You divide P(x) = x^3 - 6x^2 + 11x - 6 by (x - 3) using long division. After three rounds of subtraction the remainder comes out 0. Clean. No leftover number.

Now the question — what does that tell you? About P, and about the number 3?

It tells you two things at once, and they are really the same thing. The number 3 is a root of P: plugging x = 3 into P gives zero. And (x - 3) is a factor of P: it divides P exactly, leaving nothing behind. This is the Factor Theorem, and it is the single most-used fact in polynomial root-finding.

The Remainder Theorem, stated

Before the factor version, one step back. When you divide any polynomial P(x) by the linear divisor (x - a), the remainder is not some leftover polynomial — it is a single number, and that number is exactly P(a).

When P(x) is divided by (x - a), the remainder is P(a).

Not "a constant we can compute by continuing the division." The remainder equals the value of P evaluated at x = a. You can find it without doing any long division at all — just plug in.

Derivation

The polynomial division algorithm gives you the identity

P(x) = (x - a) \cdot Q(x) + R

where Q(x) is the quotient and R is the remainder. Because the divisor (x - a) has degree 1, the remainder must have degree strictly less than 1 — so R is a constant, a single number, not a polynomial in x.

Now substitute x = a into that identity:

P(a) = (a - a) \cdot Q(a) + R = 0 \cdot Q(a) + R = R

The first term vanishes because (a - a) = 0. Whatever Q(a) is, it gets multiplied by zero and drops out. What is left is R.

So R = P(a). The remainder equals the evaluation. That is the whole proof.

What "remainder = 0" means

Now tighten the screws. Suppose the division leaves remainder zero — R = 0. Four things follow, and they all say the same thing from different angles:

These are four vocabularies for one fact. "Root," "zero," "(x - a) divides exactly," and "(x - a) is a factor" are synonyms in this context. If a problem hands you any one of them, you get the other three for free.

Factor Theorem, stated

Wrapping those two implications together gives the Factor Theorem. This is the if-and-only-if form:

(x - a) is a factor of P(x) if and only if P(a) = 0.

The "if and only if" is doing real work — both directions are true.

Every root corresponds to a linear factor, and every linear factor corresponds to a root. The sign flips — root a gives factor (x - a), not (x + a) — but the correspondence is exact.

Worked example: is 1 a root of x^3 - 6x^2 + 11x - 6?

Take P(x) = x^3 - 6x^2 + 11x - 6. Is 1 a root?

Do not divide. Just substitute.

P(1) = 1 - 6 + 11 - 6 = 0 \; \checkmark

Yes. So by the Factor Theorem, (x - 1) is a factor of P. Divide to find the other factor:

P(x) \div (x - 1) = x^2 - 5x + 6

The quadratic x^2 - 5x + 6 factors as (x - 2)(x - 3). Putting it all together:

P(x) = (x - 1)(x - 2)(x - 3)

All three roots exposed: 1, 2, 3. Each root contributes one linear factor, exactly as the factor theorem in reverse would build it.

Using the Factor Theorem to find roots

Here is the general recipe for factoring a polynomial you cannot crack by inspection:

  1. Try small integer candidates. Start with \pm 1, \pm 2, \pm 3, and the small factors of the constant term.
  2. For each candidate a, compute P(a). This is just arithmetic — no division needed.
  3. If P(a) = 0, then (x - a) is a factor. Divide P by (x - a) to get the quotient Q(x).
  4. Repeat on Q(x). Look for roots of Q, peel off another linear factor, continue until the remaining polynomial is a quadratic (which you can solve directly) or is irreducible.

Each root found reduces the degree by one. A cubic becomes a quadratic after one root; a quartic becomes a cubic. The algorithm walks down the degree ladder one rung at a time.

Rational Root Theorem

"Try small integers" is fine for textbook cubics, but which values should you try in general? The Rational Root Theorem formalises the answer.

For a polynomial a_n x^n + a_{n-1} x^{n-1} + \cdots + a_0 with integer coefficients, any rational root \tfrac{p}{q} in lowest terms must satisfy:

So the candidates are (factors of a_0) over (factors of a_n). For P(x) = x^3 - 6x^2 + 11x - 6, a_0 = -6 and a_n = 1, so the rational-root candidates are \pm 1, \pm 2, \pm 3, \pm 6 — the factors of 6. The roots 1, 2, 3 are all on that list.

This narrows the search from "infinitely many real numbers" to "a handful of rationals." If none of the candidates is a root, the polynomial has no rational roots at all — only irrational or complex ones.

Worked example: P(x) = 2x^3 - 3x^2 - 11x + 6

Here a_0 = 6 and a_n = 2. Factors of 6: \pm 1, \pm 2, \pm 3, \pm 6. Factors of 2: \pm 1, \pm 2. Rational-root candidates:

\pm 1, \; \pm 2, \; \pm 3, \; \pm 6, \; \pm \tfrac{1}{2}, \; \pm \tfrac{3}{2}

Work through the list. Try x = 3:

P(3) = 2(27) - 3(9) - 11(3) + 6 = 54 - 27 - 33 + 6 = 0 \; \checkmark

So (x - 3) is a factor. Divide:

\frac{2x^3 - 3x^2 - 11x + 6}{x - 3} = 2x^2 + 3x - 2

The quadratic 2x^2 + 3x - 2 factors as (2x - 1)(x + 2), giving roots \tfrac{1}{2} and -2. All three roots:

x = 3, \; \tfrac{1}{2}, \; -2

And \tfrac{1}{2} was on the candidate list — the Rational Root Theorem told you to look there.

If you don't find a root easily

Sometimes the candidate list runs out with no hits. That does not mean the polynomial has no roots — it means it has no rational roots. The polynomial may have only irrational roots (like \sqrt{2}), or only complex roots (like i), or a mix.

At that point the Factor Theorem has no more easy moves. You reach for numerical methods — the Newton-Raphson iteration, the bisection method, or approximate graphical solving. For cubics, Cardano's formula gives exact roots in radicals. For higher degrees, exact formulas exist up to degree 4 and no further (that is Abel's theorem — no general formula in radicals for degree \geq 5).

Checking a claim — substitute, don't divide

A teacher asks: "Is 5 a root of x^4 - 6x^3 + 2x^2 - x + 30?" The textbook approach would be to divide by (x - 5) and check whether the remainder is zero. That is a lot of arithmetic.

The Remainder Theorem gives you the shortcut. Just substitute:

P(5) = 625 - 750 + 50 - 5 + 30 = -50

Not zero, so 5 is not a root, and (x - 5) is not a factor. Done in three lines. The theorem guarantees that substitution and division-then-remainder give the same answer — so always substitute when you only need the remainder. Only actually divide when you need the quotient as well.

Use in equation-solving

The whole point of factoring a polynomial is to solve the equation P(x) = 0. Once P is in factored form

P(x) = a(x - r_1)(x - r_2) \cdots (x - r_n),

a product is zero if and only if one of the factors is zero, so the roots are exactly r_1, r_2, \dots, r_n. Reading roots off a factored polynomial is immediate.

The Factor Theorem is the engine that takes you from unfactored to factored form:

  1. Find one root a (by inspection, rational-root candidates, or a lucky substitution).
  2. Divide by (x - a) to reduce the degree.
  3. Repeat on the quotient.

Each rung of the ladder exposes one root. After n rungs, a degree-n polynomial is fully factored and all its roots are in hand.

Recognition drill

For each, check whether the given value is a root by direct substitution. No division.

Each check is one substitution and a line of arithmetic. Faster than any division.

The takeaway

"Remainder is zero after dividing by (x - a)" is the Factor Theorem in action. It tells you three things packaged as one: a is a root, (x - a) is a factor, and you can divide out (x - a) to reduce the degree and continue factoring.

This is the fundamental trick of polynomial root-finding — find a root by substitution, peel off the corresponding linear factor by division, and repeat on a smaller polynomial. Every factoring recipe, every rational-root hunt, every equation-solving procedure for polynomials comes back to this one fact. Zero remainder means root. Root means factor. Factor means you can divide and keep going.