# How do you factor 9x^2 + 6x - 8?

Then $p = \frac{p '}{a} = - \frac{6}{9} = - \frac{2}{3} ,$and $q = \frac{q '}{a} = \frac{12}{9} = \frac{4}{3}$.
Factored form: $f \left(x\right) = \left(x - \frac{2}{3}\right) \left(x + \frac{4}{3}\right) = \left(3 x - 2\right) \left(3 x + 4\right)$
Check by developing:$f \left(x\right) = 9 {x}^{2} + 12 x - 6 x - 8.$ OK