# How do you use algebra to solve (x-1)(x+2)=18?

Mar 24, 2015

We know that a if a product is $0$, then at least one of the factors must be $0$. But, that's not what we have here.

Let's try to make it what we have:

$\left(x - 1\right) \left(x + 2\right) = 18$
$\left(x - 1\right) \left(x + 2\right) - 18 = 0$

Great!, Except -- now we don't have a product on the left. Try something and see if it helps or makes things worse.
(I have experience, I often know what will work. A less experienced person has to just try something.)

$\left[{x}^{2} + x - 2\right] - 18 = 0$ (I multiplied $\left(x - 1\right) \left(x + 2\right)$)

${x}^{2} + x - 20 = 0$ (Simplify)
$\left(x + 5\right) \left(x - 4\right) = 0$ (Factor the NEW expression)

$x + 5 = 0$ or $x - 4 = 0$

It is clear that there are 2 solutions: $- 5$ is a solution and $4$ is a solution