Remainder Theorem - Statement, Proof & Examples

What Is The Remainder Theorem?

The remainder theorem states that when a polynomial $p(x)$ is divided by a linear divisor $(x - a)$, the remainder equals $p(a)$ — the value of the polynomial at $x = a$. Because the divisor is linear (degree 1), the remainder is always a constant.

In symbols, dividing $p(x)$ by $(x - a)$ gives a quotient $q(x)$ and a remainder $r$:

$$p(x) = (x - a) , q(x) + r$$

and the theorem tells you that $r = p(a)$.

Variable Glossary:

A useful extension: when the divisor is written as $(ax + b)$ rather than $(x - a)$, the remainder is $p\left(-\tfrac{b}{a}\right)$ — you substitute the value that makes the divisor zero.

The link to synthetic division

There is a second way to find the same remainder: synthetic division, a compact shorthand for dividing by $(x - a)$ using only the coefficients. You write the dividend's coefficients in a row, bring down the leading one, multiply by $a$, add to the next coefficient, and repeat. The last number you produce is the remainder — and it equals $p(a)$, exactly what the remainder theorem promises. Synthetic division is the faster route when you also want the quotient; pure substitution is faster when you want the remainder alone. Both rest on the same identity $p(x) = (x - a)q(x) + r$.

How Do You Prove The Remainder Theorem?

The proof is short and rests on the division algorithm. Any polynomial $p(x)$ divided by $(x - a)$ can be written as:

$$p(x) = (x - a) , q(x) + r$$

where $r$ is a constant because the divisor has degree 1, so the remainder has degree 0.

This identity holds for every value of $x$, so substitute $x = a$:

$$p(a) = (a - a) , q(a) + r$$

The factor $(a - a)$ is 0, which kills the first term:

$$p(a) = 0 \cdot q(a) + r = r$$

So $r = p(a)$. The remainder equals the polynomial evaluated at $a$.

Examples of Remainder Theorem

Example 1

Find the remainder when $p(x) = x^2 + 5x + 6$ is divided by $(x - 2)$.

Substitute $x = 2$:

$$p(2) = 2^2 + 5(2) + 6$$

$$= 4 + 10 + 6 = 20$$

Final answer: the remainder is 20.

Example 2

Find the remainder when $p(x) = x^3 - 4x + 1$ is divided by $(x + 2)$.

The divisor $(x + 2)$ is the place students slip, so watch it.

Wrong attempt. A student reads "$x + 2$" and substitutes $x = 2$, getting $p(2) = 8 - 8 + 1 = 1$. But the theorem needs the value of $x$ that makes the divisor zero. Set $x + 2 = 0$ and you get $x = -2$, not $+2$. Substituting the wrong sign gives the wrong remainder.

Correct. Substitute $x = -2$:

$$p(-2) = (-2)^3 - 4(-2) + 1$$

$$= -8 + 8 + 1 = 1$$

The number happens to match here by coincidence of this polynomial, but the reasoning matters: always substitute the zero of the divisor.

Final answer: the remainder is 1.

Example 3

Find the remainder when $p(x) = 2x^3 + x^2 - 5x + 3$ is divided by $(x - 1)$.

Substitute $x = 1$:

$$p(1) = 2(1)^3 + (1)^2 - 5(1) + 3$$

$$= 2 + 1 - 5 + 3 = 1$$

Final answer: the remainder is 1.

Example 4

Find the remainder when $p(x) = x^3 + 3x^2 - 2x + 4$ is divided by $(2x - 1)$.

The divisor is $(2x - 1)$, so use the extended form: substitute the value where $2x - 1 = 0$, which is $x = \tfrac{1}{2}$:

$$p\left(\tfrac{1}{2}\right) = \left(\tfrac{1}{2}\right)^3 + 3\left(\tfrac{1}{2}\right)^2 - 2\left(\tfrac{1}{2}\right) + 4$$

$$= \tfrac{1}{8} + 3\left(\tfrac{1}{4}\right) - 1 + 4$$

$$= \tfrac{1}{8} + \tfrac{3}{4} + 3 = \tfrac{1}{8} + \tfrac{6}{8} + \tfrac{24}{8} = \tfrac{31}{8}$$

Final answer: the remainder is $\tfrac{31}{8}$.

Example 5

Use the remainder theorem to check whether $(x - 3)$ is a factor of $p(x) = x^3 - 7x + 6$.

Substitute $x = 3$:

$$p(3) = 3^3 - 7(3) + 6$$

$$= 27 - 21 + 6 = 12$$

The remainder is 12, not 0, so $(x - 3)$ is not a factor.

Final answer: $(x - 3)$ is not a factor, since $p(3) = 12 \neq 0$.

Example 6

Find the value of $k$ if the remainder is 5 when $p(x) = x^2 + kx + 2$ is divided by $(x - 3)$.

By the theorem, the remainder is $p(3)$, and we want it to equal 5:

$$p(3) = 3^2 + k(3) + 2 = 9 + 3k + 2 = 11 + 3k$$

Set equal to 5:

$$11 + 3k = 5$$

$$3k = -6$$

$$k = -2$$

Final answer: $k = -2$.

Why This Beats Long Division, And Where It Leads

The remainder theorem is the WHY behind the factor theorem, which is its most useful corollary. The factor theorem says: if $p(a) = 0$, then $(x - a)$ is a factor of $p(x)$. It is just the remainder theorem with the remainder set to zero — a zero remainder means the divisor divides evenly, which is what "factor" means.

That single idea is how you find the roots of higher-degree polynomials you cannot factor by inspection. Test candidate values; any that give a remainder of zero is a root, and $(x - \text{that value})$ is a factor. This pairs naturally with the rational root theorem, which produces the list of candidates to test, and with cubic polynomials, where this test-and-factor loop does the heavy lifting. The slow alternative, long division of polynomials, still has its place when you need the full quotient — but for the remainder alone, substitution wins.

The remainder theorem and the factor theorem differ in one line:

Where Students Trip Up On The Remainder Theorem

Mistake 1: Substituting the wrong sign for the divisor's value

Where it slips in: With a divisor like $(x + 2)$, students substitute $x = 2$ instead of the value that makes the divisor zero, $x = -2$.

Don't do this: Reading $(x + 3)$ and computing $p(3)$.

The correct way: Set the divisor equal to zero and solve. For $(x + 2)$, that is $x = -2$; for $(x - 5)$, that is $x = 5$. The first-instinct error is to lift the number straight out of the bracket with its printed sign — the fix is to always solve $\text{divisor} = 0$ first.

Mistake 2: Forgetting the theorem only works for linear divisors

Where it slips in: A student tries to apply $p(a)$ when dividing by a quadratic like $(x^2 - 1)$.

Don't do this: Claiming the remainder of $p(x) \div (x^2 - 1)$ is a single number from one substitution.

The correct way: The remainder theorem applies only when the divisor is linear, $(x - a)$. Dividing by a degree-2 divisor leaves a remainder of degree up to 1 — not a constant. The habit that fixes this is checking the divisor's degree before substituting; the silent understander who can use the theorem but never asks "is this divisor linear?" gets caught out exactly here.

Mistake 3: Confusing the remainder with the quotient

Where it slips in: Students sometimes report $p(a)$ as a root or a factor rather than as the remainder.

Don't do this: Saying "the remainder is 0, so the root is 0" when actually the root is the value $a$ that produced the zero.

The correct way: $p(a)$ is the remainder. If that remainder is zero, then $a$ is a root and $(x - a)$ is a factor — but the root is $a$, not the remainder. Keep the roles straight: substitute $a$ in, read the remainder out.

Practice Questions on the Remainder Theorem

For each one, solve $\text{divisor} = 0$ first, then substitute. Answers follow so you can check the sign of the value you plugged in.

1. Find the remainder when $p(x) = x^2 - 4x + 7$ is divided by $(x - 3)$.

2. Find the remainder when $p(x) = x^3 + 2x^2 - x + 1$ is divided by $(x + 2)$.

3. Find the remainder when $p(x) = 4x^3 - 3x + 5$ is divided by $(2x + 1)$.

4. Determine whether $(x - 2)$ is a factor of $p(x) = x^3 - 3x^2 + 4$.

5. Find the value of $k$ if the remainder is 7 when $p(x) = x^2 + kx - 1$ is divided by $(x - 2)$.

6. Find the remainder when $p(x) = x^4 - 1$ is divided by $(x - 1)$.

Answers

1. Answer to Question 1: $p(3) = 9 - 12 + 7 = 4$.

2. Answer to Question 2: $p(-2) = -8 + 8 + 2 + 1 = 3$.

3. Answer to Question 3: substitute $x = -\tfrac{1}{2}$: $p\left(-\tfrac{1}{2}\right) = -\tfrac{1}{2} + \tfrac{3}{2} + 5 = 6$.

4. Answer to Question 4: $p(2) = 8 - 12 + 4 = 0$, so $(x - 2)$ is a factor.

5. Answer to Question 5: $p(2) = 4 + 2k - 1 = 3 + 2k = 7$, so $k = 2$.

6. Answer to Question 6: $p(1) = 1 - 1 = 0$.

Key Takeaways

• The remainder theorem says the remainder of $p(x) \div (x - a)$ equals $p(a)$, found by substitution, not division.

• The proof follows from the division identity $p(x) = (x - a)q(x) + r$ evaluated at $x = a$.

• The factor theorem is the special case: $p(a) = 0$ means $(x - a)$ is a factor.

• The theorem works only for linear divisors; substitute the zero of the divisor, signs included.

• It is the engine behind finding roots of higher-degree polynomials without full long division.

A Practical Next Step

Try these three to solidify your understanding:

1. Find the remainder when $p(x) = x^3 + 2x^2 - x + 5$ is divided by $(x - 1)$.

2. Find the remainder when $p(x) = 2x^3 - 3x + 4$ is divided by $(x + 1)$.

3. Determine whether $(x - 2)$ is a factor of $p(x) = x^3 - 3x^2 + 4$.

If you get stuck choosing the value to substitute, come back to Mistake 1 and solve $\text{divisor} = 0$ first. At Bhanzu, our trainers pair the remainder theorem with the factor theorem in one lesson, so students see the substitution shortcut and its root-finding payoff together. Want a live trainer to work through more remainder theorem problems? Book a free demo class.