# How do you factor n^2+4n-12?

Mar 28, 2018

$\left(n - 2\right) \left(n + 6\right)$

#### Explanation:

By using SUM PRODUCT

= ${n}^{2} + 6 n - 2 n - 12$

= $n \left(n + 6\right) - 2 \left(n + 6\right)$

= $\left(n - 2\right) \left(n + 6\right)$

Hope this helps!

Mar 28, 2018

$\left(n + 6\right) \left(n - 2\right)$

#### Explanation:

To factor this we, have to split the middle term.
If the quadratic equation is $a {x}^{2} + b x + c$, then we need to split the $b x$ into two terms such that the ratio of $a$ to first half = second half to $c$

So, we split ${n}^{2} + 4 n - 12$ into ${n}^{2} + 6 n - 2 n - 12$
As we can see, $1 : 6$=$- 2 : - 12$

Now in the first and second half take common the biggest term possible common

=$\left(n + 6\right) n - \left(n + 6\right) 2$

Here take a look, if the terms inside brackets are same, then you're on the right track

Now take the remaining outside the bracket common and you get

$\left(n + 6\right) \left(n - 2\right)$