MathsICSE Class 10 · Chapter 7

Factorisation of Polynomials ICSE Class 10: Remainder and Factor Theorems

Updated 14 October 2026 · Fact-checked

The remainder theorem says that when f(x) is divided by (x − a), the remainder is f(a). The factor theorem says (x − a) is a factor exactly when f(a) = 0. ICSE Class 10 uses these to find remainders and unknown coefficients and to factorise cubic polynomials completely after finding one factor.

What Factorisation of Polynomials is about

This chapter gives two short theorems that save you from long division.

The remainder theorem: if a polynomial $f (x)$ is divided by $(x - a)$ , the remainder is $f (a)$ . If it is divided by $(a x - b)$ , the remainder is $f (\frac{b}{a})$ .

The factor theorem: $(x - a)$ is a factor of $f (x)$ if and only if $f (a) = 0$ .

The CISCE 2027 syllabus asks for (a) the factor theorem, (b) the remainder theorem, and (c) factorising a polynomial completely after obtaining one factor by the factor theorem, with $f (x)$ of degree at most 3. A typical Section B question gives a cubic with an unknown coefficient $k$ and one known factor: find $k$ , then factorise completely. The examiners' reports for 2024 and 2025 record the same faults year after year: arithmetic slips in finding $k$ , errors in the division, stopping after finding $k$ , and not writing the final answer as a product of three factors.

Weightage: CISCE publishes no chapter or unit weightage for ICSE Mathematics. Factorisation of Polynomials is in the 2. Algebra (iv) Factorisation of polynomials part of the 2027 syllabus and can be asked in both Section A and Section B of the 80-mark paper.Source: CISCE ICSE Mathematics syllabus, examination year 2027

Key concepts

Remainder theorem

When $f (x)$ is divided by $(x - a)$ , we can write $f (x) = (x - a) q (x) + r$ , where $r$ is a constant. Put $x = a$ : the first term vanishes, so $r = f (a)$ .

So to find a remainder:

divisor $(x - a)$ : work out $f (a)$ ;
divisor $(x + a)$ : work out $f (- a)$ ;
divisor $(a x - b)$ : work out $f (\frac{b}{a})$ (find the value of $x$ that makes the divisor zero).

Use brackets for negative and fractional values, e.g. $2 (- \frac{1}{2})^{3}$ .

Worked example

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

Try it first, then show the solutionSolution

(a) $f (2) = 16 - 12 + 8 - 5 = 7$ .
(b) $2 x + 1 = 0$ gives $x = - \frac{1}{2}$ ; $f (- \frac{1}{2}) = - \frac{1}{4} - \frac{3}{4} - 2 - 5 = - 8$ .

Answer: (a) 7 (b) −8

Factor theorem

$(x - a)$ is a factor of $f (x)$ exactly when the remainder is zero, i.e. $f (a) = 0$ . To test whether a linear expression is a factor, substitute the value that makes it zero; if you get 0, it is a factor.

A useful shortcut: if the sum of the coefficients is 0, then $f (1) = 0$ and $(x - 1)$ is a factor. If the sum of the coefficients of the even powers equals the sum for the odd powers, then $f (- 1) = 0$ and $(x + 1)$ is a factor.

Worked example

Show that $(x + 3)$ is a factor of $x^{3} + 2 x^{2} - 5 x - 6$ .

Try it first, then show the solutionSolution

$f (- 3) = - 27 + 18 + 15 - 6 = 0$ .
By the factor theorem, $(x + 3)$ is a factor.

Answer: Shown.

Finding an unknown coefficient

If $(x - a)$ is a factor, set $f (a) = 0$ . If the remainder on dividing by $(x - a)$ is $r$ , set $f (a) = r$ . Each condition gives one equation. With one unknown, one condition is enough; with two unknowns, you need two conditions, which give a pair of simultaneous linear equations.

Worked example

Find $a$ and $b$ if $(x - 2)$ is a factor of $x^{3} + a x^{2} + b x + 6$ and the remainder on division by $(x - 3)$ is 3.

Try it first, then show the solutionSolution

$f (2) = 0$ : $8 + 4 a + 2 b + 6 = 0$ , so $2 a + b = - 7$ .
$f (3) = 3$ : $27 + 9 a + 3 b + 6 = 3$ , so $3 a + b = - 10$ .
Subtract: $a = - 3$ ; then $b = - 1$ .

Answer: a = −3, b = −1

What must be added or subtracted

If the remainder on dividing $f (x)$ by $(x - a)$ is $r$ , then $f (x) - r$ is exactly divisible by $(x - a)$ . So:

subtract $r$ , or equivalently
add $- r$ .

Read the question's verb carefully: 'what must be subtracted' and 'what must be added' have answers with opposite signs. A safe method is to call the number $k$ , write $f (a) - k = 0$ (for subtracted) or $f (a) + k = 0$ (for added), and solve for $k$ .

Worked example

What must be subtracted from $2 x^{3} + x^{2} - 5 x + 7$ to make it exactly divisible by $(x - 2)$ ?

Try it first, then show the solutionSolution

Let $k$ be subtracted: $f (2) - k = 0$ .
$f (2) = 16 + 4 - 10 + 7 = 17$ , so $k = 17$ .

Answer: 17

Factorising a cubic completely

1. Find one factor by trial. Try $x = 1, - 1, 2, - 2, 3, \dots$ : the values to try are divisors of the constant term (and, when the leading coefficient is not 1, fractions like $\frac{1}{2}$ ).
2. Divide $f (x)$ by that factor (long division, or by comparing coefficients) to get a quadratic quotient.
3. Factorise the quadratic by splitting the middle term.
4. Write the answer as a product of three factors.

Check your answer by expanding mentally, or by testing another value of $x$ .

Worked example

Factorise $x^{3} - 2 x^{2} - 5 x + 6$ completely.

Try it first, then show the solutionSolution

Sum of coefficients $1 - 2 - 5 + 6 = 0$ , so $f (1) = 0$ and $(x - 1)$ is a factor.
Divide: $x^{3} - 2 x^{2} - 5 x + 6 = (x - 1) (x^{2} - x - 6)$ .
$x^{2} - x - 6 = (x - 3) (x + 2)$ .

Answer: (x − 1)(x − 3)(x + 2)

Cubics with leading coefficient other than 1

The method is the same, but the quotient has the same leading coefficient as the cubic, and the quadratic factor often splits into something like $(2 x - 1) (x + 3)$ . When dividing, line up the powers and include a $0$ coefficient for any missing power, for example $x^{3} + 0 x^{2} - 7 x + 6$ .

Worked example

Factorise $2 x^{3} + x^{2} - 13 x + 6$ completely.

Try it first, then show the solutionSolution

$f (2) = 16 + 4 - 26 + 6 = 0$ , so $(x - 2)$ is a factor.
Divide: quotient $2 x^{2} + 5 x - 3$ .
$2 x^{2} + 5 x - 3 = (2 x - 1) (x + 3)$ .

Answer: (x − 2)(2x − 1)(x + 3)

Maths in real life: Factorisation of Polynomials

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.

Factorising a volume expression

Suppose a toy company makes boxes whose volume, for a design number $x$ , is $x^{3} + 3 x^{2} + 2 x$ cm³.

Factorise the expression to find the edges of the box, and give the edges and volume for design number 4.

Try it first, then show the solutionSolution

$x^{3} + 3 x^{2} + 2 x = x (x^{2} + 3 x + 2) = x (x + 1) (x + 2)$ .
For $x = 4$ : edges 4 cm, 5 cm, 6 cm.
Volume $= 4 \times 5 \times 6 = 120$ cm³ (and $64 + 48 + 8 = 120$ ).

Answer: Edges x, x + 1, x + 2; for x = 4: 4, 5, 6 cm and 120 cm³.

Memory hook: Factors of a volume are the edges of the box.

Factor theorem to find the other edges

Imagine a water tank shaped like a cuboid whose volume is $2 x^{3} + 7 x^{2} + 7 x + 2$ m³, and you are told that one edge is $(x + 2)$ m.

Find the other two edges, and the volume when $x = 3$ .

Try it first, then show the solutionSolution

$f (- 2) = - 16 + 28 - 14 + 2 = 0$ , so $(x + 2)$ is a factor.
Divide: $2 x^{2} + 3 x + 1 = (2 x + 1) (x + 1)$ .
For $x = 3$ : edges 5 m, 7 m, 4 m; volume $= 140$ m³.

Answer: (2x + 1) m and (x + 1) m; 140 m³

Memory hook: Know one edge, divide it out, split the rest.

Roots of a cubic as break-even points

Suppose a small stall's weekly profit, in thousands of rupees, is modelled by $P (x) = x^{3} - 6 x^{2} + 11 x - 6$ , where $x$ is the number of hundreds of items made.

For which values of $x$ is the profit exactly zero?

Try it first, then show the solutionSolution

$P (1) = 1 - 6 + 11 - 6 = 0$ , so $(x - 1)$ is a factor.
$P (x) = (x - 1) (x^{2} - 5 x + 6) = (x - 1) (x - 2) (x - 3)$ .
Zero profit when $x = 1, 2$ or $3$ .

Answer: x = 1, 2 or 3 (100, 200 or 300 items)

Memory hook: Factorise to find where a cubic is zero.

Topics in this chapter

Factorisation of Polynomials formula sheet

Division identity: $f (x) = (x - a) q (x) + r$; r is a constant when the divisor is linear.
Remainder theorem: $r = f (a)$; For the divisor (x − a).
Remainder for (ax − b): $r = f (\frac{b}{a})$; Use the value of x that makes the divisor zero.
Factor theorem: $f (a) = 0 ⟺ (x - a) is a factor$; Works both ways.
Quick test for (x − 1): $f (1) = sum of coefficients$; Sum 0 means (x − 1) is a factor.
Making f exactly divisible: $f (x) - f (a)$; Subtract the remainder (or add its negative).

How the board asks Factorisation of Polynomials

The question patterns to expect, and the method that works for each.

Question type	How it is asked	Method
MCQ: remainder or number to add/subtract	Section A, Question 1: find the remainder, or what must be added or subtracted to make a polynomial divisible.	Compute f(a). Read 'added' or 'subtracted' carefully, or use k and solve.
MCQ: common factor	Which linear expression is a factor of two given polynomials?	Test each option in both polynomials with the factor theorem.
Find k, then factorise completely (3-4 marks)	A cubic with one unknown coefficient and a stated factor.	f(a) = 0 for k; divide by the factor; split the quadratic; write three factors.
Two unknowns from two conditions	One factor and one remainder (or two remainders) given.	Two equations from f(a) and f(b); solve simultaneously.
Equal remainders	Two polynomials leave the same remainder when divided by the same (x − a).	Set f(a) = g(a) and solve.
Factorise with no factor given	'Factorise completely' a cubic.	Trial values from the divisors of the constant term, then divide.

Tricks to remember Factorisation of Polynomials

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

Make the divisor zero
Whatever x makes the divisor zero is the x you put into f.
Use it when: x + 3 → f(−3); 2x − 1 → f(½).
Zero means factor
Remainder 0 ⇔ factor. 'Factor' in the question means 'set it equal to 0'.
Use it when: Finding k.
Three brackets at the end
A cubic's complete factorisation has three linear brackets (or one linear and one quadratic that will not split).
Use it when: Never stopping after finding k.
Add one, check with the sum
Sum of coefficients is f(1): zero means (x − 1) is a factor.
Use it when: Finding a first factor fast.

Factorisation of Polynomials: board-level solved questions

Ordered from board level to HOTS. Cover the solution, write your own, then compare step by step.

Example 1: A remainder · 1 mark · MCQBoard level

The remainder when $x^{3} - 4 x^{2} + 2 x + 5$ is divided by $(x - 3)$ is:
(a) 2
(b) −2
(c) 5
(d) 0

Try it first, then show the solutionSolution

$f (3) = 27 - 36 + 6 + 5 = 2$ .

Answer: (a) 2

Example 2: Factor test · 1 mark · Assertion-reasonBoard level

Assertion (A): $(x + 1)$ is a factor of $x^{3} + 1$ .
Reason (R): If $f (- 1) = 0$ , then $(x + 1)$ is a factor of $f (x)$ .
(a) (A) is true and (R) is false.
(b) (A) is false and (R) is true.
(c) Both (A) and (R) are true, and (R) is the correct explanation for (A).
(d) Both (A) and (R) are true, but (R) is not the correct explanation for (A).

Try it first, then show the solutionSolution

$(- 1)^{3} + 1 = 0$ , so by (R), $(x + 1)$ is a factor.

Answer: (c)

Example 3: Find k, then factorise · 4 marks · Short answerHarder board level

$(x - 3)$ is a factor of $2 x^{3} - 5 x^{2} + k x + 3$ . Find $k$ and factorise the polynomial completely.

Try it first, then show the solutionSolution

$f (3) = 54 - 45 + 3 k + 3 = 0$ , so $3 k = - 12$ , $k = - 4$ .
$2 x^{3} - 5 x^{2} - 4 x + 3 \div (x - 3) = 2 x^{2} + x - 1$ .
$2 x^{2} + x - 1 = (2 x - 1) (x + 1)$ .
$f (x) = (x - 3) (2 x - 1) (x + 1)$ .

Answer: k = −4; (x − 3)(2x − 1)(x + 1)

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

The polynomials $p x^{3} + 3 x^{2} - 3$ and $2 x^{3} - 5 x + p$ leave the same remainder when divided by $(x - 4)$ . Find $p$ .

Try it first, then show the solutionSolution

First remainder: $64 p + 48 - 3 = 64 p + 45$ .
Second: $128 - 20 + p = 108 + p$ .
$64 p + 45 = 108 + p \Rightarrow 63 p = 63$ .
$p = 1$ .

Answer: p = 1

Example 5: Remainder for a divisor (3x + 1) · 3 marks · Short answerHarder board level

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

Try it first, then show the solutionSolution

$3 x + 1 = 0 \Rightarrow x = - \frac{1}{3}$ .
$f (- \frac{1}{3}) = 3 (- \frac{1}{27}) - 2 (\frac{1}{9}) - \frac{1}{3} - 4 = - \frac{1}{9} - \frac{2}{9} - \frac{3}{9} - \frac{36}{9}$ .
$= - \frac{42}{9} = - \frac{14}{3}$ .

Answer: −14/3

Example 6: Factorise with no factor given · 3 marks · Long answerBoard level

Factorise completely: $x^{3} + 2 x^{2} - 5 x - 6$ .

Try it first, then show the solutionSolution

$f (- 1) = - 1 + 2 + 5 - 6 = 0$ , so $(x + 1)$ is a factor.
Quotient: $x^{2} + x - 6 = (x + 3) (x - 2)$ .
$f (x) = (x + 1) (x + 3) (x - 2)$ .

Answer: (x + 1)(x + 3)(x − 2)

Example 7: Two factors given: find the third · 4 marks · Long answerHOTS / competency

$(x - 2)$ and $(x + 3)$ are both factors of $x^{3} + a x^{2} + b x - 12$ . (a) Find $a$ and $b$ . (b) Find the third factor without long division, and explain your method. (c) Hence solve $x^{3} + a x^{2} + b x - 12 = 0$ .

Try it first, then show the solutionSolution

$f (2) = 0$ : $8 + 4 a + 2 b - 12 = 0 \Rightarrow 2 a + b = 2$ .
$f (- 3) = 0$ : $- 27 + 9 a - 3 b - 12 = 0 \Rightarrow 3 a - b = 13$ . Adding: $a = 3$ , $b = - 4$ .
(b) $f (x) = x^{3} + 3 x^{2} - 4 x - 12 = (x - 2) (x + 3) (x + c)$ . Comparing constant terms: $(- 2) (3) (c) = - 12$ , so $c = 2$ : third factor $(x + 2)$ . (Check: $f (- 2) = - 8 + 12 + 8 - 12 = 0$ .)
(c) $x = 2, - 3, - 2$ .

Answer: (a) a = 3, b = −4 (b) (x + 2) (c) x = 2, −3, −2

Example 8: Remainder involving the unknown · 3 marks · Long answerHarder board level

When $x^{3} + 3 x^{2} - m x + 4$ is divided by $(x - 2)$ , the remainder is $m + 3$ . Find $m$ and the remainder.

Try it first, then show the solutionSolution

$f (2) = 8 + 12 - 2 m + 4 = 24 - 2 m$ .
$24 - 2 m = m + 3 \Rightarrow 3 m = 21 \Rightarrow m = 7$ .
Remainder $= 7 + 3 = 10$ .

Answer: m = 7; remainder 10

Case-study practice (Section B style)

Read the situation, then answer every part. In the board paper this is a 10-mark question split into 3 + 3 + 4 marks.

Designing gift boxes

A packaging firm in Ahmedabad makes gift boxes in several sizes. For size number $x$ , the volume of the box is $V (x) = x^{3} + 6 x^{2} + 11 x + 6$ cm³, and each edge is a linear expression in $x$ .

(i) · 3 marks
Show that $(x + 1)$ is a factor of $V (x)$ .
Answer
Answer: V(−1) = 0
$V (- 1) = - 1 + 6 - 11 + 6 = 0$ , so by the factor theorem $(x + 1)$ is a factor.
(ii) · 3 marks
Factorise $V (x)$ completely and state the three edges.
Answer
Answer: (x + 1)(x + 2)(x + 3)
$V (x) \div (x + 1) = x^{2} + 5 x + 6 = (x + 2) (x + 3)$ . Edges: $(x + 1)$ , $(x + 2)$ , $(x + 3)$ cm.
(iii) · 4 marks
Find the edges and the volume for size 4, and check using $V (4)$ . Find also the remainder when $V (x)$ is divided by $(x - 1)$ and say what it represents.
Answer
Answer: 5, 6, 7 cm; 210 cm³; remainder 24
Edges 5, 6, 7 cm; volume 210 cm³. $V (4) = 64 + 96 + 44 + 6 = 210$ . Remainder $= V (1) = 1 + 6 + 11 + 6 = 24$ , which is the volume of the size-1 box (2 × 3 × 4).

A polynomial with two unknowns

A polynomial $f (x) = 2 x^{3} + a x^{2} + b x - 6$ has $(x - 2)$ as a factor, and leaves a remainder of $- 15$ when divided by $(x + 1)$ .

(i) · 3 marks
Form two equations in $a$ and $b$ .
Answer
Answer: 2a + b = −5 and a − b = −7
$f (2) = 16 + 4 a + 2 b - 6 = 0 \Rightarrow 2 a + b = - 5$ . $f (- 1) = - 2 + a - b - 6 = - 15 \Rightarrow a - b = - 7$ .
(ii) · 3 marks
Solve for $a$ and $b$ .
Answer
Answer: a = −4, b = 3
Adding: $3 a = - 12$ , $a = - 4$ ; $b = - 5 - 2 a = 3$ .
(iii) · 4 marks
Factorise $f (x)$ as far as possible and explain why it has only one linear factor.
Answer
Answer: (x − 2)(2x² + 3)
$f (x) = 2 x^{3} - 4 x^{2} + 3 x - 6 = 2 x^{2} (x - 2) + 3 (x - 2) = (x - 2) (2 x^{2} + 3)$ . For $2 x^{2} + 3$ , $D = 0 - 24 < 0$ , so it has no real roots and no linear factors (indeed $2 x^{2} + 3 > 0$ for every $x$ ).

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

$(x + 2)$ is a factor of $x^{3} + k x^{2} - x - 2$ . Find $k$ and factorise completely. (3 marks)

Step	Marks
$f (- 2) = - 8 + 4 k + 2 - 2 = 0 \Rightarrow k = 2$	1
$x^{3} + 2 x^{2} - x - 2 = x^{2} (x + 2) - (x + 2) = (x + 2) (x^{2} - 1)$	1
$= (x + 2) (x - 1) (x + 1)$	1

$(x - 3)$ is a factor of $x^{3} + k x^{2} - 9 x - 18$ . Find $k$ and factorise completely. (4 marks)

Step	Marks
$f (3) = 27 + 9 k - 27 - 18 = 0$	1
$k = 2$	1
$x^{3} + 2 x^{2} - 9 x - 18 \div (x - 3) = x^{2} + 5 x + 6$	1
$= (x - 3) (x + 2) (x + 3)$	1

$2 x^{3} + a x^{2} + b x + 4$ has $(x - 2)$ as a factor and leaves remainder $- 3$ when divided by $(x - 1)$ . Find $a$ and $b$ , and factorise completely. (5 marks)

Step	Marks
$f (2) = 0$ : $2 a + b = - 10$	1
$f (1) = - 3$ : $a + b = - 9$	1
$a = - 1$ , $b = - 8$	1
$2 x^{3} - x^{2} - 8 x + 4 \div (x - 2) = 2 x^{2} + 3 x - 2$	1
$= (x - 2) (2 x - 1) (x + 2)$	1

Where students lose marks in Factorisation of Polynomials

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

Mistake: Wrong sign for 'what must be subtracted'
CISCE examiners' comments, 2024: the remainder was 31, so 31 must be subtracted, but many gave −31 because they missed the words 'what must be subtracted'.
Do this: Let the number be $k$ ; write $f (a) - k = 0$ for 'subtracted' and $f (a) + k = 0$ for 'added', then solve.
Usually costs: 1 mark.
Mistake: Arithmetic slips when finding k
CISCE examiners' comments, 2024 and 2025: calculation errors gave a wrong k (e.g. k = −4 instead of the correct value), and everything after was wrong.
Do this: Write each term of $f (a)$ separately in brackets before adding, and recheck by substituting k back.
Usually costs: Usually all marks after the first step.
Mistake: Stopping after finding k
CISCE examiners' comments, 2024: candidates found k but did not factorise further.
Do this: The word 'hence' means the question continues: divide and factorise.
Usually costs: 2 marks of a 3-mark question.
Mistake: Errors in division
CISCE examiners' comments, 2024 and 2025: errors in dividing by $(x - 1)$ or $(x - 2)$ by long division.
Do this: Include zero coefficients for missing powers; after dividing, check that the quotient times the divisor gives back the cubic.
Usually costs: 1 to 2 marks.
Mistake: Final answer not written as a product
CISCE examiners' comments, 2024 and 2025: candidates did not write the answer as a product of three factors such as $(x - 2) (x - 1) (2 x - 1)$ .
Do this: Last line: ' $f (x) = () () ()$ '.
Usually costs: The final mark.

Practice questions on Factorisation of Polynomials

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
If $(x - 2)$ is a factor of $x^{2} + k x - 6$ , then $k$ is:
(a) −1
(b) 1
(c) 2
(d) 3
Hint
f(2) = 0.
Answer
(b) 1
Q2 · 1 mark · MCQBoard level
The remainder when $x^{3} + 1$ is divided by $(x + 2)$ is:
(a) 9
(b) −9
(c) 7
(d) −7
Hint
f(−2).
Answer
(d) −7
Q3 · 1 mark · MCQHarder board level
What must be added to $x^{3} - 3 x^{2} + 4 x - 1$ to make it exactly divisible by $(x - 1)$ ?
(a) 1
(b) −1
(c) 2
(d) 0
Hint
f(1) + k = 0.
Answer
(b) −1, since f(1) = 1.
Q4 · 1 mark · Assertion-reasonBoard level
Assertion (A): $(x - 1)$ is a factor of $3 x^{3} - 5 x^{2} + x + 1$ .
Reason (R): $f (1)$ is the sum of the coefficients of $f (x)$ .
(a) (A) is true and (R) is false.
(b) (A) is false and (R) is true.
(c) Both (A) and (R) are true, and (R) is the correct explanation for (A).
(d) Both (A) and (R) are true, but (R) is not the correct explanation for (A).
Hint
Add the coefficients.
Answer
(c): $3 - 5 + 1 + 1 = 0 = f (1)$ .
Q5 · 4 marks · Short answerHarder board level
$(x + 2)$ is a factor of $3 x^{3} + k x^{2} - 17 x + 6$ . Find $k$ and factorise the polynomial completely.
Hint
f(−2) = 0.
Answer
$- 24 + 4 k + 34 + 6 = 0 \Rightarrow k = - 4$ . Quotient $3 x^{2} - 10 x + 3 = (3 x - 1) (x - 3)$ . So $(x + 2) (3 x - 1) (x - 3)$ .
Q6 · 4 marks · Short answerHarder board level
$x^{3} + p x^{2} + q x + 4$ leaves remainder 5 when divided by $(x - 1)$ and remainder 7 when divided by $(x + 1)$ . Find $p$ and $q$ .
Hint
Two equations.
Answer
$f (1) = 5 + p + q = 5 \Rightarrow p + q = 0$ . $f (- 1) = 3 + p - q = 7 \Rightarrow p - q = 4$ . So $p = 2$ , $q = - 2$ .
Q7 · 3 marks · Long answerBoard level
Factorise completely: $2 x^{3} - 3 x^{2} - 3 x + 2$ .
Hint
Try x = −1.
Answer
$f (- 1) = 0$ ; quotient $2 x^{2} - 5 x + 2 = (2 x - 1) (x - 2)$ . So $(x + 1) (2 x - 1) (x - 2)$ .
Q8 · 4 marks · Long answerHarder board level
Show that $(2 x + 1)$ is a factor of $2 x^{3} + 3 x^{2} - 11 x - 6$ , and hence factorise it completely.
Hint
f(−½).
Answer
$f (- \frac{1}{2}) = - \frac{1}{4} + \frac{3}{4} + \frac{11}{2} - 6 = 0$ . Quotient $x^{2} + x - 6 = (x + 3) (x - 2)$ . So $(2 x + 1) (x + 3) (x - 2)$ .
Q9 · 4 marks · Long answerHOTS / competency
$(x^{2} - 1)$ is a factor of $x^{3} + a x^{2} + b x - 2$ . Find $a$ and $b$ , and the remaining factor, without long division.
Hint
x² − 1 = (x − 1)(x + 1): use f(1) = 0 and f(−1) = 0.
Answer
$f (1) = 0$ : $a + b = 1$ . $f (- 1) = 0$ : $a - b = 3$ . So $a = 2$ , $b = - 1$ . $x^{3} + 2 x^{2} - x - 2 = (x^{2} - 1) (x + c)$ ; constant terms give $- c = - 2$ , $c = 2$ : remaining factor $(x + 2)$ .
Q10 · 3 marks · Long answerHarder board level
Find the remainder when $x^{3} - 6 x^{2} + 2 x - 4$ is divided by $(1 - 2 x)$ .
Hint
1 − 2x = 0 at x = ½.
Answer
$f (\frac{1}{2}) = \frac{1}{8} - \frac{3}{2} + 1 - 4 = - \frac{35}{8}$ .
Q11 · 1 mark · MCQHOTS / competency
Which of these is NOT a factor of $x^{3} - 7 x + 6$ ?
(a) $x - 1$
(b) $x - 2$
(c) $x + 3$
(d) $x + 2$
Hint
A cubic has at most three linear factors.
Answer
(d): $f (- 2) = - 8 + 14 + 6 = 12 \neq = 0$ ; the other three give 0.
Q12 · 4 marks · Short answerHOTS / competency
A cubic polynomial $f (x)$ leaves remainder 0 when divided by $(x - 1)$ , $(x - 2)$ and $(x - 3)$ , and $f (0) = - 12$ . Find $f (x)$ .
Hint
f(x) = k(x − 1)(x − 2)(x − 3).
Answer
$f (0) = k (- 1) (- 2) (- 3) = - 6 k = - 12 \Rightarrow k = 2$ . $f (x) = 2 (x - 1) (x - 2) (x - 3) = 2 x^{3} - 12 x^{2} + 22 x - 12$ .

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.State the factor theorem.Show answer
(x − a) is a factor of f(x) if and only if f(a) = 0.
3.What value do you substitute for the divisor (2x − 3)?Show answer
x = 3/2
4.What does it mean if the sum of the coefficients is 0?Show answer
f(1) = 0, so (x − 1) is a factor.
5.Remainder when x³ − 1 is divided by x − 1?Show answer
0
6.What is the maximum degree of f(x) in the ICSE Class 10 syllabus?Show answer
3
7.After finding one factor of a cubic, what next?Show answer
Divide to get a quadratic and factorise it.
8.What must be subtracted from f(x) to make it divisible by (x − a)?Show answer
f(a)

Best order to study it

Learn the two theorems and practise substituting negative and fractional values.
Practise finding one unknown from a factor or a remainder condition.
Practise 'what must be added or subtracted'.
Practise complete factorisation, including leading coefficient 2 or 3.
Finish with two-unknown questions and equal-remainder questions.

Quick revision checklist

The night before the exam: can you do each of these without looking?

Can you find the remainder for divisors (x − a), (x + a) and (ax − b)?
Can you test whether a linear expression is a factor?
Can you find k from a factor or remainder condition?
Can you answer 'what must be added/subtracted' with the right sign?
Can you divide a cubic by a linear factor correctly?
Do you always write the final answer as a product of three factors?
Can you set up and solve two equations for two unknown coefficients?

Factorisation of Polynomials: frequently asked questions

Do I have to use long division to factorise a cubic?

No. Any correct method is fine: long division, comparing coefficients, or regrouping terms. Show enough working for the examiner to follow.

How do I choose which value to try first?

Try the divisors of the constant term: 1, −1, 2, −2 and so on. Check the sum of coefficients first, because that tells you about x = 1 straight away.

What if the quadratic factor does not split?

If its discriminant is negative, it has no linear factors. Write the answer as the linear factor times the quadratic.

Are polynomials of degree 4 in the syllabus?

No. The 2027 syllabus limits f(x) to degree 3.

Is the remainder theorem the same as finding f(a)?

Yes: the remainder on dividing by (x − a) equals f(a). That is why you can find remainders without dividing.

Based on: CISCE ICSE 2027 Mathematics syllabus; CISCE ICSE 2027 Mathematics specimen question paper and marking scheme; CISCE Analysis of Pupil Performance, ICSE Mathematics, 2025; CISCE Analysis of Pupil Performance, ICSE Mathematics, 2024