# How do you factor completely 15a^2 + 55a - 20?

Oct 26, 2016

$15 {a}^{2} + 55 a - 20 = 5 \left(3 a - 1\right) \left(a + 4\right)$

#### Explanation:

First separate out the common scalar factor $5$ to find:

$15 {a}^{2} + 55 a - 20 = 5 \left(3 {a}^{2} + 11 a - 4\right)$

Next use an AC Method to factor $3 {a}^{2} + 11 a - 4$:

Find a pair of factors of $A C = 3 \cdot 4 = 12$ which differ by $B = 11$.

The pair $12 , 1$ works, in that $12 \cdot 1 = 12$ and $12 - 1 = 11$.

Use this pair to split the middle term and factor by grouping:

$3 {a}^{2} + 11 a - 4 = 3 {a}^{2} + 12 a - a - 4$

$\textcolor{w h i t e}{3 {a}^{2} + 11 a - 4} = \left(3 {a}^{2} + 12 a\right) - \left(a + 4\right)$

$\textcolor{w h i t e}{3 {a}^{2} + 11 a - 4} = 3 a \left(a + 4\right) - 1 \left(a + 4\right)$

$\textcolor{w h i t e}{3 {a}^{2} + 11 a - 4} = \left(3 a - 1\right) \left(a + 4\right)$

So:

$15 {a}^{2} + 55 a - 20 = 5 \left(3 a - 1\right) \left(a + 4\right)$