I am sure there is a better strategy that someone smarter than me would use, but I am not that person.
I am trying to factor
$$x^2 – 3x^2 – 4x + 12$$
I don't know how so I attempt to guess with long division. I cheat and look at the answer so find out one of the factors to save myself time, so I try x – 2
I am not sure how to type out long division but I get
$$ x – 2 | x^3 – 3x^2 – 4x + 12$$
so I know I can have an $x^2$ for how many times x goes into the leading term.
Subract it all and I am left with
$$ x – 2 | -x^2 – 4x + 12$$
I know that the leading term goes into the inner leading term $-x$ times or however you say that.
$$ x – 2 | – 2x + 12$$
Now -2
$$ x – 2 | 8$$
Now I don't know what to do, how did this go so wrong? I have $x^2 – x -2$ ontop and I have a remainder of 8. This can't be right, I cheated so I know that this should be a factor.
Best Answer
Your first term, upon division: $x^2$ is correct $\large\checkmark$ (in the quotient), leaving
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, $$x^3 - 3x^2 - 4x + 12 = (x -2)(x^2 - x - 6) = (x-2)(x +2)(x - 3)$$