# How do you factor completely 3a^2+9a-12?

Apr 11, 2016

$3 \left(a - 1\right) \left(a + 4\right)$

#### Explanation:

3 is the factor common to all the terms.
Let's take it out

$3 \left({a}^{2} + 3 a - 4\right)$

Now, we find the factors of $\left({a}^{2} + 3 a - 4\right)$

Use splitting of the middle term to find the answer.

Find a pair of factors of $1 \times - 4 = - 4$ which sum up to $3$. The pair $- 1$ and $4$ works.

Write the expression, $\left({a}^{2} + 3 a - 4\right)$ as:

${a}^{2} - a + 4 a - 4$

Take out the common terms

$a \left(a - 1\right) + 4 \left(a - 1\right)$

Separate the like terms and you get

$\left(a - 1\right) \left(a + 4\right)$

The factors of $3 {a}^{2} + 9 a - 12$ = $3 \left(a - 1\right) \left(a + 4\right)$