Factoring third degree polynomials using long division

Question

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.

Answer 1

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

Answer 2

When you subtracted $(-x)(x-2)$ you should have gotten $-6x+12$. You dropped a sign there.

Answer 3

As Andrew Salmon's comment points out, the easiest thing to try first to factor a cubic polynomial is to try to pair off the terms and see if they share a common factor. If paired off in the order in which they are written, this becomes

$$x^2(x-3)-4(x-3)$$

or if the odd and even exponents are paired together

$$x(x^2-4)-3(x^2-4)$$

Then simply take out the common factor.

Of course, this won't always work, when it doesn't, the method of guessing roots is known as the rational root theorem. First you'll want to try plugging in factors of the constant term to see if you get zero. $\pm1$ is the easiest to check, then in this case you'd also have $\pm2,\pm3,\pm4,\pm6$ and $\pm12$. If those all failed, you'd want to test any of those values divided by a factor of the leading term. However, since your leading term is $1$, there isn't anything extra to test here. If you find one root, you should be able to handle the remaining quadratic. I doubt you'll be expected to factor any cubic polynomials with no rational roots, possibly excepting the form

$$x^3\pm a$$

at which point you could use the formula for the sum or difference of cubes even if $a$ is not a perfect cube.

Finally, although it's not foolproof, if you suspect a polynomial is a multiple of $x-2$, then plugging in $2$ should yield $0$. Since you're dividing by $x-2$, your remainder at each step should also be a multiple of $x-2$. Testing your polynomials at each step reveals

$$2^3-3(2)^2-4(2)+12=8-12-8+12=0$$ $$-(2)^2-4(2)+12=-4-8+12=0$$ $$-2(2)+12=-4+12=8$$

This implies there was an error in the remainder $-2x+12$