# How do you factor 10w² + 11w - 8?

May 17, 2015

$10 {w}^{2} + 11 w - 8$ is of the form $a {w}^{2} + b w + c$ with $a = 10$, $b = 11$ and $c = - 8$.

The discriminant of this quadratic is given by the formula

$\Delta = {b}^{2} - 4 a c = {11}^{2} - \left(4 \times 10 \times - 8\right) = 121 + 320 = 441 = {21}^{2}$

Being a positive square integer the equation $10 {w}^{2} + 11 w - 8 = 0$ has two distinct rational roots, given by the formula:

$w = \frac{- b \pm \sqrt{\Delta}}{2 a} = \frac{- 11 \pm 21}{20}$

That is

$w = \frac{- 11 + 21}{20} = \frac{10}{20} = \frac{1}{2}$

and

$w = \frac{- 11 - 21}{20} = - \frac{32}{20} = - \frac{8}{5}$

These give us factors $\left(2 w - 1\right)$ and $\left(5 w + 8\right)$

So $10 {w}^{2} + 11 w - 8 = \left(2 w - 1\right) \left(5 w + 8\right)$