What is Exercise 2.3 of Class 10 Maths Chapter 2 about?

Exercise 2.3 focuses on dividing one polynomial by another. It introduces the polynomial division algorithm and helps in solving division-related problems.

What is the Polynomial Division Algorithm?

The division algorithm states that for any two polynomials p(x) and g(x), there exist polynomials q(x)q and r(x) such that p(x)=g(x)⋅q(x)+r(x), where r(x) is either zero or has a lower degree than g(x).

Why is polynomial division important?

Polynomial division helps in simplifying expressions, solving equations, and finding factors of polynomials. It is also used in calculus and algebraic computations.

What happens if the remainder is zero in polynomial division?

If the remainder is zero, it means that the divisor is a factor of the dividend. This method is useful in factorization and solving polynomial equations.

What types of polynomials are covered in this exercise?

This exercise involves division of linear, quadratic, and cubic polynomials. It emphasizes the systematic approach to polynomial division.

NCERT Solutions

Class 10

Maths

Chapter 2 Polynomials

Exercise 2.3

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NCERT Solutions Class 10 Maths Chapter 2 Polynomials Exercise 2.3

(This page features Exercise 2.3 (Polynomials) from NCERT Class 10 Maths, which has been removed from the current syllabus [2025-26] but remains a useful reference.)

The NCERT Solutions Class 10 Maths Chapter 2 Exercise 2.3 is a fundamental exercise to comprehend the idea of polynomials and their division. The exercise revolves around the division algorithm for polynomials, which is the cornerstone for solving polynomial equations. The solved exercise provided here sticks to the new CBSE syllabus and examination guidelines so that students can prepare and learn important concepts with ease. Now, let us discuss the Division Algorithm for polynomials in detail.

1.0Download NCERT Solutions Class 10 Maths Chapter 2 Exercise 2.3 : Free PDF

NCERT Solutions Class 10 Maths Chapter 2 Exercise 2.3

2.0Introduction to Division Algorithm for Polynomials

The Division Algorithm for Polynomials is a technique employed to divide a polynomial by another. It says that if a polynomial p(x) is divided by any non-zero polynomial g(x). It is used to make complicated polynomial expressions simpler and to determine the quotient(or the result of division) and the remainder(leftover of polynomial after division) when a polynomial is divided by a polynomial. Which, in turn, makes the calculation for determining the zeroes of the polynomials easier.

3.0Class 10 Maths Chapter 2 Polynomials Exercise 2.3 Overview: Key Concepts

Euclidean Division Process:

The division algorithm works similar to the long division process used to divide one integer to another. It includes continuous subtraction of terms of g(x) from p(x) until we get the degree of the remainder which is less than that of g(x). In the exercise, the euclidean division theorem follows certain steps while dividing the polynomial.

Steps for Polynomial Division:

Let’s understand these steps by the following example, which focuses on dividing one polynomial by another by taking an example:

$3 x^{3} + 6 x^{2} + 3 x -5$ by $x^{2} + 2 x + 1$

Divide the First Terms: Divide the first term of the highest degree of the dividend by the first term of the highest degree of the divisor. This will be the first term of the quotient.

$\frac{3 x ^{3}}{x ^{2}} = 3 x$

Multiply and Subtract: Multiply the whole divisor by the first term of the quotient from step 2. Subtract the product from the dividend.

$(x^{2} + 2 x + 1) \times 3 x = 3 x^{2} + 6 x^{2} + 3 x$

$(3 x^{3} + 6 x^{2} + 3 x - 5) - (3 x^{3} + 6 x^{2} + 3 x) = - 5$

Repeat the Process: Repeat the process by dividing the highest degree term of the new polynomial by the highest degree term of its divisor and adding the new term to the quotient. Since the term ends here with a constant term, there is no need to repeat the process.
Proceed Until Degree of Remainder is Lower Than Divisor: Proceed with the process of division until the degree of the remaining polynomial (remainder) is lower than that of the divisor or until a remainder does not exist.

The same steps are expressed as the typical form of dividing integers, as shown below:

$x^{2} + 2 x + 1) 3 x^{3} + 6 x^{2} + 3 x -5 (3 x$

$3 x^{3} + 6 x^{2} + 3 x$

— – — — — — —

– 5

The Verification Formula for Division Algorithm

In the exercise, the verification formula is used to verify the correctness of certain divisions of polynomials. The formula also helps in calculating the roots of a polynomial with a degree higher than 2.

The formula for the division of polynomials states that if p(x) and g(x) are two polynomials, such that p(x) is divided by g(x) where $g (x) \neq = 0$ , then q(x) will be termed as the quotient and r(x) is termed as the remainder of the polynomial. The formula can mathematically be written as:

$D i v i d e n t = D i v i sor \times Q u o t i e n t + R e main d er$

$p (x) = g (x) \times q (x) + r (x)$

Note that r(x) = 0 or if $r (x) \neq = 0$ , then degree of r(x) < degree of g(x).

Also Read: CBSE Class 10 Maths Chapter 2 Polynomials

4.0NCERT Class 10 Maths Chapter 2 Exercise 2.3 : Detailed Solutions

1. Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in each of the following:

(i) p(x) = x³ - 3x² + 5x - 3, g(x) = x² - 2

(ii) p(x) = x⁴ - 3x² + 4x + 5, g(x) = x² + 1 - x

(iii) p(x) = x⁴ - 5x + 6, g(x) = 2 - x².

Sol.

(i)

Question on Class 10 Maths Chapter 2 Exercise 2.3

Hence, Quotient q(x) = x - 3 and Remainder r(x) = 7x - 9.

(ii)

Sample Questions on chapter 2 polynomials

Hence, Quotient, q(x) = x² + x - 3 and remainder, r(x) = 8.

(iii)

Hence, Quotient, q(x) = -x² - 2 Remainder, r(x) = -5x + 10.

2. Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial.

(i) t² - 3, 2t⁴ + 3t³ - 2t² - 9t - 12

(ii) x² + 3x + 1, 3x⁴ + 5x³ - 7x² + 2x + 2

(iii) x³ - 3x + 1, x⁵ - 4x³ + x² + 3x + 1

Sol.

(i)

Hence, t² - 3 is a factor of 2t⁴ + 3t³ - 2t² - 9t - 12.

(ii)

Hence, x² + 3x + 1 is a factor of 3x⁴ + 5x³ - 7x² + 2x + 2.

(iii)

Hence, x³ - 3x + 1 is not a factor of x⁵ - 4x³ + x² + 3x + 1.

3. Obtain all other zeros of 3x⁴ + 6x³ - 2x² - 10x - 5, if two of its zeros are √(5/3) and -√(5/3).

Sol.

Two of the zeros of 3x⁴ + 6x³ - 2x² - 10x - 5 are √(5/3) and -√(5/3).

⇒ (x - √(5/3))(x + √(5/3)) is a factor of the polynomial.

i.e., x² - 5/3 is a factor.

i.e., (3x² - 5) is a factor of the polynomial. Then we apply the division algorithm as below:

The other two zeros will be obtained from the quadratic polynomial q(x) = x² + 2x + 1.

Now x² + 2x + 1 = (x + 1)².

Its zeros are -1, -1.

Hence, all other zeros are -1, -1.

4. On dividing x³ - 3x² + x + 2 by a polynomial g(x), the quotient and remainder were x - 2 and -2x + 4, respectively. Find g(x).

Sol. (x³ - 3x² + x + 2) = g(x) × (x - 2) + (-2x + 4)

⇒ x³ - 3x² + x + 2 + 2x - 4 = g(x) × (x - 2)

⇒ x³ - 3x² + 3x - 2 = g(x) × (x - 2)

g(x) = (x³ - 3x² + 3x - 2) / (x - 2)

= x² - x + 1

So, g(x) = x² - x + 1.

5. Give examples of polynomials where p(x) as dividend, g(x) as divisor, q(x) as quotient and r(x) as remainder, which satisfy the division algorithm and

(i) deg p(x) = deg q(x)

(ii) deg q(x) = deg r(x)

(iii) deg r(x) = 0.

Sol.

(i) p(x) = 2x² + 2x + 8, g(x) = 2x⁰ = 2; q(x) = x² + x + 4; r(x) = 0

(ii) p(x) = 2x² + 2x + 8; g(x) = x² + x + 9; q(x) = 2; r(x) = -10

(iii) p(x) = x³ + x + 5; g(x) = x² + 1; q(x) = x; r(x) = 5.

5.0Benefits Of NCERT Solutions Class 10 Maths Chapter 2 Exercise 2.3

Improved understanding of core concepts
Better logical thinking and application
Enhanced exam performance
Simplified explanation of important topics

NCERT Class 10 Maths Ch 2 Polynomials Other Exercises:-

Exercise 2.1

Exercise 2.2

Exercise 2.3

NCERT Solutions Class 10 Maths All Chapters:-

Chapter 1 - Real Numbers

Chapter 2 - Polynomials

Chapter 3 - Linear Equations in Two Variables

Chapter 4 - Quadratic Equations

Chapter 5 - Arithmetic Progressions

Chapter 6 - Triangles

Chapter 7 - Coordinate Geometery

Chapter 8 - Introdction to Trigonometry

Chapter 9 - Some Applications of Trigonometry

Chapter 10 - Circles

Chapter 11 - Areas Related to Circles

Chapter 12 - Surface Areas and Volumes

Chapter 13 - Statistics

Chapter 14 - Probability

Frequently Asked Questions

Join ALLEN!

(Session 2026 - 27)

Name

Mobile Number

Class

Choose class

Goal

Choose your goal

Preferred Programs

Preferred Mode

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NCERT Solutions Class 10 Maths Chapter 2 Polynomials Exercise 2.3

(This page features Exercise 2.3 (Polynomials) from NCERT Class 10 Maths, which has been removed from the current syllabus [2025-26] but remains a useful reference.)

The NCERT Solutions Class 10 Maths Chapter 2 Exercise 2.3 is a fundamental exercise to comprehend the idea of polynomials and their division. The exercise revolves around the division algorithm for polynomials, which is the cornerstone for solving polynomial equations. The solved exercise provided here sticks to the new CBSE syllabus and examination guidelines so that students can prepare and learn important concepts with ease. Now, let us discuss the Division Algorithm for polynomials in detail.

1.0Download NCERT Solutions Class 10 Maths Chapter 2 Exercise 2.3 : Free PDF

NCERT Solutions Class 10 Maths Chapter 2 Exercise 2.3

2.0Introduction to Division Algorithm for Polynomials

The Division Algorithm for Polynomials is a technique employed to divide a polynomial by another. It says that if a polynomial p(x) is divided by any non-zero polynomial g(x). It is used to make complicated polynomial expressions simpler and to determine the quotient(or the result of division) and the remainder(leftover of polynomial after division) when a polynomial is divided by a polynomial. Which, in turn, makes the calculation for determining the zeroes of the polynomials easier.

3.0Class 10 Maths Chapter 2 Polynomials Exercise 2.3 Overview: Key Concepts

Euclidean Division Process:

The division algorithm works similar to the long division process used to divide one integer to another. It includes continuous subtraction of terms of g(x) from p(x) until we get the degree of the remainder which is less than that of g(x). In the exercise, the euclidean division theorem follows certain steps while dividing the polynomial.

Steps for Polynomial Division:

Let’s understand these steps by the following example, which focuses on dividing one polynomial by another by taking an example:

$3 x^{3} + 6 x^{2} + 3 x -5$ by $x^{2} + 2 x + 1$

Divide the First Terms: Divide the first term of the highest degree of the dividend by the first term of the highest degree of the divisor. This will be the first term of the quotient.

$\frac{3 x ^{3}}{x ^{2}} = 3 x$

Multiply and Subtract: Multiply the whole divisor by the first term of the quotient from step 2. Subtract the product from the dividend.

$(x^{2} + 2 x + 1) \times 3 x = 3 x^{2} + 6 x^{2} + 3 x$

$(3 x^{3} + 6 x^{2} + 3 x - 5) - (3 x^{3} + 6 x^{2} + 3 x) = - 5$

Repeat the Process: Repeat the process by dividing the highest degree term of the new polynomial by the highest degree term of its divisor and adding the new term to the quotient. Since the term ends here with a constant term, there is no need to repeat the process.
Proceed Until Degree of Remainder is Lower Than Divisor: Proceed with the process of division until the degree of the remaining polynomial (remainder) is lower than that of the divisor or until a remainder does not exist.

The same steps are expressed as the typical form of dividing integers, as shown below:

$x^{2} + 2 x + 1) 3 x^{3} + 6 x^{2} + 3 x -5 (3 x$

$3 x^{3} + 6 x^{2} + 3 x$

— – — — — — —

– 5

The Verification Formula for Division Algorithm

In the exercise, the verification formula is used to verify the correctness of certain divisions of polynomials. The formula also helps in calculating the roots of a polynomial with a degree higher than 2.

The formula for the division of polynomials states that if p(x) and g(x) are two polynomials, such that p(x) is divided by g(x) where $g (x) \neq = 0$ , then q(x) will be termed as the quotient and r(x) is termed as the remainder of the polynomial. The formula can mathematically be written as:

$D i v i d e n t = D i v i sor \times Q u o t i e n t + R e main d er$

$p (x) = g (x) \times q (x) + r (x)$

Note that r(x) = 0 or if $r (x) \neq = 0$ , then degree of r(x) < degree of g(x).

Also Read: CBSE Class 10 Maths Chapter 2 Polynomials

4.0NCERT Class 10 Maths Chapter 2 Exercise 2.3 : Detailed Solutions

1. Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in each of the following:

(i) p(x) = x³ - 3x² + 5x - 3, g(x) = x² - 2

(ii) p(x) = x⁴ - 3x² + 4x + 5, g(x) = x² + 1 - x

(iii) p(x) = x⁴ - 5x + 6, g(x) = 2 - x².

Sol.

(i)

Hence, Quotient q(x) = x - 3 and Remainder r(x) = 7x - 9.

(ii)

Hence, Quotient, q(x) = x² + x - 3 and remainder, r(x) = 8.

(iii)

Hence, Quotient, q(x) = -x² - 2 Remainder, r(x) = -5x + 10.

2. Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial.

(i) t² - 3, 2t⁴ + 3t³ - 2t² - 9t - 12

(ii) x² + 3x + 1, 3x⁴ + 5x³ - 7x² + 2x + 2

(iii) x³ - 3x + 1, x⁵ - 4x³ + x² + 3x + 1

Sol.

(i)

Hence, t² - 3 is a factor of 2t⁴ + 3t³ - 2t² - 9t - 12.

(ii)

Hence, x² + 3x + 1 is a factor of 3x⁴ + 5x³ - 7x² + 2x + 2.

(iii)

Hence, x³ - 3x + 1 is not a factor of x⁵ - 4x³ + x² + 3x + 1.

3. Obtain all other zeros of 3x⁴ + 6x³ - 2x² - 10x - 5, if two of its zeros are √(5/3) and -√(5/3).

Sol.

Two of the zeros of 3x⁴ + 6x³ - 2x² - 10x - 5 are √(5/3) and -√(5/3).

⇒ (x - √(5/3))(x + √(5/3)) is a factor of the polynomial.

i.e., x² - 5/3 is a factor.

i.e., (3x² - 5) is a factor of the polynomial. Then we apply the division algorithm as below:

The other two zeros will be obtained from the quadratic polynomial q(x) = x² + 2x + 1.

Now x² + 2x + 1 = (x + 1)².

Its zeros are -1, -1.

Hence, all other zeros are -1, -1.

4. On dividing x³ - 3x² + x + 2 by a polynomial g(x), the quotient and remainder were x - 2 and -2x + 4, respectively. Find g(x).

Sol. (x³ - 3x² + x + 2) = g(x) × (x - 2) + (-2x + 4)

⇒ x³ - 3x² + x + 2 + 2x - 4 = g(x) × (x - 2)

⇒ x³ - 3x² + 3x - 2 = g(x) × (x - 2)

g(x) = (x³ - 3x² + 3x - 2) / (x - 2)

= x² - x + 1

So, g(x) = x² - x + 1.

5. Give examples of polynomials where p(x) as dividend, g(x) as divisor, q(x) as quotient and r(x) as remainder, which satisfy the division algorithm and

(i) deg p(x) = deg q(x)

(ii) deg q(x) = deg r(x)

(iii) deg r(x) = 0.

Sol.

(i) p(x) = 2x² + 2x + 8, g(x) = 2x⁰ = 2; q(x) = x² + x + 4; r(x) = 0

(ii) p(x) = 2x² + 2x + 8; g(x) = x² + x + 9; q(x) = 2; r(x) = -10

(iii) p(x) = x³ + x + 5; g(x) = x² + 1; q(x) = x; r(x) = 5.