# How do you solve x^2+7x-18=0 by factoring?

Mar 27, 2015

While it will not always be the case, start by assuming we are only dealing with integers and the factors of
${x}^{2} + 7 x - 18$
have the form
$\left(x + a\right) \left(x - b\right)$ we know $a$ and $- b$ have different signs since their product is negative ($- 18$)
We also know that $a$ is greater than $b$ since the coefficient of $x$ is greater than zero.

There are only a limited number of possible integer values for $a$ and $b$ with $a > b$ and $a b = 18$

$\left(a , b\right) = \left(18 , 1\right)$ which would give $a - b = 17$; not what we want

$\left(a , b\right) = \left(9 , 2\right)$ which would give $a - b = 7$; this matches the $x$ coefficient of the given equation

$\left(a , b\right) = \left(6 , 3\right)$ which would give $a - b = 3$; not what we want

There are no other integer possibilities.

The factoring is
$\left(x + 9\right) \left(x - 2\right) = {x}^{2} + 7 x - 18$

Since
${x}^{2} + 7 x - 18 = 0$
either
$x + 9 = 0$ or $x - 2 = 0$

So
either $x = - 9$ or $x = 2$