Factoring third degree polynomials using long division

Your first term, upon division: $x^2$ is correct $\large\checkmark$ (in the quotient), leaving

$$ x - 2 | -x^2 - 4x + 12\quad\large\checkmark$$

I know that the leading term goes into the inner leading term $-x$ times or however you say that.

$$ x - 2 | - 2x + 12 \quad \longleftarrow \text{error}$$

Here you subtracted incorrectly: We should have $x^2 - x$ in the quotient, that's correct, but multiplying $-x(x - 2) = -x^2 + 2x$

So when we subtract, we subtract, from $(-x^2 - 4x + 12) - (-x^2 + 2x) = -6x + 12$.

Now, we have $$ x -2 \mid -6x + 12$$ and so our ongoing quotient becomes $x^2 - x {\bf - 6}$

which, leaves a zero remainder since $-6(x-2) = -6x + 12$, as desired.

So...we have that $$\frac{x^3 - 3x^2 -4x + 12}{x - 2} = x^2 - x - 6$$

Or, that is, $$x^3 - 3x^2 - 4x + 12 = (x -2)(x^2 - x - 6) = (x-2)(x +2)(x - 3)$$