Question 1

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

Accepted Answer

**Answer: (a) $14$.** By the remainder theorem the remainder is $p(2)=8+12-10+4=14$.

Question 2

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

Accepted Answer

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

Question 3

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

Accepted Answer

**Answer: (a) $6$.** $(x-1)$ a factor $\Rightarrow p(1)=0$: $1-k+11-6=0\Rightarrow 6-k=0\Rightarrow k=6$.

Question 4

The remainder when $4x^3-3x^2+2x-1$ is divided by $(2x+1)$ is:

Accepted Answer

**Answer: (a) $-\dfrac{13}{4}$.** Set $2x+1=0\Rightarrow x=-\dfrac12$. Remainder $=p\!\left(-\dfrac12\right)=4\!\left(-\dfrac18\right)-3\!\left(\dfrac14\right)+2\!\left(-\dfrac12\right)-1=-\dfrac12-\dfrac34-1-1=-\dfrac{13}{4}$.

Question 5

**Assertion (A):** $(x+2)$ is a factor of $x^3+2x^2-x-2$. **Reason (R):** $(x-a)$ is a factor of $p(x)$ if and only if $p(a)=0$.

Accepted Answer

**Answer: (a)** Here $a=-2$: $p(-2)=-8+8+2-2=0$, so by R (the factor theorem) $(x+2)$ is a factor. R correctly explains A.

Question 6

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

Accepted Answer

Remainder $=p(2)=(2)^3-3(2)^2+4(2)-4=8-12+8-4=\mathbf{0}$. Since the remainder is $0$, $(x-2)$ is a factor of the polynomial.

Question 7

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

Accepted Answer

$(x+1)$ a factor $\Rightarrow p(-1)=0$. $a(-1)^3+(-1)^2-2(-1)+4a-9=0\Rightarrow -a+1+2+4a-9=0$. $3a-6=0\Rightarrow a=\mathbf{2}$.

Question 8

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

Accepted Answer

Remainder $=p(2)=2(2)^3+3(2)^2-17(2)-30$. $=16+12-34-30=-36.$ The remainder is $\mathbf{-36}$. Since $p(2)
eq0$, $(x-2)$ is **not** a factor of the polynomial.

Question 9

Find the values of $a$ and $b$ so that $(x-1)$ and $(x+2)$ are both factors of $x^3+ax^2+bx-6$.

Accepted Answer

Let $p(x)=x^3+ax^2+bx-6$. $(x-1)$ a factor: $p(1)=1+a+b-6=0\Rightarrow a+b=5.$ ...(1) $(x+2)$ a factor: $p(-2)=-8+4a-2b-6=0\Rightarrow 4a-2b=14\Rightarrow 2a-b=7.$ ...(2) Add (1) and (2): $3a=12\Rightarrow a=\mathbf{4}$, and from (1) $b=\mathbf{1}$.

Question 10

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

Accepted Answer

Try $x=-1$: $p(-1)=-1+2+5-6=0$, so $(x+1)$ is a factor. Dividing $x^3+2x^2-5x-6$ by $(x+1)$ gives $x^2+x-6$. $x^2+x-6=(x+3)(x-2)$. Hence $x^3+2x^2-5x-6=\mathbf{(x+1)(x+3)(x-2)}$.

Question 11

Using the factor theorem, factorise completely: $2x^3+x^2-13x+6$.

Accepted Answer

The factors of the constant $6$ (over possible rational roots) are tried. Take $x=2$ :

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

Divide $2x^3+x^2-13x+6$ by $(x-2)$ :

$2x^3+x^2-13x+6=(x-2)(2x^2+5x-3).$

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

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

Question 12

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

Accepted Answer

**Remainder condition:** $f(1)=-10$: $1+a+b-8=-10\Rightarrow a+b=-3.$ ...(1) **Factor condition:** $f(-1)=0$: $-1+a-b-8=0\Rightarrow a-b=9.$ ...(2) Add (1) and (2): $2a=6\Rightarrow a=\mathbf{3}$; then from (1) $b=\mathbf{-6}$. So $f(x)=x^3+3x^2-6x-8$. Since $(x+1)$ is a factor, divide: $$x^3+3x^2-6x-8=(x+1)(x^2+2x-8).$$ $x^2+2x-8=(x+4)(x-2)$. Hence $f(x)=\mathbf{(x+1)(x-2)(x+4)}$.

Question 13

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

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

Accepted Answer

**(i)** $p(-1)=(-1)^3+6(-1)^2+11(-1)+6=-1+6-11+6=0$, so by the factor theorem $(x+1)$ is a factor. **(ii)** Dividing by $(x+1)$: $p(x)=(x+1)(x^2+5x+6)=(x+1)(x+2)(x+3)$. The three edges are $(x+1)$, $(x+2)$ and $(x+3)$. **(iii)** At $x=2$: volume $=(3)(4)(5)=\mathbf{60}$ cubic units.

Question 14

Are these Remainder and Factor Theorems important questions free?

Accepted Answer

Yes. All 13 ICSE Class 10 Maths important questions for Remainder and Factor Theorems are free, with full model answers and no login required.

Question 15

Do these Remainder and Factor Theorems questions follow the latest ICSE syllabus?

Accepted Answer

Yes — they are aligned to the CISCE latest syllabus syllabus for ICSE Class 10 Maths, so nothing here is outside the current course.

Question 16

How should I practise the Remainder and Factor Theorems important questions?

Accepted Answer

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.

Question 17

What types of questions are covered for Remainder and Factor Theorems?

Accepted Answer

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.

Question type	Count	Marks
MCQ	4	1
Assertion–Reason	1	1
Very Short	2	2
Short Answer	3	3
Long Answer	2	5
Case-based	1	4

Remainder and Factor Theorems — Important Questions

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