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Factorisation — ICSE Class 9 Maths Important Questions

13 hand-picked ICSE Class 9 Maths important questions for Factorisation, each with a full model answer — the formats and topics most likely to appear in your board exam.

13
Questions
6
Question types
32
Total marks
₹0
With answers
Quick answer

High-yield ICSE Factorisation questions use the factor theorem (if f(a)=0f(a)=0f(a)=0 then (x−a)(x-a)(x-a) is a factor), factorising by grouping, splitting the middle term for quadratics, and factorising cubic polynomials. Using given factors to find unknown constants and fully factorising a cubic appear almost every year.

Syllabus reference (2026–27): CISCE (Council for the Indian School Certificate Examinations)

About Factorisation

In the ICSE Class 9 Maths chapter Factorisation you factorise algebraic expressions by grouping, split the middle term of quadratics, apply the factor theorem and remainder theorem to polynomials, and factorise cubic expressions completely using these methods.

Factorisation by groupingSplitting the middle termRemainder theoremFactor theoremFactorising cubic polynomials

Key concepts & formulas

Factor theorem

For a polynomial f(x)f(x)f(x), (x−a)(x-a)(x-a) is a factor if and only if f(a)=0f(a)=0f(a)=0. More generally (bx−a)(bx-a)(bx-a) is a factor iff f ⁣(ab)=0f\!\left(\dfrac{a}{b}\right)=0f\!(a/b)=0.

Remainder theorem

When f(x)f(x)f(x) is divided by (x−a)(x-a)(x-a), the remainder is f(a)f(a)f(a).

Splitting the middle term

To factorise ax2+bx+cax^2+bx+cax^2+bx+c, find two numbers with product acacac and sum bbb, then group.

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Important questions with answers

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Question typeCountMarks
MCQ41
Assertion–Reason11
Very Short22
Short Answer33
Long Answer25
Case-based14

Multiple-choice questions (1 mark)

Q1MCQEasy1 mark

The factors of x2−9x^2-9x^2-9 are:

  1. (a)

    (x−3)(x−3)(x-3)(x-3)(x-3)(x-3)

  2. (b)

    (x+3)(x−3)(x+3)(x-3)(x+3)(x-3)

  3. (c)

    (x+9)(x−1)(x+9)(x-1)(x+9)(x-1)

  4. (d)

    (x−9)(x+1)(x-9)(x+1)(x-9)(x+1)

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Answer: (b) (x+3)(x−3)(x+3)(x-3)(x+3)(x-3).

Using a2−b2=(a+b)(a−b)a^2-b^2=(a+b)(a-b)a^2-b^2=(a+b)(a-b), x2−9=x2−32=(x+3)(x−3).x^2-9=x^2-3^2=(x+3)(x-3).x^2-9=x^2-3^2=(x+3)(x-3).

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Q2MCQEasy1 mark

By the factor theorem, (x−2)(x-2)(x-2) is a factor of f(x)f(x)f(x) if:

  1. (a)

    f(0)=2f(0)=2f(0)=2

  2. (b)

    f(2)=0f(2)=0f(2)=0

  3. (c)

    f(−2)=0f(-2)=0f(-2)=0

  4. (d)

    f(2)=2f(2)=2f(2)=2

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Answer: (b) f(2)=0f(2)=0f(2)=0.

The factor theorem states (x−a)(x-a)(x-a) is a factor of f(x)f(x)f(x) exactly when f(a)=0f(a)=0f(a)=0; here a=2a=2a=2.

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Q3MCQModerate1 mark

The remainder when x3−3x2+2x−5x^3-3x^2+2x-5x^3-3x^2+2x-5 is divided by (x−1)(x-1)(x-1) is:

  1. (a)

    −5-5-5

  2. (b)

    −4-4-4

  3. (c)

    −3-3-3

  4. (d)

    000

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Answer: (a) −5-5-5.

By the remainder theorem, remainder =f(1)=1−3+2−5=−5.=f(1)=1-3+2-5=-5.=f(1)=1-3+2-5=-5.

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Q4MCQHOTS1 mark

If (x−1)(x-1)(x-1) is a factor of x3−2x2+kx−2x^3-2x^2+kx-2x^3-2x^2+kx-2, then kkk equals:

  1. (a)

    111

  2. (b)

    222

  3. (c)

    333

  4. (d)

    −3-3-3

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Answer: (c) 333.

By the factor theorem f(1)=0f(1)=0f(1)=0: 1−2+k−2=0⇒k−3=0⇒k=3.1-2+k-2=0\Rightarrow k-3=0\Rightarrow k=3.1-2+k-2=0 k-3=0 k=3.

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Assertion–Reason questions (1 mark)

Q5Assertion–ReasonModerate1 mark

Assertion (A): (x+1)(x+1)(x+1) is a factor of x3+1x^3+1x^3+1.

Reason (R): For a polynomial f(x)f(x)f(x), (x−a)(x-a)(x-a) is a factor if f(a)=0f(a)=0f(a)=0.

  1. (a)

    Both A and R are true and R is the correct explanation of A

  2. (b)

    Both A and R are true but R is not the correct explanation of A

  3. (c)

    A is true but R is false

  4. (d)

    A is false but R is true

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Answer: (a) Both are true and R correctly explains A. Here a=−1a=-1a=-1 and f(−1)=(−1)3+1=0f(-1)=(-1)^3+1=0f(-1)=(-1)^3+1=0, so by the factor theorem (x+1)(x+1)(x+1) is a factor of x3+1x^3+1x^3+1.

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Very short answer questions (2 marks)

Q6Very ShortEasy2 marks

Factorise: x2+7x+12x^2+7x+12x^2+7x+12.

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Split the middle term: two numbers with product 121212 and sum 777 are 333 and 444.

x2+7x+12=x2+3x+4x+12=x(x+3)+4(x+3)=(x+3)(x+4).x^2+7x+12=x^2+3x+4x+12=x(x+3)+4(x+3)=(x+3)(x+4).x^2+7x+12=x^2+3x+4x+12=x(x+3)+4(x+3)=(x+3)(x+4).

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Q7Very ShortModerate2 marks

Factorise by grouping: ax+ay+bx+byax+ay+bx+byax+ay+bx+by.

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Group in pairs and take out common factors:

ax+ay+bx+by=a(x+y)+b(x+y)=(x+y)(a+b).ax+ay+bx+by=a(x+y)+b(x+y)=(x+y)(a+b).ax+ay+bx+by=a(x+y)+b(x+y)=(x+y)(a+b).

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Short answer questions (3 marks)

Q8Short AnswerModerate3 marks

Factorise: 6x2−x−156x^2-x-156x^2-x-15.

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Here a=6, b=−1, c=−15a=6,\ b=-1,\ c=-15a=6, b=-1, c=-15, so ac=−90ac=-90ac=-90. Two numbers with product −90-90-90 and sum −1-1-1 are −10-10-10 and 999.

6x2−x−15=6x2−10x+9x−156x^2-x-15=6x^2-10x+9x-156x^2-x-15=6x^2-10x+9x-15

=2x(3x−5)+3(3x−5)=2x(3x-5)+3(3x-5)=2x(3x-5)+3(3x-5)

=(3x−5)(2x+3).=(3x-5)(2x+3).=(3x-5)(2x+3).

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Q9Short AnswerModerate3 marks

Using the factor theorem, show that (x−3)(x-3)(x-3) is a factor of x3−19x−30x^3-19x-30x^3-19x-30 is incorrect, and instead find f(3)f(3)f(3) and f(−2)f(-2)f(-2) for f(x)=x3−19x−30f(x)=x^3-19x-30f(x)=x^3-19x-30.

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Let f(x)=x3−19x−30f(x)=x^3-19x-30f(x)=x^3-19x-30.

f(3)=27−57−30=−60≠0f(3)=27-57-30=-60\neq0f(3)=27-57-30=-60≠0, so (x−3)(x-3)(x-3) is not a factor.

f(−2)=(−2)3−19(−2)−30=−8+38−30=0f(-2)=(-2)^3-19(-2)-30=-8+38-30=0f(-2)=(-2)^3-19(-2)-30=-8+38-30=0, so (x+2)(x+2)(x+2) is a factor.

Thus by the factor theorem (x+2)(x+2)(x+2) divides f(x)f(x)f(x), while (x−3)(x-3)(x-3) does not.

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Q10Short AnswerHOTS3 marks

If (x−2)(x-2)(x-2) and (x+3)(x+3)(x+3) are both factors of x3+ax2+bx−6x^3+ax^2+bx-6x^3+ax^2+bx-6... actually find aaa and bbb if (x−2)(x-2)(x-2) and (x−1)(x-1)(x-1) are factors of x3+ax2+bx−2x^3+ax^2+bx-2x^3+ax^2+bx-2.

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Let f(x)=x3+ax2+bx−2f(x)=x^3+ax^2+bx-2f(x)=x^3+ax^2+bx-2.

Since (x−1)(x-1)(x-1) is a factor, f(1)=0f(1)=0f(1)=0: 1+a+b−2=0⇒a+b=11+a+b-2=0\Rightarrow a+b=11+a+b-2=0 a+b=1 ...(1)

Since (x−2)(x-2)(x-2) is a factor, f(2)=0f(2)=0f(2)=0: 8+4a+2b−2=0⇒4a+2b=−6⇒2a+b=−38+4a+2b-2=0\Rightarrow4a+2b=-6\Rightarrow2a+b=-38+4a+2b-2=04a+2b=-62a+b=-3 ...(2)

Subtract (1) from (2): (2a+b)−(a+b)=−3−1⇒a=−4(2a+b)-(a+b)=-3-1\Rightarrow a=-4(2a+b)-(a+b)=-3-1 a=-4.

From (1): b=1−a=1−(−4)=5b=1-a=1-(-4)=5b=1-a=1-(-4)=5.

∴ a=−4, b=5.\therefore\ a=-4,\ b=5.a=-4, b=5.

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Long answer questions (5 marks)

Q11Long AnswerModerate5 marks

Factorise completely: x3−6x2+11x−6x^3-6x^2+11x-6x^3-6x^2+11x-6.

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Let f(x)=x3−6x2+11x−6f(x)=x^3-6x^2+11x-6f(x)=x^3-6x^2+11x-6. Try small factors of the constant 666.

f(1)=1−6+11−6=0f(1)=1-6+11-6=0f(1)=1-6+11-6=0, so (x−1)(x-1)(x-1) is a factor.

Divide f(x)f(x)f(x) by (x−1)(x-1)(x-1):

x3−6x2+11x−6=(x−1)(x2−5x+6).x^3-6x^2+11x-6=(x-1)(x^2-5x+6).x^3-6x^2+11x-6=(x-1)(x^2-5x+6).

Factorise the quadratic: x2−5x+6=(x−2)(x−3).x^2-5x+6=(x-2)(x-3).x^2-5x+6=(x-2)(x-3).

∴ x3−6x2+11x−6=(x−1)(x−2)(x−3).\therefore\ x^3-6x^2+11x-6=(x-1)(x-2)(x-3).x^3-6x^2+11x-6=(x-1)(x-2)(x-3).

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Q12Long AnswerHOTS5 marks

Using the factor theorem, factorise completely: 2x3+x2−13x+62x^3+x^2-13x+62x^3+x^2-13x+6.

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Let f(x)=2x3+x2−13x+6f(x)=2x^3+x^2-13x+6f(x)=2x^3+x^2-13x+6. Test factors of 62\dfrac{6}{2}6/2.

f(2)=2(8)+4−26+6=16+4−26+6=0f(2)=2(8)+4-26+6=16+4-26+6=0f(2)=2(8)+4-26+6=16+4-26+6=0, so (x−2)(x-2)(x-2) is a factor.

Divide f(x)f(x)f(x) by (x−2)(x-2)(x-2) (long division):

2x3+x2−13x+6=(x−2)(2x2+5x−3).2x^3+x^2-13x+6=(x-2)(2x^2+5x-3).2x^3+x^2-13x+6=(x-2)(2x^2+5x-3).

Factorise 2x2+5x−32x^2+5x-32x^2+5x-3: product =2×(−3)=−6=2\times(-3)=-6=2×(-3)=-6, sum =5=5=5, so numbers are 666 and −1-1-1.

2x2+5x−3=2x2+6x−x−3=2x(x+3)−1(x+3)=(x+3)(2x−1).2x^2+5x-3=2x^2+6x-x-3=2x(x+3)-1(x+3)=(x+3)(2x-1).2x^2+5x-3=2x^2+6x-x-3=2x(x+3)-1(x+3)=(x+3)(2x-1).

∴ 2x3+x2−13x+6=(x−2)(x+3)(2x−1).\therefore\ 2x^3+x^2-13x+6=(x-2)(x+3)(2x-1).2x^3+x^2-13x+6=(x-2)(x+3)(2x-1).

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Case-based questions (4 marks)

Q13Case-basedModerate4 marks

A teacher writes the polynomial f(x)=x3+2x2−5x−6f(x)=x^3+2x^2-5x-6f(x)=x^3+2x^2-5x-6 and asks the class to factorise it using the factor theorem.

(i) Evaluate f(−1)f(-1)f(-1).

(ii) State one linear factor from part (i).

(iii) Divide f(x)f(x)f(x) by that factor to get the quadratic factor.

(iv) Factorise f(x)f(x)f(x) completely.

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(i) f(−1)=(−1)3+2(−1)2−5(−1)−6=−1+2+5−6=0.f(-1)=(-1)^3+2(-1)^2-5(-1)-6=-1+2+5-6=0.f(-1)=(-1)^3+2(-1)^2-5(-1)-6=-1+2+5-6=0.

(ii) Since f(−1)=0f(-1)=0f(-1)=0, by the factor theorem (x+1)(x+1)(x+1) is a factor.

(iii) Dividing f(x)f(x)f(x) by (x+1)(x+1)(x+1) gives x3+2x2−5x−6=(x+1)(x2+x−6).x^3+2x^2-5x-6=(x+1)(x^2+x-6).x^3+2x^2-5x-6=(x+1)(x^2+x-6).

(iv) Factorise x2+x−6=(x+3)(x−2)x^2+x-6=(x+3)(x-2)x^2+x-6=(x+3)(x-2).

∴ f(x)=(x+1)(x+3)(x−2).\therefore\ f(x)=(x+1)(x+3)(x-2).f(x)=(x+1)(x+3)(x-2).

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All ICSE Class 9 Maths Chapters

Rational and Irrational NumbersCompound Interest (Without Using Formula)Compound Interest (Using Formula)ExpansionsFactorisationSimultaneous (Linear) EquationsIndices (Exponents)LogarithmsTriangles (Congruency in Triangles)Isosceles TrianglesInequalitiesMid-point and Its Converse (Including Intercept Theorem)Pythagoras TheoremRectilinear Figures (Quadrilaterals)Construction of PolygonsArea TheoremsCircleStatistics (Classification of Data, Tabulation)Mean and MedianArea and Perimeter of Plane FiguresSolids (Surface Area and Volume of 3-D Solids)Trigonometrical RatiosTrigonometrical Ratios of Standard AnglesSolution of Right TrianglesComplementary AnglesCo-ordinate GeometryGraphical Solution (Simultaneous Linear Equations, Graphically)Distance Formula

Frequently asked questions

  • Are these Factorisation important questions free?
    Yes. All 13 ICSE Class 9 Maths important questions for Factorisation are free, with full model answers and no login required.
  • Do these Factorisation questions follow the latest ICSE syllabus?
    Yes — they are aligned to the CISCE latest syllabus syllabus for ICSE Class 9 Maths, so nothing here is outside the current course.
  • How should I practise the Factorisation important questions?
    Attempt each question on paper first, then reveal the model answer to check your method — not just the final result. Re-do anything you got wrong the same day.
  • What types of questions are covered for Factorisation?
    A full mix — multiple-choice questions, assertion–reason questions, very short answer questions, short answer questions, long answer questions, case-based questions — so every format in the ICSE paper is covered.

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