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

1 Answer
Jim H
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:

#(x-1)(x+2)=18#
#(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# (I multiplied #(x-1)(x+2)#)

#x^2+x-20=0# (Simplify)
#(x+5)(x-4)=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

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