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c^2-c-2 c^2+3c+2
___________ / ____________
c^2-1 c^2-3c+2
A) (c+1)(c+2)
________
(c-1)^2
B) c-2
__________
c-1
C) (c-2)^2
___________
(c+1)(c+2)
D) c-2
__________
c+1
Consider c^2-c-2. We want factors of the form (c + a)(c + b) where ab=-2 and a+b=-1. A moment of thought reveals that a=-2 and b=+1.
c^2 - c - 2 = (c - 2)(c + 1)
You could also use the quadratic formula to find that.
Do that for each quadratic term and then assess which terms cancel each other. It's a lot of pencil work, but it's fairly simple.
There are four binomials to factorise.
c^2 - c - 2 = (c - 2)(c + 1)
c^2 - 1 = (c - 1)(c + 1)
c^2 + 3c + 2 = (c + 2)(c + 1)
c^2 - 3c + 2 = (c - 2)(c - 1)
------- The whole thing now looks like this:
(((c - 2)(c + 1)) / ((c - 1)(c + 1))) / (((c + 2)(c + 1)) / ((c - 2)(c - 1)))
divide fractions by multiplying the inverse
(((c - 2)(c + 1)) / ((c - 1)(c + 1))) * (((c - 2)(c - 1)) / ((c + 2)(c + 1)))
((c - 2)(c + 1)(c - 2)(c - 1)) / ((c - 1)(c + 1)(c + 2)(c + 1))
------ (c - 1)(c + 1) cancels
((c - 2)(c - 2)) / ((c + 2)(c + 1))
(c - 2)^2 / ((c + 2)(c + 1)) .
It's answer C). Of course!! (bit of sick humour, isn't it? ;)
