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