Here is the equation: $$\frac{-3(x+h)^2-6(x+h)+4+3x^2+6x-4}{h}$$
I am getting the wrong answer every time. \begin{align*} f & = \frac{-3(x^2+3xh+h^2)-6x+6h+4+3x^2+6x-4}{h}\\ f & = \frac{-3x^2-6xh-3h^2-6x+6h+4+3x^2+6x-4}{h}\\ f & = \frac{-6xh-3h^2}{h}\\ f & = 3(-2x-h^2) \end{align*}
Two questions, what am I doing wrong when factoring down this problem and where is the best place to really learn factoring super well?
it looks like you are differentiating $-(3x^2 + 6x-4)$
$$ -3(x+h)^2 -6(x+h)+4 = -3x^2 -6xh-3h^2 -6x -6h+4 $$
from which $-3x^2 -6x +4 $ cancel out with $3x^2 +6x -4$
$$ -6xh-3h^2-6h$$
you are missing the $-6h$ term,
and $\frac{-6xh-3h^2-6h}{h}=-6x-3h-6= -3(2x+h+2)$
Furthermore, taking the limit as $h \rightarrow 0$, $-3(2x+2)=-6x-6$.
Check:
$\frac{d}{dx}[-3x^2-6x+4] = (-3)\cdot 2 x -6 +0= -6x-6$.
Included the limit and derivative comments as a bonus, you tagged the question as pre-calculus, I’m not sure what level you’re at, if they confuse you for the moment, you can disregard them.