How do you factor 9x ^ { 2} + 21x - 18?

Jul 21, 2017

To factor 9x2+21x−18

Factor out the $3$: $3 \left(3 {x}^{2} + 7 x - 6\right)$
Continue factoring: $3 \left(3 x - 2\right) \left(x + 3\right)$

Sep 10, 2017

$3 \left(3 x - 2\right) \left(x + 3\right)$

Explanation:

$f \left(x\right) = 9 {x}^{2} + 21 x - 18 = 3 \left(3 {x}^{2} + 7 x - 6\right)$

Factor the trinomial in parentheses by the new AC Method (Socratic, Google Search)

$y = 3 {x}^{2} + 7 x - 6 = 3 \left(x + p\right) \left(x + q\right)$

Factor converted trinomial:
$y ' = {x}^{2} + 7 x - 18 =$ (x + p')(x + q').

Find $p ' \mathmr{and} q '$

knowing the sum $\left(b = 7\right)$ and the product $\left(a c = - 18\right) .$

They are: $p ' = - 2 \mathmr{and} q ' = 9 \rightarrow \left[\text{sum " = 7 and " product } = - 18\right]$

Back to y:

$p = \frac{p '}{a} = - \frac{2}{3}$ , and $q = \frac{q '}{a} = \frac{9}{3} = 3$

Factored form of $f \left(x\right) :$

$f \left(x\right) = 3 \left(3\right) \left(x - \frac{2}{3}\right) \left(x + 3\right) = 3 \left(3 x - 2\right) \left(x + 3\right)$