MathsICSE Class 10 · Factorisation of Polynomials

Remainder Theorem: Method, Typical Questions and Solved Examples (ICSE Class 10)

Updated 15 October 2026 · Fact-checked

The remainder theorem says that when a polynomial f(x) is divided by (x − a), the remainder is f(a). For a divisor (px − q), the remainder is f(q/p). So you find a remainder by substitution, without long division, and you can use it to find unknown coefficients.

Put x = a to make the (x − a) term vanish, so the remainder is f(a); for 2x − 1 use x = 1/2, and a zero remainder means a factor.

Illustration drawn for QuestionBuddy.

Understand Remainder Theorem

Dividing a polynomial $f (x)$ by a linear divisor $(x - a)$ gives a quotient $q (x)$ and a remainder $r$ , which is a constant because it must have lower degree than the divisor:

$f (x) = (x - a) q (x) + r .$

Put $x = a$ : the first term vanishes, leaving $f (a) = r$ . That is the remainder theorem:

$Remainder when f (x) is divided by (x - a) = f (a) .$

For a divisor $(p x - q)$ , put $p x - q = 0$ , so $x = \frac{q}{p}$ , and the remainder is $f (\frac{q}{p})$ . For $(x + a)$ the value is $- a$ .

What the theorem is used for.

1. Finding a remainder quickly, especially for divisors like $(2 x + 1)$ where long division is slow.
2. Finding unknown coefficients when the remainder is given: set $f (a)$ equal to that remainder.
3. 'Equal remainder' questions: two polynomials leave the same remainder on division by $(x - a)$ , so set $f (a) = g (a)$ .
4. 'What must be added or subtracted' questions. If $f (a) = r$ , then $f (x) - r$ is exactly divisible by $(x - a)$ . So **subtract $r$ , or equivalently add $- r$ **. In 2024 Q1(ii), $f (x) = x^{3} + x^{2} - 2 x + 1$ and $f (3) = 31$ ; the number to be subtracted was 31. The examiners reported candidates who answered $- 31$ , missing the word 'subtracted'. A safe method: let the number subtracted be $k$ , then solve $f (a) - k = 0$ .

Connection with the factor theorem. If the remainder $f (a)$ is 0, then $(x - a)$ is a factor. So the factor theorem is a special case.

Checking with division. Synthetic division by $a$ (bring down, multiply by $a$ , add) gives both the quotient and the remainder in one line; its last number must equal $f (a)$ .

Remainder Theorem: formulas to remember

Division statement: $f (x) = (x - a) q (x) + r$; $r$ is a constant for a linear divisor.
Remainder theorem: $r = f (a)$; Divisor $(x - a)$ .
Divisor with a coefficient: $(p x - q) : r = f (\frac{q}{p})$; Set the divisor to zero.
Making a polynomial divisible: $f (x) - f (a) is divisible by (x - a)$; Subtract $f (a)$ , or add $- f (a)$ .
Equal remainders: $f (a) = g (a)$; Two polynomials, same divisor $(x - a)$ .

How to solve Remainder Theorem questions

Every remainder question is a substitution followed by simple algebra.

1Set the divisor equal to zero to find the value of $x$ to substitute.
2Write $f (value)$ term by term and simplify; that number is the remainder.
3If the remainder is given and a coefficient is unknown, equate and solve.
4For 'same remainder', write $f (a) = g (a)$ .
5For 'what must be added/subtracted', let the number be $k$ and solve $f (a) \pm k = 0$ , reading the words carefully.
6Check if needed with synthetic division.

Remainder Theorem in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Remainder theorem as a fast evaluator

Suppose a workshop models its daily cost (in hundreds of rupees) as $C (x) = x^{3} - 2 x^{2} + 5 x + 10$ , where $x$ is the number of machines running.

Use the remainder theorem idea to find the cost when 4 machines run, and state what the remainder would be on dividing $C (x)$ by $(x - 4)$ .

Try it first, then show the solutionSolution

$C (4) = 64 - 32 + 20 + 10 = 62$ .
By the remainder theorem, dividing $C (x)$ by $(x - 4)$ leaves remainder $C (4) = 62$ .

Answer: ₹6,200; the remainder is 62.

Memory hook: Remainder on dividing by (x − a) is just the value at a.

Adjusting a polynomial to make it divisible

Imagine a puzzle designer who wants the expression $x^{3} - 4 x^{2} + 2 x + c$ to be exactly divisible by $(x - 3)$ , so that a pattern 'closes' neatly.

Find the constant $c$ .

Try it first, then show the solutionSolution

$f (3) = 27 - 36 + 6 + c = c - 3$ .
$c - 3 = 0 \Rightarrow c = 3$ .

Answer: $c = 3$ .

Memory hook: Make the remainder zero and divisibility follows.

Tricks to remember Remainder Theorem

Short memory aids. Say each one aloud once, then use it on a question the same day.

Divide by (x − a)? Plug in a.
No long division needed: the remainder is the value at $a$ .
Use it when: Every remainder question.
Let it be k
For 'what must be added/subtracted', call the number $k$ , write $f (a) - k = 0$ (subtracted) or $f (a) + k = 0$ (added), and solve.
Use it when: Avoiding the sign slip in 2024 Q1(ii).
Factor theorem is remainder zero
Remainder 0 ⇔ factor.
Use it when: Linking the two theorems.

Remainder Theorem: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: A direct remainder · 1 mark · MCQBoard level

The remainder when $x^{3} - 3 x^{2} + 4 x - 5$ is divided by $(x - 2)$ is:
(a) 1
(b) −1
(c) −5
(d) 3

Try it first, then show the solutionSolution

$f (2) = 8 - 12 + 8 - 5 = - 1$ .

Answer: (b) −1

Example 2: A divisor with a coefficient · 3 marks · Short answerBoard level

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

Try it first, then show the solutionSolution

$2 x + 1 = 0 \Rightarrow x = - \frac{1}{2}$ .
$f (- \frac{1}{2}) = 2 (- \frac{1}{8}) + 3 (\frac{1}{4}) + \frac{5}{2} + 4 = - \frac{1}{4} + \frac{3}{4} + \frac{5}{2} + 4$ .
$= 7$ .

Answer: 7

Example 3: What must be added · 3 marks · Short answerHarder board level

What must be added to $x^{3} - 4 x^{2} + 2 x$ so that the result is exactly divisible by $(x - 3)$ ?

Try it first, then show the solutionSolution

Let $k$ be added: $f (x) + k$ must vanish at $x = 3$ .
$27 - 36 + 6 + k = 0 \Rightarrow - 3 + k = 0$ .
$k = 3$ .

Answer: 3

Example 4: Equal remainders · 3 marks · Short answerHarder board level

The polynomials $x^{3} + a x^{2} - 3 x + 5$ and $2 x^{3} - x^{2} + a x - 1$ leave the same remainder when divided by $(x - 2)$ . Find $a$ .

Try it first, then show the solutionSolution

First: $8 + 4 a - 6 + 5 = 4 a + 7$ .
Second: $16 - 4 + 2 a - 1 = 2 a + 11$ .
$4 a + 7 = 2 a + 11 \Rightarrow a = 2$ .

Answer: $a = 2$

Example 5: Two remainders, two unknowns · 4 marks · Short answerHOTS / competency

$f (x) = x^{3} + a x^{2} + b x + 4$ leaves a remainder of 6 when divided by $(x - 1)$ and a remainder of 10 when divided by $(x + 1)$ . Find $a$ and $b$ .

Try it first, then show the solutionSolution

$f (1) = 1 + a + b + 4 = 6 \Rightarrow a + b = 1$ .
$f (- 1) = - 1 + a - b + 4 = 10 \Rightarrow a - b = 7$ .
$a = 4$ , $b = - 3$ .
Check: $f (1) = 1 + 4 - 3 + 4 = 6$ ; $f (- 1) = - 1 + 4 + 3 + 4 = 10$ ✓.

Answer: $a = 4$ , $b = - 3$

Example 6: Case study: remainder, quotient and a correction · 4 marks · Case-basedHOTS / competency

Let $f (x) = 3 x^{3} - 4 x^{2} + 2 x - 7$ .
(a) Use the remainder theorem to find the remainder on dividing $f (x)$ by $(x - 2)$ .
(b) Confirm it by synthetic division and write the quotient.
(c) What must be subtracted from $f (x)$ so that $(x - 2)$ becomes a factor?

Try it first, then show the solutionSolution

(a) $f (2) = 24 - 16 + 4 - 7 = 5$ .
(b) Synthetic division by 2: bring down 3; $3 \times 2 - 4 = 2$ ; $2 \times 2 + 2 = 6$ ; $6 \times 2 - 7 = 5$ . Quotient $3 x^{2} + 2 x + 6$ , remainder 5 ✓.
(c) Subtract 5: $f (x) - 5 = (x - 2) (3 x^{2} + 2 x + 6)$ .

Answer: (a) 5 (b) quotient $3 x^{2} + 2 x + 6$ , remainder 5 (c) subtract 5.

Where students lose marks in Remainder Theorem

The slips examiners see most often, and what to write instead.

Mistake: Answering with the wrong sign in 'what must be subtracted'
If $f (a) = r$ , then $r$ must be subtracted (or $- r$ added).
Wrong working
$f (3) = 31$ , so $- 31$ must be subtracted.
Right working
31 must be subtracted.
Do this: Let the number be $k$ and write $f (a) - k = 0$ .
Usually costs: The MCQ in 2024 Q1(ii).
Mistake: Substituting the wrong value for (x + a) or (px − q)
The value is the zero of the divisor.
Wrong working
Divisor $(2 x + 1)$ : substitute $x = 1$ or $x = - 2$ .
Right working
$x = - \frac{1}{2}$ .
Do this: Always solve 'divisor = 0' first.
Usually costs: All the marks.
Mistake: Arithmetic slips with fractions
Powers of fractions shrink quickly.
Wrong working
$2 (- \frac{1}{2})^{3} = - 1$ .
Right working
$2 \times (- \frac{1}{8}) = - \frac{1}{4}$ .
Do this: Write each term separately before adding.
Usually costs: The answer mark.

Exam tips

Write 'By the remainder theorem, remainder $= f (a)$ ' before substituting.
In 2024 Q1(ii) the answer was 31 (to be subtracted); the examiners reported candidates who chose −31.
Remainder-theorem questions are usually 1 mark in Section A, but the method is reused inside 3- and 4-mark factorisation questions.
Synthetic division is a quick check that also gives the quotient.

Practice questions on Remainder Theorem

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The remainder when $x^{2} + 3 x + 7$ is divided by $(x + 1)$ is:
(a) 11
(b) 5
(c) 7
(d) 3
Hint
Substitute $x = - 1$ .
Answer
(b) 5
Q2 · 3 marks · Short answerBoard level
Find the remainder when $4 x^{3} - 3 x^{2} + 2 x - 4$ is divided by $(x + 2)$ .
Hint
$f (- 2)$ .
Answer
$- 32 - 12 - 4 - 4 = - 52$ .
Q3 · 3 marks · Short answerHarder board level
$x^{3} + k x^{2} - 2 x + 3$ leaves a remainder of 11 when divided by $(x - 2)$ . Find $k$ .
Hint
$8 + 4 k - 4 + 3 = 11$ .
Answer
$4 k + 7 = 11 \Rightarrow k = 1$ .
Q4 · 3 marks · Short answerHarder board level
What must be subtracted from $2 x^{3} - 3 x^{2} + 4 x + 5$ so that the result is divisible by $(x - 1)$ ?
Hint
Find $f (1)$ .
Answer
$f (1) = 2 - 3 + 4 + 5 = 8$ , so subtract 8.
Q5 · 1 mark · MCQHarder board level
When $f (x)$ is divided by $(3 x - 1)$ , the remainder is:
(a) $f (3)$
(b) $f (- 3)$
(c) $f (\frac{1}{3})$
(d) $f (- \frac{1}{3})$
Hint
Set $3 x - 1 = 0$ .
Answer
(c)
Q6 · 3 marks · Short answerHarder board level
$a x^{3} + 3 x^{2} - 3$ and $2 x^{3} - 5 x + a$ leave the same remainder when divided by $(x - 4)$ . Find $a$ .
Hint
$64 a + 48 - 3 = 128 - 20 + a$ .
Answer
$63 a = 63 \Rightarrow a = 1$ .
Q7 · 4 marks · Short answerHOTS / competency
$x^{3} + 2 x^{2} + k x + 3$ leaves a remainder of 21 when divided by $(x - 3)$ . Find $k$ , and then the remainder when the polynomial is divided by $(x + 1)$ .
Hint
First find $k$ from $f (3) = 21$ .
Answer
$27 + 18 + 3 k + 3 = 21 \Rightarrow k = - 9$ . Then $f (- 1) = - 1 + 2 + 9 + 3 = 13$ .

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.State the remainder theorem.Show answer
When $f (x)$ is divided by $(x - a)$ , the remainder is $f (a)$ .
2.Remainder when $x^{2} - 5 x + 6$ is divided by $(x - 1)$ ?Show answer
$1 - 5 + 6 = 2$ .
3.Value to substitute for divisor $(4 x - 3)$ ?Show answer
$\frac{3}{4}$ .
4. $f (2) = 9$ . What must be subtracted from $f (x)$ to make it divisible by $(x - 2)$ ?Show answer
9.
5.Remainder 0 means what?Show answer
The divisor is a factor.

Remainder Theorem: frequently asked questions

What is the remainder theorem?

If a polynomial f(x) is divided by (x − a), the remainder equals f(a). It lets you find a remainder by substitution instead of long division.

How do I find what must be added to make a polynomial divisible by (x − a)?

Find f(a). The number to be added is −f(a); equivalently, f(a) must be subtracted. Letting the number be k and solving avoids sign errors.

Can the remainder theorem be used with divisors like (2x + 3)?

Yes. Set 2x + 3 = 0 to get x = −3/2, and the remainder is f(−3/2).