# How do you factor 12c^2+23c-9?

Sep 23, 2016

$12 {c}^{2} + 23 c - 9 = \left(3 c - 1\right) \left(4 c + 9\right)$

#### Explanation:

$12 {c}^{2} + 23 c - 9$

Use an AC method:

Find a pair of factors of $A C = 12 \cdot 9 = 108 = {2}^{2} \cdot {3}^{3}$ which differ by $23$

Since $23$ is not divisible by $3$ or $2$, one of the factors is divisible by ${3}^{3} = 27$ and one factor (possibly the same one) is divisible by ${2}^{2} = 4$. Note that $27 - 4 = 23$, so we have found the required pair in $27 , 4$

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

$12 {c}^{2} + 23 c - 9 = \left(12 {c}^{2} + 27 c\right) - \left(4 c + 9\right)$

$\textcolor{w h i t e}{12 {c}^{2} + 23 c - 9} = 3 c \left(4 c + 9\right) - 1 \left(4 c + 9\right)$

$\textcolor{w h i t e}{12 {c}^{2} + 23 c - 9} = \left(3 c - 1\right) \left(4 c + 9\right)$