reduce a rational expression to lowest terms $$\frac{x^2 + 4x + 3}{x^2 - 2x - 3}$$
I'm coming up with $\frac{x  + 3}{ x - 3}$ however it seems wrong.
$$x^2 + 4x + 3 = (x + 3) ( x + 1)$$
$$x^2 - 2x - 3 = (x - 3 )(x + 1) $$
cancel out $x + 1$  and left with $\frac{x + 3}{x - 3}$

It's correct, 
$$\require{cancel} \frac{x^2 + 4x + 3} {x^2 - 2x - 3} = \frac{(x+3)\cancel{(x+1)}} {(x-3) \cancel{(x+1)}}$$
Provided $x \neq 3$, $x \neq -1$. 
