Polynomial remainder theorem

Statement

Formally: for any polynomial P(x) with coefficients in a field (such as the real or complex numbers) and any element a of that field, the remainder of dividing P(x) by x − a is the constant P(a). Because x − a has degree 1, the remainder must be a degree‑0 polynomial (a constant), and the theorem identifies that constant explicitly.

Proof idea

Write P(x) = (x − a)Q(x) + r for some quotient Q(x) and constant r. Substitute x = a to obtain P(a) = r, because (a − a)Q(a) = 0. This short argument shows both existence of the remainder as a constant and its equality to P(a). The argument works over any commutative ring where substitution and division by monic linear polynomials are defined.

Examples

• Example 1: P(x) = x^3 − 2x + 5 divided by x − 1. Evaluate P(1) = 1 − 2 + 5 = 4, so the remainder is 4.
• Example 2: If P(2) = 0, then x − 2 divides P(x) exactly; this is the factor theorem, an immediate corollary.

Uses and importance

The theorem is used to test whether a given value is a root (P(a) = 0), to compute remainders quickly without performing full polynomial long division, and to facilitate methods such as synthetic division. It also appears in modular arithmetic with polynomials and in algorithms for polynomial factorization and interpolation.

Remarks and extensions

Key points to remember:

1. The divisor must be linear of the form x − a; for higher-degree divisors the remainder is generally a polynomial of lower degree than the divisor.
2. The same reasoning establishes the factor theorem: P(a) = 0 iff x − a is a factor of P(x).
3. The statement extends to polynomials over arbitrary commutative rings, though care is needed when substitution and division properties differ from fields.

For practical computation, synthetic division offers a compact implementation of the theorem and of obtaining the full quotient when needed.