1. Factoring equations

(12-x-x^2)/(x^2-4+3)

Can someone help me with this?

2. It's not magic. You have to look at it and encourage it to make sense.

The numerator is a little funny, so switch it around so that it looks like you expect:

12-x-x^2 = -x^2 - x + 12 = -(x^2 + x - 12)

That's better. Now it is easy to see that we need 4*(-3) = 12 and 4 + (-3) = +1

The denominator should be an eyeball problem.

(-1)*(-3) = +3 and (-1)+(-3) = -4

3. Once you do what TK said, look for binomials that are (a + b) to cancel.

like this: $\displaystyle \frac{\rlap{---------}(x + 5)(x - 2)}{\rlap{---------}(x + 5)} = x - 2$

but this doesn't cancel: $\displaystyle \frac{4 - {\color {red}{\not x}}}{{\color {red} {\not x}}}$

4. Do i keep the negative 1 factored out?

-1[(x+4)(x-3)
____________
(x-1)(x-3)

I cancel out the (x-3) right?

5. Do i keep the negative 1 factored out?
It doesn't really matter. However, if you do you need not write the 1 and can just say $\displaystyle \frac{-(x+4)}{x-1}$

I cancel out the (x-3) right?
right