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MasterMath

Synthetic Division Calculator

Divide a polynomial by (x − r) in a single row of arithmetic. The remainder is the value of the polynomial at r, which is the remainder theorem in action.

Quotient

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Quotient—
Remainder—
Is it a root of the polynomial?—
Degree of the quotient—

The synthetic division table

CoefficientRunning totalWhat passes to the next

How it was solved

    The formula

    P(x) ÷ (x − r) = Q(x) + R, with R = P(r)

    Why it works

    Synthetic division — Ruffini's rule — is long division stripped to the coefficients. Bring down the first one, multiply by r, add to the next, and repeat. What you get is the quotient's coefficients and, at the end, the remainder.

    How to solve it by hand

    1. Write the coefficients in order, skipping none — use 0 for missing powers
    2. Bring down the first coefficient unchanged
    3. Multiply it by r and add the result to the next coefficient
    4. Repeat to the end: the last number is the remainder

    What is worth knowing

    The remainder theorem is what makes this more than a shortcut: dividing P(x) by (x − r) always leaves P(r) as the remainder, so a remainder of zero proves r is a root. That gives you a fast root test and, when it works, a factorisation — the polynomial becomes (x − r) × quotient, and the quotient is one degree lower. Repeating the process is the standard way to factor a cubic or quartic by hand. The one thing that trips people up is missing powers: x³ + 1 has coefficients 1, 0, 0, 1, and forgetting the zeros gives nonsense.

    Frequently asked questions

    What is the remainder theorem?

    Dividing a polynomial by (x − r) leaves P(r) as the remainder. A remainder of zero proves r is a root.

    What if a power is missing?

    Write a zero for it. For x³ + 1 the coefficients are 1, 0, 0, 1, and skipping the zeros gives the wrong answer.

    Can I divide by something other than x − r?

    This form handles linear divisors. For higher-degree divisors you need full polynomial long division.

    How do I find candidate roots?

    The rational root theorem: try the divisors of the constant term over the divisors of the leading coefficient.