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

May 19, 2015

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

We can Split the Middle Term of this expression to factorise it
In this technique, if we have to factorise an expression like $a {n}^{2} + b n + c$, we need to think of 2 numbers such that:

${N}_{1} \cdot {N}_{2} = a \cdot c = 2 \times \left(- 9\right) = - 18$
and
${N}_{1} + {N}_{2} = b = 3$

After trying out a few numbers we get ${N}_{1} = 6$ and ${N}_{2} = - 3$

$6 \times \left(- 3\right) = - 18$ and $6 + \left(- 3\right) = 3$

$2 {n}^{2} + 3 n - 9 = 2 {n}^{2} + 6 n - 3 n - 9$

$= 2 n \left(n + 3\right) - 3 \left(n + 3\right)$

=color(green)((2n - 3)(n +3)