Remainder Theorem Calculator
Calculate the remainder of a polynomial division without carrying out the division. Enter f(x) and a linear divisor x − a, and the calculator returns f(a) — the remainder — along with the quotient polynomial and the synthetic division table behind it.
- Synthetic division shown in full
- Runs entirely in your browser
Polynomial Remainder Theorem
Live^ for powers. Fractions like
1/2x^2 are accepted.
x - 3,
x + 2 or
2x - 3. A bare number is read as
a in x − a.
The remainder f(a), the quotient and the factor check.
- Value of a
- —
- Degree of f(x)
- —
The synthetic division, step by step
Every panel updates as you type. The table is the synthetic division you would write by hand, the list narrates each multiplication and addition, and the candidate list applies the Rational Root Theorem to the same polynomial.
Synthetic division table
Bring down, multiply by a, add. The last entry of the bottom row is the remainder.
Narrated steps
Rational root candidates
What is Remainder Theorem Calculator?
A Remainder Theorem Calculator is an online tool that returns the remainder when a polynomial is divided by a linear polynomial. Entering f(x) and the divisor produces the remainder in a single step, since the Remainder Theorem replaces the whole division with one substitution.
The theorem itself, also called little Bézout's theorem, states that dividing a polynomial f(x) by x − a leaves the remainder f(a). Dividing x³ − x² + 2x + 7 by x − 3 leaves 31, and evaluating the same polynomial at x = 3 gives 27 − 9 + 6 + 7 = 31 — the same number, reached without dividing anything.
The reason sits in the division algorithm. A divisor of degree 1 forces a remainder of degree 0, meaning a plain constant:
f(x) = (x − a) · q(x) + r with r a constant set x = a: f(a) = (a − a) · q(a) + r f(a) = 0 · q(a) + r f(a) = rSubstituting a collapses the first term to zero and leaves r = f(a). That single line is the entire theorem, and every result the calculator produces follows from it.
How to Use Remainder Theorem Calculator?
To use the Remainder Theorem Calculator, enter the polynomial, enter the divisor, and read the remainder. The answer and the working appear together in 4 steps:
- Enter the polynomial in the first field, written
the way you would write it:
x^3 - x^2 + 2x + 7. Terms may come in any order, and gaps in the powers are filled with zero coefficients automatically. - Enter the divisor in the second field.
x - 3sets a = 3, andx + 2sets a = −2. - Read the remainder in the headline, with the identity f(x) = (x − a)·q(x) + r written out underneath using your own numbers.
- Open the working panel for the synthetic division table, the narrated steps, and the rational root candidates.
The calculator reports 5 results for every input: the
remainder f(a), the value of a taken from the divisor, the degree of
f(x), the quotient polynomial, and a Factor Theorem verdict saying
whether x − a divides exactly. Coefficients may be whole numbers,
decimals or simple fractions such as
1/2, and powers up to x²⁰ are handled.
Several other things share the name remainder theorem, and this calculator covers 1 of them. The table separates them:
Dividing f(x) by x − a leaves f(a). Algebra, class 9 onward.
Reconstructs a whole number from its remainders against several coprime moduli. Solve a system of congruences.
The division algorithm for integers: a = bq + r. Divide two whole numbers.
Bound the error of a truncated series in Calc 2 and Calc BC. Different theorems, not covered here.
Remainder Theorem Formula
The Remainder Theorem formula is r = f(a), sitting inside the division algorithm for polynomials:
Dividend = (Divisor × Quotient) + Remainder f(x) = (x − a) · q(x) + r and r = f(a) x³ − x² + 2x + 7 = (x − 3)(x² + 2x + 8) + 31Applying the formula takes 3 steps. Set the divisor to zero and solve for x to get a. Substitute a into f(x). Read the value as the remainder.
Synthetic division reaches the same 31 and hands over the quotient at the same time. Write the coefficients in descending order, bring down the first, then multiply by a and add, once per column:
| a = 3 | 1 | −1 | 2 | 7 |
| 3 | 6 | 24 | ||
| 1 | 2 | 8 | 31 |
Bottom row: 1, 2, 8 are the quotient coefficients — q(x) = x² + 2x + 8 — and the final 31 is the remainder.
Three more worked examples, each solved by substitution alone:
| Polynomial p(x) | Divisor | a | Substitution | Remainder |
|---|---|---|---|---|
| 3x⁵ − x⁴ + x³ − 4x² + 2 | x − 1 | 1 | 3 − 1 + 1 − 4 + 2 | 1 |
| x⁵ − 2x⁴ − 3x³ + 2x² + 7 | x − 2 | 2 | 32 − 32 − 24 + 8 + 7 | −9 |
| x² + 2x + 5 | x − 1 | 1 | 1 + 2 + 5 | 8 |
| 3x³ − 2x² + 4x − 1 | x + 5 | −5 | −375 − 50 − 20 − 1 | −446 |
The formula also runs backwards to find a missing coefficient. Asked for the value of k that makes x³ + kx² − 3x + 4 leave a remainder of 10 on division by x − 2, substitute a = 2 and solve: f(2) = 8 + 4k − 6 + 4 = 6 + 4k, and setting 6 + 4k = 10 gives k = 1.
Remainder Theorem and the Factor Theorem
The Factor Theorem is the Remainder Theorem with the remainder set to zero. Whenever f(a) = 0, the identity f(x) = (x − a)·q(x) + r loses its last term, leaving f(x) = (x − a)·q(x) — so x − a divides f(x) exactly and a is a root.
x − a is a factor of f(x), and a is a root. Divide it out and factor the smaller quotient.
f(3) = 0 for x³ − 6x² + 11x − 6 x − a is not a factor, and the value of f(a) is the remainder. Try the next candidate.
f(3) = 31 for x³ − x² + 2x + 7 That branch is the standard first move when factoring a cubic or a quartic by hand. The Rational Root Theorem narrows the search first: every rational root p/q of a polynomial with whole-number coefficients has p dividing the constant term and q dividing the leading coefficient. For 2x³ − 5x² + 3x − 7 that gives 8 candidates — ±1, ±7, ±1/2, ±7/2 — and testing each with the Remainder Theorem shows none returns 0, so the cubic has no rational roots at all. The candidate panel in the working section runs that test on whatever polynomial you enter.
Factoring x³ − 6x² + 11x − 6 goes the other way. The candidates are ±1, ±2, ±3, ±6, and f(1) = 0 straight away. Dividing out gives x² − 5x + 6, which factors to (x − 2)(x − 3), so the full factorisation is (x − 1)(x − 2)(x − 3).
Common Remainder Theorem Mistakes
There are 5 common Remainder Theorem mistakes, and 4 of them are caught by multiplying the quotient back out.
Substitute −a for a divisor of x + a. The divisor x + 2 is x − (−2), so the value to use is −2 rather than 2. Getting the sign wrong changes the answer entirely and misses factors.
x⁴ − 16 ÷ (x + 2) f(2) = 16 − 16 = 0 "so x − 2 is the factor"
x⁴ − 16 ÷ (x + 2) x + 2 = 0 → a = −2 f(−2) = 16 − 16 = 0 ✓
Write a 0 for every missing power before starting. x⁴ − 16 has no x³, x² or x terms, so the coefficient row reads 1, 0, 0, 0, −16. Leaving the gaps out shifts every later column and corrupts the quotient.
x⁴ − 16 → 1, −16
x⁴ − 16 → 1, 0, 0, 0, −16
Use polynomial long division for a divisor of degree 2 or higher. The Remainder Theorem covers linear divisors only, since a divisor of degree 2 leaves a remainder of degree up to 1 rather than a constant. Dividing by x² + 1 has no single f(a) to report.
Divide the quotient coefficients by b for a divisor of bx − c. The remainder f(c/b) is already correct, and the quotient from synthetic division belongs to x − c/b instead. Dividing 6x³ + 5x² − 17x − 6 by 2x − 3 means evaluating at 1.5 and then halving the quotient.
Report f(a) as the remainder and a as the root. A remainder of 31 means x − 3 is not a factor; it does not mean 31 is a root. The 2 numbers answer different questions, and only a remainder of 0 makes a a root.
Check any Remainder Theorem answer in 1 multiplication: expand (x − a)·q(x) + r and confirm it reproduces f(x). For the worked example that reads (x − 3)(x² + 2x + 8) + 31 = x³ − x² + 2x + 7, which matches. The remainder calculator handles the same question for whole numbers, and the long division tableau tool draws the arithmetic version of the identity.