The Polynomial Remainder Theorem states that when a polynomial f(x) is divided by the linear expression x − a, the remainder equals f(a). Dividing f(x) = 2x³ + 5x² − 7x + 4 by x − 2 leaves a remainder of 26, because f(2) = 16 + 20 − 14 + 4 = 26. No division is carried out at any point.
The theorem replaces a full polynomial division with a single substitution. That is its entire value: a question that would take four rounds of long division collapses into one evaluation. The Remainder Theorem Calculator performs the same substitution and shows the quotient alongside it.
This guide states the theorem precisely, proves it in four lines, works through divisors of the form x − a, x + a and 2x − 3, shows how to recover an unknown coefficient from a known remainder, and ends with ten practice problems and full answers.
What the Polynomial Remainder Theorem Says
The theorem applies to one specific situation: a polynomial divided by a linear divisor.
If a polynomial f(x) is divided by x − a, where a is any constant, then the remainder is the constant f(a).
Three pieces of vocabulary matter here. f(x) is the polynomial being divided, called the dividend. x − a is the divisor. f(a) means the number you get by replacing every x in f(x) with the number a.
The theorem is sometimes called little Bézout’s theorem, and it holds for a polynomial of any degree. A degree-30 polynomial divided by x − 5 still leaves a single constant, because the size of the remainder is controlled by the divisor, not by the dividend.
Why the Remainder Is Always a Constant
Polynomial division obeys the same identity as whole-number division:
Dividend = Divisor × Quotient + Remainder
f(x) = (x − a) · q(x) + r(x)For integers, the remainder has to be smaller than the divisor. For polynomials, the rule is about degree: the remainder must have a lower degree than the divisor.
The divisor x − a has degree 1. Anything of lower degree than 1 is degree 0, which is a plain constant. So r(x) is not a polynomial in x at all. It is a fixed number, and we can write it as r.
Why f(a) Gives That Remainder
Start from the identity, then substitute x = a into both sides:
f(x) = (x − a) · q(x) + r
set x = a:
f(a) = (a − a) · q(a) + r
f(a) = 0 · q(a) + r
f(a) = rThe factor (a − a) is zero, so the entire quotient term vanishes no matter what q(x) looks like. Whatever is left is the remainder, and it equals f(a).
That is the whole proof. The theorem works because substituting the root of the divisor deletes the part of the identity you do not know.
How to Use the Remainder Theorem in 3 Steps
- Set the divisor equal to zero and solve for x. This gives the value a. For x − 2, solve x − 2 = 0 to get a = 2.
- Substitute a into f(x). Replace every x with that number and simplify.
- Read the result as the remainder. The number you land on is r.
Step 1 is where most errors happen, so it is worth slowing down there. The divisor x − 7 gives a = 7. The divisor x + 7 gives a = −7, because x + 7 is the same expression as x − (−7).
Worked Example 1: Dividing by x − a
Problem. Find the remainder when f(x) = 2x³ + 5x² − 7x + 4 is divided by x − 2.
Step 1. x − 2 = 0, so a = 2.
Step 2. Substitute 2 for every x:
f(2) = 2(2)³ + 5(2)² − 7(2) + 4
= 2(8) + 5(4) − 14 + 4
= 16 + 20 − 14 + 4
= 26Step 3. The remainder is 26.
Verification. The actual quotient is 2x² + 9x + 11. Multiplying back:
(x − 2)(2x² + 9x + 11) + 26
= 2x³ + 9x² + 11x − 4x² − 18x − 22 + 26
= 2x³ + 5x² − 7x + 4 ✓The identity reproduces the original polynomial exactly, which confirms both the quotient and the remainder.
Worked Example 2: Dividing by x + a
Problem. Find the remainder when f(x) = x⁴ − 3x³ + 2x − 5 is divided by x + 3.
Step 1. x + 3 = 0, so a = −3. This sign flip is the single most common slip in the whole topic.
Step 2. Substitute −3, keeping careful track of the signs of the odd powers:
f(−3) = (−3)⁴ − 3(−3)³ + 2(−3) − 5
= 81 − 3(−27) − 6 − 5
= 81 + 81 − 6 − 5
= 151Step 3. The remainder is 151.
Notice that f(x) has no x² term. That does not matter for the substitution, since a missing term contributes nothing. It does matter for synthetic division and for long division, where a zero placeholder has to be written in.
Worked Example 3: A Fifth-Degree Polynomial
Problem. Find the remainder when f(x) = x⁵ − 4x³ + 6x − 9 is divided by x − 2.
The degree does not change the method, which is the point of the theorem.
f(2) = (2)⁵ − 4(2)³ + 6(2) − 9
= 32 − 4(8) + 12 − 9
= 32 − 32 + 12 − 9
= 3The remainder is 3. Doing this by polynomial long division would take five rounds of divide, multiply and subtract. The substitution takes one line.
Divisors That Are Not x − a
The theorem is stated for x − a, but a divisor such as 2x − 3 still has a single root, so the same substitution works.
Problem. Find the remainder when f(x) = 4x³ − 8x² + 5x − 1 is divided by 2x − 3.
Step 1. Solve 2x − 3 = 0, giving x = 3/2.
Step 2. Substitute 3/2:
f(3/2) = 4(3/2)³ − 8(3/2)² + 5(3/2) − 1
= 4(27/8) − 8(9/4) + 15/2 − 1
= 27/2 − 18 + 15/2 − 1
= 13.5 − 18 + 7.5 − 1
= 2The remainder is 2.
One caution: the remainder is correct, but the quotient is not what a plain x − 3/2 division would give. Dividing by 2x − 3 produces a quotient of 2x² − x + 1, while dividing by x − 3/2 produces 4x² − 2x + 2, exactly twice as large. Check the identity to see why:
(2x − 3)(2x² − x + 1) + 2
= 4x³ − 2x² + 2x − 6x² + 3x − 3 + 2
= 4x³ − 8x² + 5x − 1 ✓The remainder survives the rescaling untouched. The quotient does not.
Finding an Unknown Coefficient
The theorem also runs in reverse. If you know the remainder, you can solve for a missing number inside the polynomial.
Problem. Find k so that f(x) = 2x³ + kx² − 5x + 6 leaves a remainder of 12 when divided by x − 1.
Substitute a = 1 and set the result equal to 12:
f(1) = 2(1)³ + k(1)² − 5(1) + 6
= 2 + k − 5 + 6
= 3 + k
3 + k = 12
k = 9Check. With k = 9 the polynomial is 2x³ + 9x² − 5x + 6, and f(1) = 2 + 9 − 5 + 6 = 12. ✓
This pattern appears constantly in exam questions, usually phrased as “given that the remainder is …, find the value of the constant”.
The Remainder Theorem and the Factor Theorem
The two theorems describe the same identity, read for two different purposes.
| Remainder Theorem | Factor Theorem | |
|---|---|---|
| Question asked | What is the remainder? | Is x − a a factor? |
| Statement | Dividing f(x) by x − a leaves f(a) | x − a is a factor exactly when f(a) = 0 |
| Answer type | Any number | Yes or no |
| Used for | Evaluating, checking work | Factoring, finding roots |
The Factor Theorem is the single case of the Remainder Theorem where the remainder happens to be zero. When f(a) = 0, the identity f(x) = (x − a)·q(x) + r loses its final term and becomes f(x) = (x − a)·q(x), which is what “x − a is a factor” means.
For testing factors, factoring cubics and locating roots, see the full guide to the Factor Theorem.
The Remainder Theorem and Synthetic Division
Substitution gives you one number. Synthetic division gives you that same number plus the quotient polynomial, from a comparable amount of arithmetic.
Running 2x³ + 5x² − 7x + 4 through synthetic division with a = 2:
| a = 2 | 2 | 5 | −7 | 4 |
|---|---|---|---|---|
| 4 | 18 | 22 | ||
| 2 | 9 | 11 | 26 |
The bottom row reads 2, 9, 11, 26. The first three numbers are the coefficients of the quotient 2x² + 9x + 11, and the final number is the remainder 26, matching f(2) exactly.
Use the theorem when only the remainder matters. Use synthetic division with remainders when you need the quotient as well, for example to keep factoring. When the divisor is quadratic or higher, neither shortcut applies and you need polynomial long division.
Common Remainder Theorem Mistakes
- Using the wrong sign of a. Dividing by x + 4 means evaluating f(−4), not f(4). Always solve divisor = 0 rather than reading the number off the page.
- Substituting into the divisor instead of the dividend. The value a comes from the divisor; the substitution happens in f(x).
- Dropping a negative on an even power. (−3)⁴ is +81, not −81. Bracket the negative number before raising it to a power.
- Applying the theorem to a quadratic divisor. Dividing by x² − 1 leaves a remainder of the form bx + c, not a constant, so a single substitution cannot produce it.
- Reporting the quotient as the answer. The theorem returns the remainder alone.
- Forgetting to rescale after a non-monic divisor. The remainder from f(3/2) is right for 2x − 3, but the quotient needs dividing by 2.
Remainder Theorem Practice Problems
Work these by substitution, then check against the answers below.
- Find the remainder when f(x) = x³ + 4x² − x − 10 is divided by x − 2.
- Find the remainder when f(x) = 3x⁴ − 2x² + x − 8 is divided by x + 1.
- Find the remainder when f(x) = x³ − 2x² − 5x + 6 is divided by x − 1.
- Find the remainder when f(x) = 5x³ − x + 12 is divided by x − 3.
- Find the remainder when f(x) = x⁴ + x³ − 7x + 2 is divided by x + 2.
- Find the remainder when f(x) = 2x⁵ − 3x² + 4 is divided by x − 1.
- Find the remainder when f(x) = 9x³ + 6x² − 4x + 5 is divided by 3x − 1.
- Find k if f(x) = x³ + kx − 6 leaves a remainder of 4 when divided by x − 2.
- Find k if f(x) = 2x³ + 3x² + kx + 7 leaves a remainder of 1 when divided by x + 1.
- A polynomial f(x) leaves remainder 5 on division by x − 1 and remainder 11 on division by x − 3. Find the remainder when f(x) is divided by (x − 1)(x − 3).
Answers
- f(2) = 8 + 16 − 2 − 10 = 12
- f(−1) = 3 − 2 − 1 − 8 = −8
- f(1) = 1 − 2 − 5 + 6 = 0, so x − 1 is a factor.
- f(3) = 135 − 3 + 12 = 144
- f(−2) = 16 − 8 + 14 + 2 = 24
- f(1) = 2 − 3 + 4 = 3
- Solve 3x − 1 = 0 to get x = 1/3. f(1/3) = 9(1/27) + 6(1/9) − 4/3 + 5 = 1/3 + 2/3 − 4/3 + 5 = 14/3
- f(2) = 8 + 2k − 6 = 2 + 2k, and 2 + 2k = 4 gives k = 1
- f(−1) = −2 + 3 − k + 7 = 8 − k, and 8 − k = 1 gives k = 7
- The divisor has degree 2, so the remainder has the form bx + c. From f(1) = 5: b + c = 5. From f(3) = 11: 3b + c = 11. Subtracting gives 2b = 6, so b = 3 and c = 2. The remainder is 3x + 2.
Problem 10 shows the boundary of the theorem. A quadratic divisor needs two substitutions and a small system of equations, because the remainder is no longer a single number.
Polynomial Remainder Theorem FAQ
What is the Remainder Theorem formula?
The formula is r = f(a), sitting inside the polynomial division identity f(x) = (x − a)·q(x) + r. To use it, solve divisor = 0 for a, then evaluate f(a). Dividing x³ + 4x² − x − 10 by x − 2 gives a = 2 and f(2) = 12, so the remainder is 12.
Does the Remainder Theorem work for any degree of polynomial?
Yes. The dividend may be any degree from 1 upward. The restriction is on the divisor, which must be linear. Dividing a degree-50 polynomial by x − 3 still leaves the single constant f(3), because a degree-1 divisor forces a degree-0 remainder.
What does it mean if the remainder is 0?
A remainder of 0 means x − a divides f(x) exactly, so x − a is a factor of f(x) and a is a root of the equation f(x) = 0. That is the Factor Theorem. It is the polynomial version of what a remainder of zero means for whole numbers.
How is the Remainder Theorem different from finding a remainder in ordinary division?
Ordinary division works on whole numbers and produces a whole-number leftover smaller than the divisor, which is covered in how to find the remainder. The Remainder Theorem works on polynomials and produces a constant whose degree is lower than the divisor’s. Both follow the same identity, but the size rule changes from magnitude to degree.
Can the Remainder Theorem be used with a quadratic divisor?
Not directly. A quadratic divisor leaves a remainder of the form bx + c, which one substitution cannot determine. You can substitute both roots of the quadratic and solve the resulting pair of equations, as in practice problem 10, or divide properly using polynomial long division.
Is the Remainder Theorem the same as the Chinese Remainder Theorem?
No. They share a word and nothing else. The Remainder Theorem is an algebra result about dividing polynomials by x − a. The Chinese Remainder Theorem is a number theory result that rebuilds a whole number from its remainders against several coprime moduli.
Why does dividing by 2x − 3 still work?
Because 2x − 3 has exactly one root, x = 3/2, and substituting it still collapses the quotient term to zero. The remainder f(3/2) is correct as written. The quotient, however, comes out twice too large if you divide by x − 3/2 instead, so halve it to get the quotient for 2x − 3.
