# How do you factor 5x^2 + 9x - 81?

May 31, 2015

$5 {x}^{2} + 9 x - 81$ is of the form $a {x}^{2} + b x + c$
with $a = 5$, $b = 9$, $c = - 81$

The discriminant $\Delta$ is given by the formula:

$\Delta = {b}^{2} - 4 a c = {9}^{2} - \left(4 \times 5 \times - 81\right) = {9}^{2} \cdot 21$

$\Delta > 0$ so $5 {x}^{2} + 9 x - 81 = 0$ has two distinct real roots. Unfortunately $\Delta$ is not a perfect square, so those roots are irrational.

The roots of $5 {x}^{2} + 9 x - 81 = 0$ are given by the formula:

$x = \frac{- b \pm \sqrt{\Delta}}{2 a}$

$= \frac{- 9 \pm 9 \sqrt{21}}{10}$

$= \frac{9}{10} \left(- 1 \pm \sqrt{21}\right)$

Hence

$5 {x}^{2} + 9 x - 81$

$= 5 \left(x - \frac{9}{10} \left(- 1 + \sqrt{21}\right)\right) \left(x - \frac{9}{10} \left(- 1 - \sqrt{21}\right)\right)$

$= 5 \left(x + \frac{9}{10} \left(1 - \sqrt{21}\right)\right) \left(x + \frac{9}{10} \left(1 + \sqrt{21}\right)\right)$