# Thread: Another basic factoring question

1. ## Another basic factoring question

If im presented with
$\displaystyle \frac{(x+1)(x-3)}{(x+5)(x+1)}$we can reduce that to $\displaystyle \frac{x-3}{x+5}$ canceling out the x+1 from the numerator/ denominator

however if we were to take this same problem without it being joined by multiplication the above would not be possible e.g.

$\displaystyle \frac{(x+1)+(x-3)}{(x+5)(x+1)}$ then cancellation is no longer possible

instead here we have the following after combing like terms in the numerator
$\displaystyle \frac{(x+1)+(x-3)}{(x+5)(x+1)} =\frac{2x-2}{(x+5)(x+1)}$

correct?

2. Originally Posted by allyourbass2212
If im presented with
$\displaystyle \frac{(x+1)(x-3)}{(x+5)(x+1)}$we can reduce that to $\displaystyle \frac{x-3}{x+5}$ canceling out the x+1 from the numerator/ denominator

however if we were to take this same problem without it being joined by multiplication the above would not be possible e.g.

$\displaystyle \frac{(x+1)+(x-3)}{(x+5)(x+1)}$ then cancellation is no longer possible

instead here we have the following after combing like terms in the numerator
$\displaystyle \frac{(x+1)+(x-3)}{(x+5)(x+1)} =\frac{2x-2}{(x+5)(x+1)}$

correct?
yes

3. Yes but you can factor further....