The Factor and Remainder Theorems

The exact moments a polynomial hits zero often carry real meaning — the instant a projectile lands, the price where a business breaks even, the setting where a design sits in perfect balance. Finding those points means factorising, and factorising a stubborn cubic starts with one quick test.

Suppose someone hands you the cubic f(x) = 2x^3 - 3x^2 + x - 5 and asks: what's the remainder when you divide it by (x - 2)? You could set up the whole polynomial long division — several lines of careful bookkeeping — just to read off one number at the bottom.

Or you could use a shortcut so good it feels like cheating. The remainder theorem says that remainder is simply f(2):

f(2) = 2(8) - 3(4) + 2 - 5 = 16 - 12 + 2 - 5 = 1.

One substitution, done. And here's the real prize: if that value ever comes out zero, then (x - 2) divides in perfectly — it's a factor. That single observation, the factor theorem, is the key that cracks open cubics and quartics that would otherwise refuse to factorise.

The two theorems

Why do they hold? Write the division as f(x) = (x - a)\,q(x) + r, where q(x) is the quotient and r the (constant) remainder. Substitute x = a: the (x - a) term collapses to 0, so f(a) = r. The remainder is the value of the polynomial at x = a — and it is nothing left over precisely when f(a) = 0. The factor theorem is just the remainder theorem with the remainder switched off.

See it built

Step through evaluating f(a) for a chosen quadratic and a chosen a. The value you land on is exactly the remainder — and when it hits zero, you have found a factor.

Worked example — test a factor

Is (x - 1) a factor of f(x) = x^3 - 6x^2 + 11x - 6? By the factor theorem we don't divide — we just check whether f(1) = 0:

f(1) = 1 - 6 + 11 - 6 = 0.

It's zero, so (x - 1) is a factor. If instead we try (x - 4), we get f(4) = 64 - 96 + 44 - 6 = 6 \neq 0, so (x - 4) is not a factor — and, as a bonus, we now also know that dividing by (x - 4) would leave remainder 6.

Worked example — factorise a cubic completely

Now use the factor we found to crack the whole cubic f(x) = x^3 - 6x^2 + 11x - 6. We know (x - 1) is a factor, so divide it out:

x^3 - 6x^2 + 11x - 6 = (x - 1)(x^2 - 5x + 6).

The leftover quadratic is easy — x^2 - 5x + 6 = (x - 2)(x - 3) — so the cubic splits completely into

f(x) = (x - 1)(x - 2)(x - 3),

with roots x = 1, 2, 3. One clever guess plus one easy quadratic beat a mountain of algebra. That is the whole strategy for higher-degree polynomials: find one root by the factor theorem, divide it out, and repeat on what's left until only a quadratic (or simpler) remains. It is exactly what powers solving by factorising beyond degree two.

The single most common mistake is the sign of the value you plug in. To test the factor (x - a) you evaluate f(a) — the value that makes the bracket zero. So for the factor (x + 3), the bracket is zero when x + 3 = 0, i.e. x = -3 — you must evaluate f(-3), not f(3). The sign flips. A quick rule: rewrite the factor as (x - a) first — (x + 3) = (x - (-3)) — and the correct a = -3 stares right back at you. Get this sign wrong and a genuine factor will look like a non-factor, and vice versa.

Because the factor theorem tells you exactly where to dig. Any nice whole-number root of a polynomial has to divide the constant term (the rational root idea), so you don't guess blindly — you try the small factors of the constant: \pm 1, \pm 2, \pm 3, \dots For x^3 - 6x^2 + 11x - 6 the constant is -6, so the sensible values to test are \pm 1, \pm 2, \pm 3, \pm 6 — and the very first, x = 1, strikes gold. A handful of substitutions, and the buried factor is yours.

This treasure hunt is the practical engine behind solving cubic and quartic equations. And it gestures at something deep: the fundamental theorem of algebra promises that a polynomial of degree n has exactly n roots — provided you're willing to count repeats and to step beyond the real number line into the complex numbers. Every polynomial factors completely; the factor theorem is how you start prising it apart.

See it explained

Sal Khan introduces the polynomial remainder theorem and shows why f(a) is the remainder.