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:
Symbol | Meaning |
|---|---|
$p(x)$ | the polynomial being divided (the dividend) |
$(x - a)$ | the linear divisor |
$q(x)$ | the quotient |
$r$ | the remainder, a constant |
$a$ | the number that makes the divisor zero |
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:
Remainder theorem | Factor theorem | |
|---|---|---|
States | remainder of $p(x) \div (x-a)$ is $p(a)$ | if $p(a) = 0$, then $(x-a)$ is a factor |
Used for | finding any remainder | finding factors and roots |
Relationship | the general case | the special case where the remainder is 0 |
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.
Find the remainder when $p(x) = x^2 - 4x + 7$ is divided by $(x - 3)$.
Find the remainder when $p(x) = x^3 + 2x^2 - x + 1$ is divided by $(x + 2)$.
Find the remainder when $p(x) = 4x^3 - 3x + 5$ is divided by $(2x + 1)$.
Determine whether $(x - 2)$ is a factor of $p(x) = x^3 - 3x^2 + 4$.
Find the value of $k$ if the remainder is 7 when $p(x) = x^2 + kx - 1$ is divided by $(x - 2)$.
Find the remainder when $p(x) = x^4 - 1$ is divided by $(x - 1)$.
Answers
Answer to Question 1: $p(3) = 9 - 12 + 7 = 4$.
Answer to Question 2: $p(-2) = -8 + 8 + 2 + 1 = 3$.
Answer to Question 3: substitute $x = -\tfrac{1}{2}$: $p\left(-\tfrac{1}{2}\right) = -\tfrac{1}{2} + \tfrac{3}{2} + 5 = 6$.
Answer to Question 4: $p(2) = 8 - 12 + 4 = 0$, so $(x - 2)$ is a factor.
Answer to Question 5: $p(2) = 4 + 2k - 1 = 3 + 2k = 7$, so $k = 2$.
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:
Find the remainder when $p(x) = x^3 + 2x^2 - x + 5$ is divided by $(x - 1)$.
Find the remainder when $p(x) = 2x^3 - 3x + 4$ is divided by $(x + 1)$.
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.
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