Polynomial division

For quadratic equations there is a formula. For $x^3 - 2x^2 - 5x + 6 = 0$ there is none anybody wants to memorise. The way out is a detour: if you find one solution by trying, you can lower the degree by one and end up with a quadratic equation you have long since mastered. The tool for that is polynomial division.

A root means a factor

The whole trick sits in one sentence: if $x = a$ is a root, then the factor $(x - a)$ is contained in the polynomial. For $x^2 - 5x + 6$ you can still see it directly, because it equals $(x - 2)(x - 3)$, and the roots are 2 and 3. Backwards it works just as well: whoever knows a root knows a factor. And a factor can be divided out.

The procedure, step by step

Polynomial division works just like long division, except that you look at the highest power instead of place values. Example: $(x^3 - 2x^2 - 5x + 6) : (x - 1)$. Each time you only ask: what do I have to multiply the divisor by so that exactly the current leading term appears? First question: $x \cdot x^2 = x^3$, so the first part of the answer is $x^2$.

Then you multiply back and subtract: $x^2 \cdot (x - 1) = x^3 - x^2$. Subtracted from $x^3 - 2x^2$ this leaves $-x^2$. Now you bring down the next term and ask again: $x \cdot (-x) = -x^2$, so $-x$ joins the answer. And once more: at the end $-6$ remains, and $x \cdot (-6) = -6x$ fits, so $-6$ is added. The answer is $x^2 - x - 6$, with remainder 0.

Now you are on familiar ground: $x^2 - x - 6 = 0$ gives $x = 3$ and $x = -2$, either by the quadratic formula or by factorising. Together with the root 1 you guessed, the equation therefore has three solutions: $-2$, 1 and 3.

Guessing the first root

Without a first root you cannot start, and you have to find it yourself. Luckily the guessing is not blind: if the polynomial has whole-number coefficients and a 1 in front, then the only possible whole-number roots are divisors of the last term. So for $\dots + 6$ you try $\pm 1, \pm 2, \pm 3, \pm 6$, and usually one of the first two already works.