Posted by anonymous on Saturday, September 22, 2007 at 2:39am.
yes, the z is a typo
Factor 1/(x+2) out of each term and see what you get.
[1/(x+2)][12/(x-2) - (3-x)/x]
=[1/(x+2)][12x/x(x-2) - (3-x)(x-2)/x(x-2)]
={1/[(x-2)(x+2)]}[12x/x -(3-x)(x-2)/x]
={1/[(x-2)(x+2)]}[(x^2+ 7x +6)/x]
That is in the form of your answer (except for the typo "z" for 2), but it can be factored further.
= (x+6)(x+1)/[x(x-2)(x+2)]
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