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I'm self-studying. I have this problem I can't wrap my head around.

$\frac{x}{x+1} + \frac{4}{1-x} + \frac{x^2-5x-8}{x^2-1}$

The answer is $\frac{2\left(x-6\right)}{x-1}$

How do I get that answer? I keep getting $2(x^2-5x-2)$...

This is how I solved it:

$\frac{x}{x+1}-\frac{4}{x+1}+\frac{x^2-5x-8}{x^2-1}$

$=\frac{x-4+x^2-5x-8}{(x+1)(x-1)}$

$=\frac{x(x-1)-4(x-1)+x^2-5x-8}{(x+1)(x-1)}$

= $2(x^2−5x−2)$

I see the problem now!

$\frac{4}{1-x}$ = $\frac{4}{x-1}$ ...not $\frac{4}{x+1}$

because when I multiply the denominator by -1, the result is... $-(1-x) = -1+x = x-1$

_thus_$\,\, \frac{4}{1-x}$ _becomes_$ \frac{4}{x-1}$

and then I changed the sign of the fraction from plus to minus $-\frac{4}{x-1}$ to keep the fraction equal to the original.

Solved!

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  • $\begingroup$ How do you get $2(x^2-5x-2)$? $\endgroup$ – evil999man May 10 '14 at 16:17
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    $\begingroup$ You didn't obtain the correct numerator when you merged the fractions into one using the common denominator. You need $x(x-1)+ 4(x+1)+ x^2 - 5x + 8$ in the numerator. Please note that you are adding three rational functions, at the start of your post. $\endgroup$ – amWhy May 10 '14 at 16:30
  • $\begingroup$ Please look at the second piece. It is not the same at the beginning and in the body $4/(1-x)$ became $4/(1+x)$. $\endgroup$ – Claude Leibovici May 10 '14 at 16:32
  • $\begingroup$ Don't I need to multiply the denominator by -1 and change the fraction to a negative from $\frac{4}{1-x}$ to $-\frac{4}{x+1}$ ? $\endgroup$ – prexcel2215 May 10 '14 at 16:34
  • $\begingroup$ No, if your original post is what you intended. $\endgroup$ – amWhy May 10 '14 at 16:35
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Find the common denominator, and simplify. (You'll be able to utilize the fact that $x^2-1$ is a difference of squares: $$x^2 - 1= (x+1)(x-1)$$

$$\begin{align} \frac{x}{x+1} + \frac{4}{1-x} + \frac{x^2-5x-8}{x^2-1}& = \dfrac {x}{x+1} -\frac{4}{x-1} + \frac{x^2 - 5x-8}{x^2-1}\\ \\& =\dfrac{x(x-1)-4(x+1) +x^2 - 5x -8}{x^2-1}\\ \\ &= \dfrac{x^2 - x + -4x - 4 + x^2 +x^2 -5x - 8}{x^2-1} \\ \\ &= \dfrac{2x^2 -10x -12}{(x^2-1)}\\ \\ &= \dfrac{2(x^2 - 5x-6)}{(x+1)(x-1)}\\ \\ &= \dfrac{2(x+1)(x-6)}{(x + 1)(x-1)}\\ \\ & = \dfrac{2(x-6)}{x-1}, \quad x\neq -1\end{align}$$

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  • $\begingroup$ $\frac{x}{x+1}-\frac{4}{x+1}+\frac{x^2-5x-8}{x^2-1}$<br> $\frac{x-4+x^2-5x-8}{(x+1)(x-1)}$<br> $\frac{x(x-1)-4(x-1)+x^2-5x-8}{(x+1)(x-1)}<br> = $2(x2−5x−2)$ $\endgroup$ – prexcel2215 May 10 '14 at 16:24
  • $\begingroup$ Nice amWhy! This equality given in the the difference of squares we call it in the French language remarkable product. Do you have a similar appellation in English? $\endgroup$ – user63181 May 12 '14 at 17:02
  • $\begingroup$ I like that! (No, I don't think we have much of anything except "difference of squares".) $\endgroup$ – amWhy May 12 '14 at 17:04

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