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Hello everyone I have the following question.

I have the following fraction

$f(x)=-\frac{4}{x^2}+\frac{1}{(x-1)^2}$

But how would I reduce it I know I have to use the multiply the opposite numerator by denominator and I got.

$(x-1)^2(4)=4x^2-8x+4$

$1(x^2)=x^2$

So I got got $f(x)=-\frac{4x^2-8x+4+x^2}{x^2(x-1)^2}$

But this is incorrect.What am I doing wrong.

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Yes sorry I forgot. –  Fernando Martinez May 13 '13 at 22:48

2 Answers 2

up vote 2 down vote accepted

You are close. Your mistake was applying the "-" to both terms instead of just one.

You wrote $f(x)=-\frac{4x^2-8x+4+x^2}{x(x-1)^2}$.

What you should have is $f(x)=\frac{-(4x^2-8x+4)+x^2}{x(x-1)^2} =\frac{-4x^2+8x-4+x^2}{x(x-1)^2} =\frac{-3x^2+8x-4}{x(x-1)^2} $.

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Yes that makes sense now I have to find what x could be... –  Fernando Martinez May 13 '13 at 22:51
    
Thanks for the help I see now $(-3x+2)(1x-2)$ –  Fernando Martinez May 13 '13 at 22:54

Already answered, but I would like to point out the reduction:

$$f(x) = -{(3 x-2)(x-2)\over x^2 (x-1)^2}$$

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