The Remainder Theorem
Dividing polynomials by hand takes time — but what if you just need the remainder? The Remainder Theorem gives you a shortcut: substitute a single value into $P(x)$ and the answer is immediate. In this lesson you'll prove why it works and use it to find unknown coefficients.
In the previous lesson, dividing $x^3 - 2x^2 + 3x - 4$ by $x - 1$ gave remainder $-2$. Without using long division, calculate $P(1)$ for this polynomial. Is there a connection between your answer and the remainder?
There are two core applications of the Remainder Theorem — and both flow from a single formula. Lock $R = P(a)$ when dividing by $(x - a)$, and $R = P(b/a)$ when dividing by $(ax - b)$, into your toolkit.
Every Remainder Theorem question asks you to do one of two things: find the remainder by substituting a value, or find an unknown coefficient by using a known remainder to form an equation.
Key facts
- When $P(x)$ is divided by $(x - a)$, the remainder is $P(a)$
- When divided by $(ax - b)$, the remainder is $P(b/a)$
- The theorem follows directly from the division algorithm
Concepts
- Why the remainder must be a constant when dividing by a linear factor
- The algebraic proof via the division algorithm
- How to extend the theorem to non-monic linear divisors
Skills
- Use the Remainder Theorem to find remainders without long division
- Set up and solve equations to find unknown coefficients
- Apply the theorem to divisors of the form $(ax - b)$
Proof: By the division algorithm, dividing $P(x)$ by $(x - a)$ gives:
where $R$ is a constant. (Since the divisor has degree 1, the remainder must have degree less than 1 — that is, degree 0, so it's a constant.)
Substituting $x = a$:
Therefore $R = P(a)$. ∎
Remainder Theorem: P(x) (x - a) has remainder P(a); Proof sketch: from P(x) = (x-a)Q(x) + R, sub x = a to get R = P(a)
Pause — copy the Remainder Theorem into your book: dividing $P(x)$ by $(x-a)$ gives remainder $P(a)$; proof — from $P(x) = (x-a)Q(x) + R$, substitute $x = a$ to get $R = P(a)$.
Quick check: What is the remainder when $P(x) = x^3 - 2x + 1$ is divided by $(x - 2)$?
We just saw the Remainder Theorem: dividing $P(x)$ by $(x-a)$ leaves remainder $P(a)$. That raises a question: what if the divisor is not monic — e.g. $(2x-3)$? Can you still evaluate directly without full long division? This card answers it → set $2x - 3 = 0$ to get $x = 3/2$; the remainder is $P(3/2)$; the substitution value is always the zero of the divisor.
When the divisor is not monic — for example $(2x - 3)$ — we still use the same idea: set the divisor equal to zero and solve for $x$.
If $P(x) = (ax - b) \cdot Q(x) + R$, substituting $x = \dfrac{b}{a}$:
So the remainder when dividing by $(ax - b)$ is $P\!\left(\dfrac{b}{a}\right)$.
Example: Find the remainder when $P(x) = 2x^3 + x - 5$ is divided by $(2x - 1)$.
Set $2x - 1 = 0 \Rightarrow x = \tfrac{1}{2}$. Remainder $= P\!\left(\tfrac{1}{2}\right) = 2 \cdot \tfrac{1}{8} + \tfrac{1}{2} - 5 = \tfrac{1}{4} + \tfrac{1}{2} - 5 = -\tfrac{17}{4}$.
For (ax - b): set ax - b = 0, so x = b/a; remainder = P(b/a); Mnemonic: the substitution value is always the zero of the divisor
Pause — copy the non-monic extension into your book: for divisor $(ax-b)$, remainder $= P(b/a)$; the substitution value is the zero of the divisor — always set divisor $= 0$ and solve for $x$.
Did you get this? True or false: the remainder when $P(x)$ is divided by $(x + 5)$ is $P(5)$.
Worked examples · 3 in a row, reveal as you go
Find the remainder when $P(x) = x^3 - 4x^2 + 5x - 2$ is divided by $(x - 3)$.
When $P(x) = 2x^3 + kx^2 - 3x + 5$ is divided by $(x + 2)$, the remainder is $-5$. Find $k$.
Find the remainder when $2x^3 - x + 5$ is divided by $(2x - 1)$.
Fill the gap: When $P(x) = x^3 + ax^2 + 2x - 1$ is divided by $(x + 1)$, the remainder is $-4$. Then $P(-1) =$ , which gives the equation to find $a$.
Misconceptions to fix · the 3 traps that cost marks
Did you get this? True or false: the remainder when $P(x)$ is divided by $(3x + 6)$ is $P(-2)$.
Activities · practice with the ideas
Find the remainder when $P(x) = x^3 + 2x^2 - 5x + 3$ is divided by $(x - 2)$. Show your substitution clearly.
Find the remainder when $P(x) = 3x^3 - 2x^2 + x - 4$ is divided by $(x + 1)$. Remember the sign!
When $P(x) = x^3 + ax^2 + 2x - 1$ is divided by $(x + 1)$, the remainder is $-4$. Find $a$.
Find the remainder when $2x^3 - x + 5$ is divided by $(2x - 1)$. (Hint: substitute $x = \frac{1}{2}$.)
Explain in your own words why the Remainder Theorem only produces a constant remainder when dividing by a linear factor.
At the start, you were asked to find $P(1)$ for $x^3 - 2x^2 + 3x - 4$ and compare with the remainder $-2$.
$P(1) = 1 - 2 + 3 - 4 = -2$. Yes — $P(1)$ equals the remainder exactly! This is the Remainder Theorem. The proof shows this is guaranteed by the division algorithm: substituting $x = 1$ into $P(x) = (x-1)Q(x) + R$ gives $P(1) = 0 \cdot Q(1) + R = R$.
The Remainder Theorem only works for linear divisors because only then does substituting a single value kill the $(x - a)$ factor. For quadratic divisors, you'd need two substitutions — and the remainder would be a linear expression, not a constant.
Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.
Q1. Find the remainder when $P(x) = x^3 + 2x^2 - 5x + 3$ is divided by $(x - 2)$. (2 marks)
Q2. When $P(x) = x^3 + ax^2 + 2x - 1$ is divided by $(x + 1)$, the remainder is $-4$. Find $a$. (2 marks)
Q3. Find the remainder when $2x^3 - x + 5$ is divided by $(2x - 1)$. (2 marks)
Comprehensive answers (click to reveal)
Activities: 1. $P(2) = 8 + 8 - 10 + 3 = 9$ · 2. $P(-1) = -3 - 2 - 1 - 4 = -10$ · 3. $P(-1) = -4$: $-1 + a - 2 - 1 = -4 \Rightarrow a = 0$ · 4. $P(\frac{1}{2}) = 2 \cdot \frac{1}{8} - \frac{1}{2} + 5 = \frac{1}{4} - \frac{2}{4} + \frac{20}{4} = \frac{19}{4}$ · 5. Division by a degree-1 polynomial requires remainder of degree less than 1, i.e. degree 0 (constant).
Q1 (2 marks): $P(2) = 8 + 8 - 10 + 3 = 9$ [1]. Remainder $= 9$ [1].
Q2 (2 marks): $P(-1) = -4$: $-1 + a - 2 - 1 = -4 \Rightarrow a - 4 = -4 \Rightarrow a = 0$ [2].
Q3 (2 marks): $x = \frac{1}{2}$: $P(\frac{1}{2}) = 2 \cdot \frac{1}{8} - \frac{1}{2} + 5 = \frac{1}{4} - \frac{2}{4} + \frac{20}{4} = \frac{19}{4}$ [2].
Odd one out: Which of these is the odd one out? All divide $P(x)$ by a linear factor except one — which changes the substitution rule?
Five timed questions. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.
⚔ Enter the arenaClimb platforms by answering Remainder Theorem questions. Lighter alternative to the boss.
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