# How do you factor #2x^2+9x-5?

May 5, 2015

$2 {x}^{2} + 9 x - 5$

Hoping for an integer factoring, we look for factors of $2$ and factors of $5$ which when cross multiplied then subtracted give $9$
(we know we need to subtract since the final term ($5$) is negative).

Fortunately there are not a lot of possible options for the factors and we soon come up with

$\left.\begin{matrix}\null & 2 & \times & 5 & = & 10 \\ \null & 1 & \times & - 1 & = - & 1 \\ \null & \null & \null & \null & \null & 9\end{matrix}\right.$

So the factors of $2 {x}^{2} + 9 x - 5$
are $\left(2 x - 1\right) \left(x + 5\right)$