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Remainder and Factor Theorems — ICSE Class 10 Maths Important Questions

13 ICSE Class 10 Maths practice questions on Remainder and Factor Theorems, each with a full model answer, covering 5 topics from the chapter in the question formats used in the exam.

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Key concepts
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Remainder and Factor Theorems — ICSE Class 10 Maths Important Questions

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Quick answer

Core ICSE Remainder and Factor Theorem questions use the remainder theorem (remainder on dividing p(x)p(x)p(x) by (x−a)(x-a)(x-a) is p(a)p(a)p(a)) and the factor theorem ((x−a)(x-a)(x-a) is a factor iff p(a)=0p(a)=0p(a)=0) to find remainders, determine unknown constants, and factorise cubic polynomials completely. Two-condition problems fixing aaa and bbb appear almost every year.

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

About Remainder and Factor Theorems

Within the ICSE Class 10 Maths chapter Remainder and Factor Theorems you find the remainder when a polynomial p(x)p(x)p(x) is divided by a linear divisor using p(a)p(a)p(a), test and use factors with the factor theorem, find unknown coefficients from given conditions, and factorise cubic polynomials completely.

Remainder theoremFactor theoremFinding unknown constantsFactorising cubic polynomialsDivisors of the form $(ax\pm b)$

Key concepts & formulas

Remainder theorem

When p(x)p(x)p(x) is divided by (x−a)(x-a)(x-a), the remainder is p(a)p(a)p(a). For divisor (ax−b)(ax-b)(ax-b), evaluate p ⁣(ba)p\!\left(\dfrac{b}{a}\right)p\!(b/a).

Factor theorem

(x−a)(x-a)(x-a) is a factor of p(x)p(x)p(x) if and only if p(a)=0p(a)=0p(a)=0.

Factorising a cubic

Find one root aaa by trial (a factor of the constant term), so (x−a)(x-a)(x-a) is a factor; divide to get a quadratic and factorise that.

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Important questions with answers

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Multiple-choice questions (1 mark)

Q1MCQEasy1 mark

The remainder when x3+3x2−5x+4x^3+3x^2-5x+4x^3+3x^2-5x+4 is divided by (x−2)(x-2)(x-2) is:

  1. (a)

    141414

  2. (b)

    000

  3. (c)

    101010

  4. (d)

    −6-6-6

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Answer: (a) 141414.

By the remainder theorem the remainder is p(2)=8+12−10+4=14p(2)=8+12-10+4=14p(2)=8+12-10+4=14.

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Q2MCQEasy1 mark

Which of the following is a factor of x2−5x+6x^2-5x+6x^2-5x+6?

  1. (a)

    (x−2)(x-2)(x-2)

  2. (b)

    (x+2)(x+2)(x+2)

  3. (c)

    (x−1)(x-1)(x-1)

  4. (d)

    (x+3)(x+3)(x+3)

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Answer: (a) (x−2)(x-2)(x-2).

p(2)=4−10+6=0p(2)=4-10+6=0p(2)=4-10+6=0, so by the factor theorem (x−2)(x-2)(x-2) is a factor.

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Q3MCQModerate1 mark

If (x−1)(x-1)(x-1) is a factor of x3−kx2+11x−6x^3-kx^2+11x-6x^3-kx^2+11x-6, then k=k=k=

  1. (a)

    666

  2. (b)

    555

  3. (c)

    444

  4. (d)

    111111

Show model answer

Answer: (a) 666.

(x−1)(x-1)(x-1) a factor ⇒p(1)=0\Rightarrow p(1)=0p(1)=0: 1−k+11−6=0⇒6−k=0⇒k=61-k+11-6=0\Rightarrow 6-k=0\Rightarrow k=61-k+11-6=0 6-k=0 k=6.

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Q4MCQHOTS1 mark

The remainder when 4x3−3x2+2x−14x^3-3x^2+2x-14x^3-3x^2+2x-1 is divided by (2x+1)(2x+1)(2x+1) is:

  1. (a)

    −134-\dfrac{13}{4}-13/4

  2. (b)

    −54-\dfrac{5}{4}-5/4

  3. (c)

    134\dfrac{13}{4}13/4

  4. (d)

    −3-3-3

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Answer: (a) −134-\dfrac{13}{4}-13/4.

Set 2x+1=0⇒x=−122x+1=0\Rightarrow x=-\dfrac122x+1=0 x=-12. Remainder =p ⁣(−12)=4 ⁣(−18)−3 ⁣(14)+2 ⁣(−12)−1=−12−34−1−1=−134=p\!\left(-\dfrac12\right)=4\!\left(-\dfrac18\right)-3\!\left(\dfrac14\right)+2\!\left(-\dfrac12\right)-1=-\dfrac12-\dfrac34-1-1=-\dfrac{13}{4}=p\!(-12)=4\!(-18)-3\!(14)+2\!(-12)-1=-12-34-1-1=-13/4.

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Assertion–Reason questions (1 mark)

Q5Assertion–ReasonModerate1 mark

Assertion (A): (x+2)(x+2)(x+2) is a factor of x3+2x2−x−2x^3+2x^2-x-2x^3+2x^2-x-2.

Reason (R): (x−a)(x-a)(x-a) is a factor of p(x)p(x)p(x) if and only if p(a)=0p(a)=0p(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) Here a=−2a=-2a=-2: p(−2)=−8+8+2−2=0p(-2)=-8+8+2-2=0p(-2)=-8+8+2-2=0, so by R (the factor theorem) (x+2)(x+2)(x+2) is a factor. R correctly explains A.

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Very short answer questions (2 marks)

Q6Very ShortEasy2 marks

Using the remainder theorem, find the remainder when x3−3x2+4x−4x^3-3x^2+4x-4x^3-3x^2+4x-4 is divided by (x−2)(x-2)(x-2).

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Remainder =p(2)=(2)3−3(2)2+4(2)−4=8−12+8−4=0=p(2)=(2)^3-3(2)^2+4(2)-4=8-12+8-4=\mathbf{0}=p(2)=(2)^3-3(2)^2+4(2)-4=8-12+8-4=0.

Since the remainder is 000, (x−2)(x-2)(x-2) is a factor of the polynomial.

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Q7Very ShortModerate2 marks

Find the value of aaa if (x+1)(x+1)(x+1) is a factor of ax3+x2−2x+4a−9ax^3+x^2-2x+4a-9ax^3+x^2-2x+4a-9.

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(x+1)(x+1)(x+1) a factor ⇒p(−1)=0\Rightarrow p(-1)=0p(-1)=0.

a(−1)3+(−1)2−2(−1)+4a−9=0⇒−a+1+2+4a−9=0a(-1)^3+(-1)^2-2(-1)+4a-9=0\Rightarrow -a+1+2+4a-9=0a(-1)^3+(-1)^2-2(-1)+4a-9=0 -a+1+2+4a-9=0.

3a−6=0⇒a=23a-6=0\Rightarrow a=\mathbf{2}3a-6=0 a=2.

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Short answer questions (3 marks)

Q8Short AnswerModerate3 marks

Using the remainder theorem, find the remainder when 2x3+3x2−17x−302x^3+3x^2-17x-302x^3+3x^2-17x-30 is divided by (x−2)(x-2)(x-2), and state whether (x−2)(x-2)(x-2) is a factor.

Show model answer

Remainder =p(2)=2(2)3+3(2)2−17(2)−30=p(2)=2(2)^3+3(2)^2-17(2)-30=p(2)=2(2)^3+3(2)^2-17(2)-30.

=16+12−34−30=−36.=16+12-34-30=-36.=16+12-34-30=-36.

The remainder is −36\mathbf{-36}-36. Since p(2)≠0p(2)\neq0p(2)≠0, (x−2)(x-2)(x-2) is not a factor of the polynomial.

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Q9Short AnswerModerate3 marks

Find the values of aaa and bbb so that (x−1)(x-1)(x-1) and (x+2)(x+2)(x+2) are both factors of x3+ax2+bx−6x^3+ax^2+bx-6x^3+ax^2+bx-6.

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Let p(x)=x3+ax2+bx−6p(x)=x^3+ax^2+bx-6p(x)=x^3+ax^2+bx-6.

(x−1)(x-1)(x-1) a factor: p(1)=1+a+b−6=0⇒a+b=5.p(1)=1+a+b-6=0\Rightarrow a+b=5.p(1)=1+a+b-6=0 a+b=5. ...(1)

(x+2)(x+2)(x+2) a factor: p(−2)=−8+4a−2b−6=0⇒4a−2b=14⇒2a−b=7.p(-2)=-8+4a-2b-6=0\Rightarrow 4a-2b=14\Rightarrow 2a-b=7.p(-2)=-8+4a-2b-6=0 4a-2b=14 2a-b=7. ...(2)

Add (1) and (2): 3a=12⇒a=43a=12\Rightarrow a=\mathbf{4}3a=12 a=4, and from (1) b=1b=\mathbf{1}b=1.

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Q10Short AnswerModerate3 marks

Using the factor theorem, factorise completely: x3+2x2−5x−6x^3+2x^2-5x-6x^3+2x^2-5x-6.

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Try x=−1x=-1x=-1: p(−1)=−1+2+5−6=0p(-1)=-1+2+5-6=0p(-1)=-1+2+5-6=0, so (x+1)(x+1)(x+1) is a factor.

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

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

Hence x3+2x2−5x−6=(x+1)(x+3)(x−2)x^3+2x^2-5x-6=\mathbf{(x+1)(x+3)(x-2)}x^3+2x^2-5x-6=(x+1)(x+3)(x-2).

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Long answer questions (5 marks)

Q11Long AnswerModerate5 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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The factors of the constant 666 (over possible rational roots) are tried. Take x=2x=2x=2:

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

Divide 2x3+x2−13x+62x^3+x^2-13x+62x^3+x^2-13x+6 by (x−2)(x-2)(x-2):

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

Hence 2x3+x2−13x+6=(x−2)(2x−1)(x+3)2x^3+x^2-13x+6=\mathbf{(x-2)(2x-1)(x+3)}2x^3+x^2-13x+6=(x-2)(2x-1)(x+3).

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Q12Long AnswerHOTS5 marks

The polynomial f(x)=x3+ax2+bx−8f(x)=x^3+ax^2+bx-8f(x)=x^3+ax^2+bx-8 leaves a remainder of −10-10-10 when divided by (x−1)(x-1)(x-1), and (x+1)(x+1)(x+1) is a factor of f(x)f(x)f(x). Find aaa and bbb, and then factorise f(x)f(x)f(x) completely.

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Remainder condition: f(1)=−10f(1)=-10f(1)=-10: 1+a+b−8=−10⇒a+b=−3.1+a+b-8=-10\Rightarrow a+b=-3.1+a+b-8=-10 a+b=-3. ...(1)

Factor condition: f(−1)=0f(-1)=0f(-1)=0: −1+a−b−8=0⇒a−b=9.-1+a-b-8=0\Rightarrow a-b=9.-1+a-b-8=0 a-b=9. ...(2)

Add (1) and (2): 2a=6⇒a=32a=6\Rightarrow a=\mathbf{3}2a=6 a=3; then from (1) b=−6b=\mathbf{-6}b=-6.

So f(x)=x3+3x2−6x−8f(x)=x^3+3x^2-6x-8f(x)=x^3+3x^2-6x-8. Since (x+1)(x+1)(x+1) is a factor, divide:

x3+3x2−6x−8=(x+1)(x2+2x−8).x^3+3x^2-6x-8=(x+1)(x^2+2x-8).x^3+3x^2-6x-8=(x+1)(x^2+2x-8).

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

Hence f(x)=(x+1)(x−2)(x+4)f(x)=\mathbf{(x+1)(x-2)(x+4)}f(x)=(x+1)(x-2)(x+4).

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Case-based questions (4 marks)

Q13Case-basedModerate4 marks

The volume of a rectangular box (in cubic units) is modelled by the polynomial p(x)=x3+6x2+11x+6p(x)=x^3+6x^2+11x+6p(x)=x^3+6x^2+11x+6, where xxx is a positive integer.

(i) Show that (x+1)(x+1)(x+1) is a factor of p(x)p(x)p(x).
(ii) Factorise p(x)p(x)p(x) completely to express the three edge lengths.
(iii) Find the volume of the box when x=2x=2x=2.

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(i) p(−1)=(−1)3+6(−1)2+11(−1)+6=−1+6−11+6=0p(-1)=(-1)^3+6(-1)^2+11(-1)+6=-1+6-11+6=0p(-1)=(-1)^3+6(-1)^2+11(-1)+6=-1+6-11+6=0, so by the factor theorem (x+1)(x+1)(x+1) is a factor.

(ii) Dividing by (x+1)(x+1)(x+1): p(x)=(x+1)(x2+5x+6)=(x+1)(x+2)(x+3)p(x)=(x+1)(x^2+5x+6)=(x+1)(x+2)(x+3)p(x)=(x+1)(x^2+5x+6)=(x+1)(x+2)(x+3).

The three edges are (x+1)(x+1)(x+1), (x+2)(x+2)(x+2) and (x+3)(x+3)(x+3).

(iii) At x=2x=2x=2: volume =(3)(4)(5)=60=(3)(4)(5)=\mathbf{60}=(3)(4)(5)=60 cubic units.

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All ICSE Class 10 Maths Chapters

GST (Goods and Services Tax)Banking (Recurring Deposit Accounts)Shares and DividendsLinear Inequations (In One Variable)Quadratic EquationsSolving (Simple) Problems Based on Quadratic EquationsRatio and ProportionRemainder and Factor TheoremsMatricesArithmetic ProgressionGeometric ProgressionReflectionSection and Mid-Point FormulaEquation of a LineSimilarity (With Applications to Maps and Models)Loci (Locus and Its Constructions)CirclesTangents and Intersecting ChordsConstructions (Circles)Cylinder, Cone and Sphere (Surface Area and Volume)Trigonometrical IdentitiesHeights and DistancesGraphical Representation (Histograms and Ogives)Measures of Central Tendency (Mean, Median, Quartiles and Mode)Probability

Frequently asked questions

  • Do these Remainder and Factor Theorems questions follow the latest ICSE syllabus?
    Yes — they are aligned to the CISCE 2026–27 syllabus for ICSE Class 10 Maths, so nothing here is outside the current course.

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