# How to solve this? (adding polynomial fractions)

I'm having trouble solving this expression: $$\frac{(x - 1)(7x + 6)}{(x - 1)(x + 1)^2 }-\frac{ 7}{ (x + 1)}$$

What's the steps to solve this?

I know you expand $(x + 1)^2$ to $(x + 1)(x + 1)$,

and that you need to find a common denominator before adding the numerators together.

The final answer is $\quad\displaystyle\frac{-1}{x^2 + 2x + 1}.$

Thanks.

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I think someone got the problem wrong: looks like it should be a $-$, not a $+$. –  Ron Gordon Feb 7 '13 at 4:39

We want to simplify the following:

$$\frac{(x-1)(7x + 6)}{(x-1)(x+1)^2} - \frac {7}{(x+1)}$$

First, we cancel the common factor $(x-1)$ in the left-hand fraction:

$$\frac{(x-1)(7x + 6)}{(x-1)(x+1)^2} - \frac {7}{(x+1)} = \frac{(7x + 6)}{(x+1)^2} - \frac {7}{(x+1)}$$

Now we find a common denominator to add the fractions, and see that we want both denominators to be $(x+1)^2$. To accomplish this, we can multiply the numerator and the denominator of the second fraction by the factor of $(x+1)$:

$$\frac{(7x + 6)}{(x+1)^2} - \frac{7}{(x+1)} \cdot \frac{(x+1)}{(x+1)} = \frac{(7x + 6)}{(x+1)^2} - \frac {7(x+1)}{(x+1)^2}$$ $$= \frac{(7x + 6 - 7(x+1))}{(x+1)^2} \;=\;\frac{ -1}{(x+1)^2}$$ $$\;=\;\frac{-1}{x^2 + 2x + 1}$$

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How did you get (x+1)^2 on the right hand side? –  Cypras Feb 7 '13 at 4:46
Because both denominators: $(x+1)$ and $(x+1)^2$ - divide $(x+1)^2$. Notice that $\dfrac{7(x+1)}{(x+1)^2} = \dfrac{7}{(x+1)}$ –  amWhy Feb 7 '13 at 4:47
Ohh okay. Sort of starting to understand. –  Cypras Feb 7 '13 at 4:50
Why not by (x+1)^2? –  Cypras Feb 7 '13 at 4:53
Yeah I see you times by (x+1) to the right hand fraction, but shouldn't it be (x+1)^2, you times 7 by (x+1) not (x+1)^2, I don't understand why. –  Cypras Feb 7 '13 at 5:00

If x$\neq$1, you can cancel $(x-1)$ in the numerator and denominator of $\frac{(x - 1)(7x + 6)}{(x - 1)(x + 1)^2 }$ so, $$\frac{(x - 1)(7x + 6)}{(x - 1)(x + 1)^2 }-\frac{ 7}{ (x + 1)}=\frac{7x + 6}{(x + 1)^2 }-\frac{ 7}{ (x + 1)}$$

Now, if x$\neq$-1 you can multiply and divide $\frac{7}{x + 1}$ by $(x+1)$ and add the fractions:

$$\frac{7x + 6}{(x + 1)^2 }-\frac{ 7}{ (x + 1)}=\frac{7x + 6}{(x + 1)^2 }-\frac{7(x+1)}{ (x + 1)^2}=\frac{7x + 6-7(x+1)}{(x + 1)^2}=\frac{-1}{(x + 1)^2}$$

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Why do you not times the right fraction by (x+1)^2? –  Cypras Feb 7 '13 at 4:57
@Cypras, because i need have a denominator in the two fractions such that both are equal, this, because you know that if i have two fractions $\frac{A}{B}$ and $\frac{C}{B}$ then, the sum of the 2 fractions is $\frac{A+C}{B}$ and so it is easier to add, (it is equivalent to searching the common denominator, which is the same thing amWhy said) –  dwarandae Feb 7 '13 at 5:06
@amWhy, sorry, I started writing before, but when I published the answer I had not yet seen another answer. –  dwarandae Feb 7 '13 at 5:09