# How do you factor by grouping 2n^2 + 3n - 9?

May 31, 2015

$2 {n}^{2} + 3 n - 9$

first try extracting the sum and difference of squares:
$\textcolor{w h i t e}{\text{XXXXX}}$$= \left({n}^{2} - 9\right) + \left({n}^{2} + 3 n\right)$

factor the two resulting terms:
$\textcolor{w h i t e}{\text{XXXXX}}$$\left(n - 3\right) \left(n + 3\right) + n \left(n + 3\right)$

extract the common term:
$\textcolor{w h i t e}{\text{XXXXX}}$$\left(n + 3\right) \cdot \left(\left(n - 3\right) + \left(n\right)\right)$

simplify
$\textcolor{w h i t e}{\text{XXXXX}}$$\left(n + 3\right) \left(2 n - 3\right)$