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Remainder theorem

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The Remainder Theorem

The remainder theorem says that dividing a polynomial f(x) by (x-a) leaves a remainder equal to f(a). Learn why this shortcut works, see it verified against synthetic division, and learn its special case: when f(a)=0, (x-a) is a factor (the factor theorem).

What the remainder theorem says

The remainder theorem is a shortcut: to find the remainder when a polynomial f(x) is divided by (x − a), you don't have to divide at all — just evaluate f(a). Whatever number comes out is the remainder.

Remainder theorem verified by synthetic division f(x) = x^3 - 2x^2 + 3x - 5. Evaluating f(2) directly gives 1. Synthetic division of f(x) by (x-2) gives quotient coefficients 1, 0, 3 and remainder 1 -- matching f(2) exactly. f(x) = x³ − 2x² + 3x − 5 Divide by (x − 2): bring down 1, multiply and add across 2 | 1−23−5 103 1 1·2=2, −2+2=00·2=0, 3+0=3 remainder = 1 = f(2)
Evaluating f(2) directly gives 1; synthetic division of f(x) by (x−2) gives the same remainder, 1.

Dividing f(x) by (x − a) always produces a quotient q(x) and a remainder r, so f(x) = (x−a)·q(x) + r. Substituting x = a makes the first term zero, since (a−a) = 0, leaving f(a) = r. The remainder is just the function evaluated at the divisor's root.

For f(x) = x³ − 2x² + 3x − 5, dividing by (x − 2): evaluate f(2) = 2³ − 2(2²) + 3(2) − 5 = 8 − 8 + 6 − 5 = 1. Long division (or synthetic division) of the same polynomial by (x−2) confirms the remainder is exactly 1.

When f(a) = 0, the remainder is zero, which means (x − a) divides evenly into f(x) with no remainder — so a is a root and (x−a) is a factor. That special case is the factor theorem, and it's the fastest way to test whether a specific number is a root of a polynomial before you commit to factoring or solving it.

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