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Factorisation Class 9 ICSE Important Questions with Answers

The 14 questions below are Factorisation Class 9 ICSE important questions, each with a full step-by-step answer: extra questions beyond the textbook exercises, covering 5 topics of ICSE Class 9 Maths in the question formats used in the exam.

By The Classmate AI Editorial Team

Reviewed by Classmate AI Team · 17 October 2026

Last updated 25 October 2026

14
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Key concepts
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Quick answer

ICSE Class 9 Factorisation questions ask you to factorise using the three forms in the syllabus: the difference of two squares a2−b2a^2-b^2a²-b², the sum and difference of two cubes a3±b3a^3\pm b^3a³± b³, and ax2+bx+cax^2+bx+cax²+bx+c by splitting the middle term. Take out any common factor first, and check each answer by expanding it.

Syllabus reference (2026–27): ICSE 2028 Mathematics syllabus (PDF) · CISCE (Council for the Indian School Certificate Examinations)

About Factorisation

In the ICSE Class 9 Maths chapter Factorisation you write algebraic expressions as products of their factors. The CISCE syllabus lists three forms: the difference of two squares, the sum and difference of two cubes, and quadratic trinomials ax2+bx+cax^2+bx+cax²+bx+c factorised by splitting the middle term. Taking out a common factor and grouping terms are the tools you use along the way. The Factor and Remainder Theorems are not part of Class 9; they come in ICSE Class 10.

Common factors and groupingDifference of two squaresSum and difference of two cubesSplitting the middle termFactorising by substitution

Key concepts & formulas

Difference of two squares

a2−b2=(a+b)(a−b)a^2-b^2=(a+b)(a-b)a²-b²=(a+b)(a-b), e.g. 25x2−9=(5x+3)(5x−3)25x^2-9=(5x+3)(5x-3)25x²-9=(5x+3)(5x-3). A sum of two squares such as x2+9x^2+9x²+9 does not factorise.

Sum and difference of two cubes

a3+b3=(a+b)(a2−ab+b2)a^3+b^3=(a+b)(a^2-ab+b^2)a³+b³=(a+b)(a²-ab+b²) and a3−b3=(a−b)(a2+ab+b2)a^3-b^3=(a-b)(a^2+ab+b^2)a³-b³=(a-b)(a²+ab+b²).

Splitting the middle term

For ax2+bx+cax^2+bx+cax²+bx+c, find two numbers with product acacac and sum bbb, split bxbxbx into those two terms, then group.

Common factor first

Always take out the HCF of the terms before using an identity, e.g. 50x2−18=2(25x2−9)50x^2-18=2(25x^2-9)50x²-18=2(25x²-9).

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Factorisation Questions and Answers

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MCQs with answers (1 mark)

Q1MCQEasy1 mark

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

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Q2MCQEasy1 mark

The factors of 8x3+278x^3+278x³+27 are:

  1. (a)

    (2x+3)(4x2−6x+9)(2x+3)(4x^2-6x+9)(2x+3)(4x²-6x+9)

  2. (b)

    (2x+3)(4x2+6x+9)(2x+3)(4x^2+6x+9)(2x+3)(4x²+6x+9)

  3. (c)

    (2x−3)(4x2+6x+9)(2x-3)(4x^2+6x+9)(2x-3)(4x²+6x+9)

  4. (d)

    (2x+3)(4x2−12x+9)(2x+3)(4x^2-12x+9)(2x+3)(4x²-12x+9)

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Answer: (a) (2x+3)(4x2−6x+9)(2x+3)(4x^2-6x+9)(2x+3)(4x²-6x+9).

Write it as a sum of two cubes: 8x3+27=(2x)3+338x^3+27=(2x)^3+3^38x³+27=(2x)³+3³. Using a3+b3=(a+b)(a2−ab+b2)a^3+b^3=(a+b)(a^2-ab+b^2)a³+b³=(a+b)(a²-ab+b²) with a=2x, b=3a=2x,\ b=3a=2x, b=3:

(2x+3)((2x)2−(2x)(3)+32)=(2x+3)(4x2−6x+9).(2x+3)\big((2x)^2-(2x)(3)+3^2\big)=(2x+3)(4x^2-6x+9).(2x+3)((2x)²-(2x)(3)+3²)=(2x+3)(4x²-6x+9).

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Q3MCQModerate1 mark

4a2−(b+c)24a^2-(b+c)^24a²-(b+c)² factorises as:

  1. (a)

    (2a+b+c)(2a−b−c)(2a+b+c)(2a-b-c)(2a+b+c)(2a-b-c)

  2. (b)

    (2a+b−c)(2a−b+c)(2a+b-c)(2a-b+c)(2a+b-c)(2a-b+c)

  3. (c)

    (4a+b+c)(a−b−c)(4a+b+c)(a-b-c)(4a+b+c)(a-b-c)

  4. (d)

    (2a−b−c)2(2a-b-c)^2(2a-b-c)²

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Answer: (a) (2a+b+c)(2a−b−c)(2a+b+c)(2a-b-c)(2a+b+c)(2a-b-c).

It is a difference of two squares with A=2aA=2aA=2a and B=b+cB=b+cB=b+c: A2−B2=(A+B)(A−B)=(2a+b+c)(2a−b−c).A^2-B^2=(A+B)(A-B)=(2a+b+c)(2a-b-c).A²-B²=(A+B)(A-B)=(2a+b+c)(2a-b-c). Keep the bracket together: the minus sign applies to both bbb and ccc.

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Q4MCQModerate1 mark

To factorise 3x2−10x+83x^2-10x+83x²-10x+8 by splitting the middle term, −10x-10x-10x should be written as:

  1. (a)

    −6x−4x-6x-4x-6x-4x

  2. (b)

    −8x−2x-8x-2x-8x-2x

  3. (c)

    −12x+2x-12x+2x-12x+2x

  4. (d)

    −5x−5x-5x-5x-5x-5x

Show model answer

Answer: (a) −6x−4x-6x-4x-6x-4x.

We need two numbers whose product is 3×8=243\times8=243×8=24 and whose sum is −10-10-10: they are −6-6-6 and −4-4-4.

3x2−6x−4x+8=3x(x−2)−4(x−2)=(x−2)(3x−4).3x^2-6x-4x+8=3x(x-2)-4(x-2)=(x-2)(3x-4).3x²-6x-4x+8=3x(x-2)-4(x-2)=(x-2)(3x-4).

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Assertion–Reason questions (1 mark)

Q5Assertion–ReasonModerate1 mark

Assertion (A): x3−8=(x−2)(x2+2x+4)x^3-8=(x-2)(x^2+2x+4)x³-8=(x-2)(x²+2x+4).

Reason (R): a3−b3=(a−b)(a2+ab+b2)a^3-b^3=(a-b)(a^2+ab+b^2)a³-b³=(a-b)(a²+ab+b²).

  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

Show model answer

Answer: (a) Both are true and R correctly explains A. With a=x, b=2a=x,\ b=2a=x, b=2: x3−23=(x−2)(x2+2x+4).x^3-2^3=(x-2)(x^2+2x+4).x³-2³=(x-2)(x²+2x+4).

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Very short answer questions (2 marks)

Q6Very ShortEasy2 marks

Factorise: x2+7x+12x^2+7x+12x²+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).x²+7x+12=x²+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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Q8Very ShortEasy2 marks

Factorise: 50x2−1850x^2-1850x²-18.

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Take out the common factor 222 first, then use a2−b2=(a+b)(a−b)a^2-b^2=(a+b)(a-b)a²-b²=(a+b)(a-b):

50x2−18=2(25x2−9)=2((5x)2−32)=2(5x+3)(5x−3).50x^2-18=2(25x^2-9)=2\big((5x)^2-3^2\big)=2(5x+3)(5x-3).50x²-18=2(25x²-9)=2((5x)²-3²)=2(5x+3)(5x-3).

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Short answer questions (3 marks)

Q9Short AnswerModerate3 marks

Factorise: 6x2−x−156x^2-x-156x²-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-156x²-x-15=6x²-10x+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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Q10Short AnswerModerate3 marks

Factorise completely: x4−81y4x^4-81y^4x⁴-81y⁴.

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Use the difference of two squares twice.

x4−81y4=(x2)2−(9y2)2=(x2+9y2)(x2−9y2)x^4-81y^4=(x^2)^2-(9y^2)^2=(x^2+9y^2)(x^2-9y^2)x⁴-81y⁴=(x²)²-(9y²)²=(x²+9y²)(x²-9y²)

x2−9y2=(x+3y)(x−3y)x^2-9y^2=(x+3y)(x-3y)x²-9y²=(x+3y)(x-3y), but x2+9y2x^2+9y^2x²+9y² (a sum of squares) does not factorise.

∴ x4−81y4=(x2+9y2)(x+3y)(x−3y).\therefore\ x^4-81y^4=(x^2+9y^2)(x+3y)(x-3y).∴ x⁴-81y⁴=(x²+9y²)(x+3y)(x-3y).

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Q11Short AnswerHOTS3 marks

Factorise: (x+y)2−5(x+y)+6(x+y)^2-5(x+y)+6(x+y)²-5(x+y)+6.

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Put p=x+yp=x+yp=x+y. The expression becomes p2−5p+6p^2-5p+6p²-5p+6.

Two numbers with product 666 and sum −5-5-5 are −2-2-2 and −3-3-3, so p2−5p+6=(p−2)(p−3)p^2-5p+6=(p-2)(p-3)p²-5p+6=(p-2)(p-3).

Put p=x+yp=x+yp=x+y back: (x+y)2−5(x+y)+6=(x+y−2)(x+y−3).(x+y)^2-5(x+y)+6=(x+y-2)(x+y-3).(x+y)²-5(x+y)+6=(x+y-2)(x+y-3).

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Long answer questions (5 marks)

Q12Long AnswerHOTS5 marks

Factorise completely: a6−b6a^6-b^6a⁶-b⁶.

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Treat it first as a difference of two squares: a6−b6=(a3)2−(b3)2=(a3+b3)(a3−b3).a^6-b^6=(a^3)^2-(b^3)^2=(a^3+b^3)(a^3-b^3).a⁶-b⁶=(a³)²-(b³)²=(a³+b³)(a³-b³).

Now use the sum and difference of two cubes:

a3+b3=(a+b)(a2−ab+b2)a^3+b^3=(a+b)(a^2-ab+b^2)a³+b³=(a+b)(a²-ab+b²) and a3−b3=(a−b)(a2+ab+b2).a^3-b^3=(a-b)(a^2+ab+b^2).a³-b³=(a-b)(a²+ab+b²).

∴ a6−b6=(a+b)(a−b)(a2−ab+b2)(a2+ab+b2).\therefore\ a^6-b^6=(a+b)(a-b)(a^2-ab+b^2)(a^2+ab+b^2).∴ a⁶-b⁶=(a+b)(a-b)(a²-ab+b²)(a²+ab+b²).

Starting with the difference of squares gives all four factors directly; starting with (a2)3−(b2)3(a^2)^3-(b^2)^3(a²)³-(b²)³ leaves a4+a2b2+b4a^4+a^2b^2+b^4a⁴+a²b²+b⁴, which needs an extra step to split.

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Q13Long AnswerModerate5 marks

Factorise: (i) 2x3+54y32x^3+54y^32x³+54y³ (ii) a3−b3−a+ba^3-b^3-a+ba³-b³-a+b.

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(i) Take out the common factor 222: 2x3+54y3=2(x3+27y3)=2(x3+(3y)3)2x^3+54y^3=2(x^3+27y^3)=2\big(x^3+(3y)^3\big)2x³+54y³=2(x³+27y³)=2(x³+(3y)³).

Using a3+b3=(a+b)(a2−ab+b2)a^3+b^3=(a+b)(a^2-ab+b^2)a³+b³=(a+b)(a²-ab+b²): 2(x+3y)(x2−3xy+9y2).2(x+3y)(x^2-3xy+9y^2).2(x+3y)(x²-3xy+9y²).

(ii) Group the cube terms and the linear terms: a3−b3−a+b=(a3−b3)−(a−b).a^3-b^3-a+b=(a^3-b^3)-(a-b).a³-b³-a+b=(a³-b³)-(a-b).

=(a−b)(a2+ab+b2)−(a−b)=(a-b)(a^2+ab+b^2)-(a-b)=(a-b)(a²+ab+b²)-(a-b)

=(a−b)(a2+ab+b2−1).=(a-b)(a^2+ab+b^2-1).=(a-b)(a²+ab+b²-1).

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Case study questions (4 marks)

Q14Case-basedModerate4 marks

A school garden is a rectangle of area (6x2+7x−20)(6x^2+7x-20)(6x²+7x-20) square metres.

(i) Factorise 6x2+7x−206x^2+7x-206x²+7x-20 to find expressions for the two sides of the garden.

(ii) When x=4x=4x=4, find the length (the longer side), the breadth and the area.

(iii) Next to it is a square plot of side (2x+5)(2x+5)(2x+5) m with a square pond of side (x+1)(x+1)(x+1) m in the middle. Write the area of the plot outside the pond as a product of factors.

(iv) Find that area when x=4x=4x=4.

Show model answer

(i) ac=6×(−20)=−120ac=6\times(-20)=-120ac=6×(-20)=-120; two numbers with product −120-120-120 and sum 777 are 151515 and −8-8-8.

6x2+15x−8x−20=3x(2x+5)−4(2x+5)=(2x+5)(3x−4).6x^2+15x-8x-20=3x(2x+5)-4(2x+5)=(2x+5)(3x-4).6x²+15x-8x-20=3x(2x+5)-4(2x+5)=(2x+5)(3x-4). The sides are (2x+5)(2x+5)(2x+5) m and (3x−4)(3x-4)(3x-4) m.

(ii) At x=4x=4x=4 the sides are 2(4)+5=132(4)+5=132(4)+5=13 m and 3(4)−4=83(4)-4=83(4)-4=8 m, so the length is 131313 m, the breadth 888 m, and the area 13×8=104 m213\times8=104\ \text{m}^213×8=104 m² (check: 6(16)+28−20=1046(16)+28-20=1046(16)+28-20=104).

(iii) Area outside the pond =(2x+5)2−(x+1)2=((2x+5)+(x+1))((2x+5)−(x+1))=(2x+5)^2-(x+1)^2=\big((2x+5)+(x+1)\big)\big((2x+5)-(x+1)\big)=(2x+5)²-(x+1)²=((2x+5)+(x+1))((2x+5)-(x+1))

=(3x+6)(x+4)=3(x+2)(x+4)=(3x+6)(x+4)=3(x+2)(x+4)=(3x+6)(x+4)=3(x+2)(x+4) square metres.

(iv) At x=4x=4x=4: 3(6)(8)=144 m23(6)(8)=144\ \text{m}^23(6)(8)=144 m² (check: 132−52=169−25=14413^2-5^2=169-25=14413²-5²=169-25=144).

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Continues in Class 10: Remainder and Factor Theorems Class 10 ICSE questions

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

  • What types of factorisation are in ICSE Class 9?
    Taking out a common factor, grouping, the difference of two squares a2−b2a^2-b^2a²-b², the sum and difference of two cubes a3±b3a^3\pm b^3a³± b³, and splitting the middle term of ax2+bx+cax^2+bx+cax²+bx+c. This page has 14 questions with answers covering all of them, ending with a school-garden case study.
  • How do you factorise by splitting the middle term?
    Find two numbers whose product is acacac and whose sum is bbb. For 6x2−x−156x^2-x-156x²-x-15, ac=−90ac=-90ac=-90 and the pair is −10-10-10 and 999: 6x2−10x+9x−15=2x(3x−5)+3(3x−5)=(2x+3)(3x−5)6x^2-10x+9x-15=2x(3x-5)+3(3x-5)=(2x+3)(3x-5)6x²-10x+9x-15=2x(3x-5)+3(3x-5)=(2x+3)(3x-5). Expand the answer to check it.
  • How do you factorise x⁴ − 81y⁴ completely?
    Use the difference of squares twice: x4−81y4=(x2+9y2)(x2−9y2)=(x2+9y2)(x+3y)(x−3y)x^4-81y^4=(x^2+9y^2)(x^2-9y^2)=(x^2+9y^2)(x+3y)(x-3y)x⁴-81y⁴=(x²+9y²)(x²-9y²)=(x²+9y²)(x+3y)(x-3y). The sum of squares x2+9y2x^2+9y^2x²+9y² does not factorise further, which is why the word 'completely' matters in these questions.
  • What are the sum and difference of cubes formulas?
    a3+b3=(a+b)(a2−ab+b2)a^3+b^3=(a+b)(a^2-ab+b^2)a³+b³=(a+b)(a²-ab+b²) and a3−b3=(a−b)(a2+ab+b2)a^3-b^3=(a-b)(a^2+ab+b^2)a³-b³=(a-b)(a²+ab+b²). So 8x3+27=(2x+3)(4x2−6x+9)8x^3+27=(2x+3)(4x^2-6x+9)8x³+27=(2x+3)(4x²-6x+9). Watch the signs: the middle term in the quadratic factor has the opposite sign to the one in the binomial factor.
  • Which ICSE Class 9 chapter should I revise before Factorisation?
    Expansions (Chapter 4), because factorising is expanding in reverse; Factorisation is Chapter 5 and Simultaneous (Linear) Equations is Chapter 6. Expansions and Simultaneous Equations both have important-questions pages here.
  • What is the most common mistake in ICSE factorisation?
    Forgetting to take out the common factor first. 50x2−1850x^2-1850x²-18 is not a difference of squares until you write it as 2(25x2−9)=2(5x+3)(5x−3)2(25x^2-9)=2(5x+3)(5x-3)2(25x²-9)=2(5x+3)(5x-3). Students who skip that step either get stuck or leave the answer incomplete.
  • Can I download factorisation questions for Class 9 with answers?
    Yes. Enter your email in the free PDF box on this page and all 14 factorisation questions, from x2−9x^2-9x²-9 to a6−b6a^6-b^6a⁶-b⁶, open with worked answers as a printable worksheet to save as a PDF.

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