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

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

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Answer 1 · 2015-03-24T22:24:04

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

Let's try to make it what we have:

$(x - 1) (x + 2) = 18$ $(x-1)(x+2)=18$
$(x - 1) (x + 2) - 18 = 0$ $(x-1)(x+2)-18=0$

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.)

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

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

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

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

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

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