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

Question

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 multiply the opposite numerator by denominator and I got: \begin{gather*} (x-1)^2(4)=4x^2-8x+4\\ 1(x^2)=x^2 \end{gather*}

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

But this is incorrect. What am I doing wrong?

Answer 1

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} $.

Answer 2

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

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