# How do you factor 9q ^ { 2} - 7q - 18?

Jul 16, 2017

$9 {q}^{2} - 7 q - 18$

Well, $7$ is a prime number, so we can't factor any of the coefficients ($9 , - 7 , - 18$)

What we need to do is find two numbers that multiply to $- 162$ ($9 \times - 18$) and adds to $- 7$:

$\textcolor{w h i t e}{0} \times$ $162$ let's leave the sign out of it for now
$\textcolor{w h i t e}{0} +$ $7$
............
$1 \times 162$
$2 \times 81$
$3 \times 54$
$6 \times 27$
$9 \times 18$

Well, that didn't work. We should check to make sure we even can factor this.

To do that, we need to graph the equation and see the roots ($x$-intercepts)

graph{y=9x^2-7x-18}

So this does have $x$-intercepts, but the roots are pretty weird numbers, with crazy decimals. That means we can't factor it.
However, we can still find solutions, using the quadratic formula:

$- \frac{b}{2 a} \pm \frac{\sqrt{{b}^{2} - 4 a c}}{2 a}$

$\textcolor{red}{a} = \textcolor{red}{9}$

$\textcolor{b l u e}{b} = \textcolor{b l u e}{- 7}$

$\textcolor{g r e e n}{c} = \textcolor{g r e e n}{- 18}$

$\frac{- \left(- 7\right)}{2 \left(9\right)} \pm \frac{\sqrt{{\left(- 7\right)}^{2} - 4 \times \left(9\right) \times \left(- 18\right)}}{2 \left(9\right)}$

$\frac{7}{18} \pm \frac{\sqrt{49 - 4 \times \left(9\right) \times \left(- 18\right)}}{18}$

$\frac{7}{18} \pm \frac{\sqrt{49 - - 648}}{18}$

$\frac{7}{18} \pm \frac{\sqrt{697}}{18}$

$\frac{7 \pm \sqrt{697}}{18}$

We can't simplify $\sqrt{697}$, because it simplifies to $17 \times 41$, both of which are prime.

So, we have two solutions:

number $1$

$7 + \frac{\sqrt{697}}{18} \approx 18.47$

number $2$

$7 - \frac{\sqrt{697}}{18} \approx 15.53$

Jul 16, 2017

can't be factored.

#### Explanation:

$y = 9 {q}^{2} - 7 q - 18$
$D = {b}^{2} - 4 a c = 49 + 648 = 697$
Since D is not a perfect square, this trinomial can't be factored.

Jul 16, 2017

$9 {q}^{2} - 7 q - 18 = \frac{1}{36} \left(18 q - 7 - \sqrt{697}\right) \left(18 q - 7 + \sqrt{697}\right)$

#### Explanation:

The difference of squares identity can be written:

${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

Use this with $a = \left(18 q - 7\right)$ and $b = \sqrt{697}$ as follows:

$36 \left(9 {q}^{2} - 7 q - 18\right) = 324 {q}^{2} - 252 q - 648$

$\textcolor{w h i t e}{36 \left(9 {q}^{2} - 7 q - 18\right)} = {\left(18 q\right)}^{2} - 2 \left(18 q\right) 7 + 49 - 697$

$\textcolor{w h i t e}{36 \left(9 {q}^{2} - 7 q - 18\right)} = {\left(18 q - 7\right)}^{2} - {\left(\sqrt{697}\right)}^{2}$

$\textcolor{w h i t e}{36 \left(9 {q}^{2} - 7 q - 18\right)} = \left(\left(18 q - 7\right) - \sqrt{697}\right) \left(\left(18 q - 7\right) + \sqrt{697}\right)$

$\textcolor{w h i t e}{36 \left(9 {q}^{2} - 7 q - 18\right)} = \left(18 q - 7 - \sqrt{697}\right) \left(18 q - 7 + \sqrt{697}\right)$

So:

$9 {q}^{2} - 7 q - 18 = \frac{1}{36} \left(18 q - 7 - \sqrt{697}\right) \left(18 q - 7 + \sqrt{697}\right)$

Note that $\sqrt{697}$ does not simplify, since $697 = 17 \cdot 41$ has no square factors.