Subtracting $\frac{(x+3)}{(x^2-1)} - \frac{(x-2)}{(x^2+2x+1)}$ $\frac{(x+3)}{(x^2-1)} - \frac{(x-2)}{(x^2+2x+1)}$
To solve the problem I first dissembled the equation on the denominator
 $ \frac{(x+3)}{(x-1)*(x+1)} - \frac{(x-2)}{(x+1)^2}$
I multiplied the denominator together and to do this, I think I have to multiply the top part as well right? This is where i get confused, I forgot how to do this problem as its been a long time. How would i go on to solve this? 
 A: $$ \frac{x+3}{(x-1)(x+1)} - \frac{x-2}{(x+1)^2}=\frac{(x+3)\color{red}{(x+1)}-(x-2)\color{red}{(x-1)}}{(x-1)(x+1)^2}$$
A: $a=(x^2-1)=(x+1)(x-1)$
$b=(x^2+2x+1)=(x+1)^2=(x+1)(x+1)$
So, we have:
$$\frac{(x+3)}{a}-\frac{(x-2)}{b}$$
$$\frac{b(x+3)-a(x-2)}{ab}$$
$ab=(x+1)^3(x-1)$
$b(x+3)=(x+3)(x+1)(x+1)$
$a(x-2)=(x-2)(x+1)(x-1)$
Then,
$$\frac{(x+3)(x+1){(x+1)}-(x-2){(x+1)}(x-1)}{{(x+1)^3}(x-1)}$$
$$\frac{{(x+1)}[(x+3)(x+1)-(x-2)(x-1)]}{{(x+1)^{3}}(x-1)}$$
$$\frac{(x+3)(x+1)-(x-2)(x-1)}{(x+1)^2(x-1)}$$
Expanding:
$$\frac{7x+1}{x^3+x^2-x-1}$$
A: Let's start here and simplify the problem.
$$\frac{(x+3)}{(x-1)(x+1)} - \frac{(x-2)}{(x+1)(x+1)}$$
I'm going to call $(x+1)=a$ and $(x-1)=b$
$$\frac{(x+3)}{(b)(a)} - \frac{(x-2)}{(a)(a)}$$
So, on the right side we have (a)(a) in the denominator, and only one in the denominator on the right. So we have to multiply the term on the right by $\frac{a}{a}$.
The term on the left has (b) in the denominator, while the one on the right doesn't. So we're going to to multiply the one on the right by $\frac{b}{b}$.
Let's take those steps, and we get:
$$\frac{(x+3)(a)}{(b)(a)(a)} - \frac{(x-2)(b)}{(a)(a)(b)}$$
See how both denominators are identical!? Great! Now we can put both of the numerators on top of the denominator we worked so hard to get.
$$\frac{(x+3)(a)-(x-2)(b)}{(b)(a)(a)}$$
Now all you have to do is plug back in $a=(x+1)$ and $b=(x-1)$, open up the brackets and collect like terms.
A: The common factor can be taken to front as coefficient in the numerator as well as in the denominator.
$$ \frac{x+3}{(x-1)(x+1)} - \frac{x-2}{(x+1)^2}=\frac{1}{(x+1)} \cdot [\frac{x+3}{(x-1)} - \frac{x-2}{(x+1)}] $$
$$=\frac{1}{(x+1)} \cdot \frac{(x^2+4 x+ 3) - (x^2- 3 x + 2)}  {(x^2-1)} $$
$$=\frac{1}{(x+1)}  \frac{(7 x + 1)}  {(x^2-1)}. $$
