^{2} + 12 v - 64 = 3 \left({v}^{2} - 4 v - 21\right)$

To factor the remaining quadratic, we can find a pair of factors of $21$ which differ by $4$. Note that $7 \cdot 3 = 21$ and $7 - 3 = 4$.

Hence we find:

${v}^{2} - 4 v - 21 = \left(v - 7\right) \left(v + 3\right)$

So:

$3 {x}^{2} + 12 v - 63 = 3 \left(v - 7\right) \left(v + 3\right)$

Mar 20, 2018

$3 \left(v + 7\right) \left(v - 4\right)$

#### Explanation:

First you take $3$ out of all the number:
$3 \left({v}^{2} + 4 v - 21\right)$

Then factorise the middle:
so $7 - 3 = 4$
and $7 \times - 3 = - 21$

so the final answer would be

$3 \left(v + 7\right) \left(v - 4\right)$