Remainder and Factor Theorem Class 10 ICSE — Bansal Ch 8

Written by Greya LakshmiPublished on 18 September 2026Updated on 28 September 2026

Chapter 8 of the Bansal Class 10 ICSE Mathematics book covers the remainder and factor theorem class 10 icse syllabus. It shows how to find a remainder without long division, how to test whether a divisor is a factor, and how to factorise cubic polynomials completely. This page walks through each section in book order, with the formulas and worked patterns you need. You can find this book in the Class 10 ICSE Maths book list.

ICSE Class 10 Mathematics Bansal Chapter 8 Remainder and Factor Theorems PDF

What is inside Chapter 8

10 pages · 4 sections · 16 worked examples · 24 exercise questions

Section	Title
8.1	A Basic Concept
8.2	Remainder Theorem
8.3	Factor Theorem
8.4	Using the Factor Theorem to factorise the given polynomial

EXERCISE 8(B): 24 questions

8.1 A Basic Concept

The chapter opens by defining function notation. In $f(x) = 2x^2 - 5x - 7$ , the symbol $f(x)$ is a function of the variable $x$ . Its value depends on the value chosen for $x$ . For example, $f(3) = 2(3)^2 - 5(3) - 7 = -4$ , while $f(-3) = 26$ . The same idea works for $f(y) = 2y^3 - 3y + 1$ , where $f(5) = 236$ and $f(-2) = -9$ .

The section then compares long division with substitution. Dividing $f(x) = x^2 - 5x + 8$ by $x - 2$ leaves remainder 2. Substituting gives $f(2) = (2)^2 - 5(2) + 8 = 2$ , the same value. Dividing $f(x) = 6x^3 - 3x^2 + 8x - 5$ by $x + 3$ leaves remainder $-218$ , and $f(-3) = 6(-3)^3 - 3(-3)^2 + 8(-3) - 5 = -218$ .

The pattern is clear. Dividing by $x - 2$ means evaluating $f(2)$ . Dividing by $x + 3$ means evaluating $f(-3)$ . This works because a linear divisor leaves a constant remainder, and at the point where the divisor equals zero, the polynomial equals that remainder. That observation leads directly to the Remainder Theorem.

You can predict a remainder without dividing. Take $f(x) = x^3 - 3x^2 + 5x - 7$ and divide by $x - 2$ . Put $x - 2 = 0$ , so $x = 2$ . Then $f(2) = (2)^3 - 3(2)^2 + 5(2) - 7 = 8 - 12 + 10 - 7 = -1$ . The remainder is $-1$ , with no long division needed.

8.2 Remainder Theorem

The Remainder Theorem states that when a polynomial $f(x)$ is divided by $x - a$ , the remainder equals $f(a)$ . For instance, dividing by $x - 3$ gives remainder $f(3)$ . The theorem holds because a linear divisor produces a constant remainder; substituting $x = a$ makes the divisor zero and leaves that constant by itself.

The chapter gives two steps for using the theorem. First, put the divisor equal to zero and solve for the variable. Second, substitute that value into the polynomial and simplify. When the divisor is $2x + 1$ , you solve $2x + 1 = 0$ to get $x = -\frac{1}{2}$ , then evaluate the polynomial at that point.

Here is the method with new numbers. Find $k$ so that $3x^3 - kx + 4$ leaves remainder 10 when divided by $x - 2$ . Step 1 gives $x - 2 = 0$ , so $x = 2$ . Step 2 substitutes and solves:
$3(2)^3 - k(2) + 4 = 10 \Rightarrow 24 - 2k + 4 = 10 \Rightarrow 28 - 2k = 10 \Rightarrow k = 9.$
So $k = 9$ .

The section also covers two harder patterns. One finds $a$ when two polynomials leave the same remainder under the same divisor. The other finds the number to add so a set remainder appears: add $k$ , evaluate at the divisor’s root, and solve for $k$ . In the chapter’s example, adding $-3$ to $2x^3 - 3x^2 + x$ makes division by $x - 2$ leave remainder 3.

Watch two sign traps when applying the theorem:

Divisor $x + 3$ means $x = -3$ , not 3. Substituting the wrong sign flips the answer.
Divisor $2x + 1$ has root $x = -\frac{1}{2}$ . Solve fully before substituting.

8.3 Factor Theorem

The Factor Theorem is the Remainder Theorem in the special case where the remainder is zero. If $f(x)$ is divided by $x - a$ and $f(a) = 0$ , then $x - a$ is a factor of $f(x)$ . This makes sense because a factor divides a polynomial exactly, so nothing is left over.

The chapter confirms this with $f(x) = x^2 - 5x + 6$ . Dividing by $x - 3$ gives $f(3) = (3)^2 - 5(3) + 6 = 0$ . Since the remainder is zero, $x - 3$ is a factor of the polynomial.

Test a new case. Is $x - 1$ a factor of $x^3 - 6x^2 + 11x - 6$ ? Set $x - 1 = 0$ , so $x = 1$ . Evaluate: $f(1) = (1)^3 - 6(1)^2 + 11(1) - 6 = 0$ . The remainder is zero, so $x - 1$ is a factor.

The section uses this test in three ways: deciding whether a given expression is a factor, finding a constant $a$ or $k$ when a divisor is known to be a factor, and checking a second factor after finding $k$ . Question 9 of the chapter carries the tag 2011: find $k$ so that $x - 2$ is a factor of $x^3 + 2x^2 - kx + 10$ , then test whether $x + 5$ also divides the polynomial.

The two theorems sit side by side like this:

Theorem	What it gives	When it applies
Remainder Theorem	The remainder $f(a)$ when $f(x)$ is divided by $x - a$	The remainder can be any number
Factor Theorem	The factor $x - a$ of $f(x)$	Only when $f(a) = 0$

The Factor Theorem is the Remainder Theorem applied at the special value where the remainder vanishes.

8.4 Using the Factor Theorem to factorise the given polynomial

This section turns the Factor Theorem into a complete method for factorising cubics. The trial value method tests small values like $x = 1$ , $x = -1$ , and $x = 2$ . When the polynomial equals zero at one of these values, the matching linear factor is known.

The full method has four steps:

Test simple values until $f(a) = 0$ . Then $x - a$ is a factor.
Divide the cubic by $x - a$ to get a quadratic quotient.
Factorise the quadratic, using middle-term splitting if needed.
Write the cubic as the product of the three linear factors.

The chapter applies this method to $2x^3 + x^2 - 2x - 1$ . Testing $x = 1$ gives $2 + 1 - 2 - 1 = 0$ , so $x - 1$ is a factor. Division leaves the quadratic $2x^2 + 3x + 1$ , which factorises as $(x + 1)(2x + 1)$ . The full factorisation is $(x - 1)(x + 1)(2x + 1)$ .

Work one complete example with new numbers. Factorise $x^3 - 6x^2 + 11x - 6$ . Test $x = 1$ : $1 - 6 + 11 - 6 = 0$ , so $x - 1$ is a factor. Divide to get the quotient $x^2 - 5x + 6$ . This factorises as $(x - 2)(x - 3)$ . Therefore $x^3 - 6x^2 + 11x - 6 = (x - 1)(x - 2)(x - 3)$ .

Several questions here carry board year tags. The 2006 paper asked to show $2x + 7$ is a factor and factorise completely. The 2008 paper asked for a value of $p$ , then a complete factorisation. The 2004, 2010, 2012, and 2014 papers asked for complete factorisations directly. The repeated tags show this method reappears often in ICSE board papers.

What to practise from this chapter

The chapter has one practice block, EXERCISE 8(B), with 24 questions. The questions cover the same patterns the sections taught:

Showing a divisor is a factor with the Factor Theorem, then factorising the expression completely.
Factorising a cubic completely using the Remainder Theorem or Factor Theorem.
Solving a cubic equation after factorising the expression.
Finding unknown constants such as $a$ , $b$ , $k$ , $m$ , and $p$ from factor and remainder conditions.

The board year tags printed in this chapter map to the question types like this:

Question type	Board paper years
Show a divisor is a factor, then factorise completely	2006, 2008
Factorise a cubic completely	2004, 2010, 2012, 2014
Find a constant so a given divisor is a factor	2003, 2011
Factor condition plus a set remainder	2005, 2013
Two polynomials leave the same remainder	2010, 2015

Use the year tags to decide what to revise. The factorise-completely pattern appeared four times, so it deserves the most practice. For the official syllabus and exam format, check the official CISCE website. In the Bansal book, Chapter 7 covers Ratio and Proportion and Chapter 9 covers Matrices. You can also browse the Class 10 resources, the book collection, or the wider ICSE hub.

Frequently asked questions

What is the Remainder Theorem in ICSE Class 10 Maths?

The Remainder Theorem says that when a polynomial $f(x)$ is divided by $x - a$ , the remainder equals $f(a)$ . You set the divisor to zero, solve for $x$ , and substitute that value into the polynomial. This removes the need for long division.

How do I find the remainder when the divisor is 2x + 1 instead of x – a?

Solve $2x + 1 = 0$ to get $x = -\frac{1}{2}$ . Then evaluate the polynomial at $x = -\frac{1}{2}$ . The chapter’s example, $x^2 - 8x + 4$ , leaves the remainder $8\frac{1}{4}$ .

What is the difference between the Remainder Theorem and the Factor Theorem?

The Remainder Theorem gives the remainder $f(a)$ , which can be any number. The Factor Theorem is the special case where $f(a) = 0$ , which tells you that $x - a$ is a factor. One finds a remainder; the other tests for exact division.

How do I factorise a cubic polynomial completely using the Factor Theorem?

Test small values like 1, -1, and 2 until the polynomial equals zero. That gives you a linear factor. Divide the cubic by that factor to obtain a quadratic, then factorise the quadratic. The result is the product of three linear factors.

How do I find the number to add to a polynomial so it leaves a set remainder?

Add an unknown constant $k$ to the polynomial and apply the Remainder Theorem at the divisor’s root. Set the result equal to the required remainder and solve for $k$ . In the chapter’s example, adding $-3$ makes division by $x - 2$ leave remainder 3.

Reference: ICSE Class 10 Mathematics textbook material, chapter 8: Remainder and Factor Theorems — CISCE-aligned.

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