# NCERT Solutions for Class 10 Maths (Chapter 2, Exercise 2.3)

NCERT Solutions for Class 10 Maths Chapter 2 is provided in this section. The Chapter 2 of Class 10 Mathematics under the NCERT Syllabus is about the Polynomials. As per the Textbook published by NCERT, the chapter discusses the concepts of the Geometrical Meaning of the Zeros of a Polynomial, the Relationship between Zeros and Coefficients of a Polynomial, and the Division Algorithm for Polynomials.

This post contains the step-by-step solutions of Exercise 2.3, Chapter 2 of Class 10 Mathematics designed in the easiest possible way. There is a total of 5 questions in the Ex. 2.3 and all the questions are covered in this section.

### NCERT Solutions for Class 10 Maths, Chapter 2 – Polynomials (Ex. 2.3)

NCERT Solutions for Class 10 Maths, Chapter 2 (Polynomials), Exercise 2.3 are described below. The Exercise 2.3 is based on Division Algorithm for Polynomials and each and every question of Ex. 2.3 are solved (step-by-step) below.

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

### i. $p(x)=x^3-3x^2+5x-3$, $g(x)=x^2-2$

Solution:

Dividing the polynomial $p(x)$ by $g(x)$,

Here, the Quotient is $x-3$ and the Remainder is $7x-9$.

### ii. $p(x)=x^4-3x^2+4x+5$, $g(x)=x^2+1-x$

Solution:

Here, $p(x)$ is in standard form but $g(x)$ is not in the standard form.

The standard form of $g(x)$ would be, $g(x)=x^2-x+1$.

Now, dividing the polynomial $p(x)$ by $g(x)$,

Here, the Quotient is $x^2+x-3$ and the Remainder is $8$.

### iii.$p(x)=x^4-5x+6$, $g(x)=2-x^2$

Solution:

Here, $p(x)$ is in standard form but $g(x)$ is not in the standard form.

The standard form of $g(x)$ would be, $g(x)=-x^2+2$.

Now, dividing the polynomial $p(x)$ by $g(x)$,

Here, the Quotient is $-x^2-2$ and the Remainder is $-5x+10$.

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

### i.$t^2-3$, $2t^4+3t^3-2t^2-9t-12$

Solution:

Dividing the second polynomial by the first polynomial,

Here, the remainder is $0$. Therefore, the first polynomial is a factor of the second polynomial.

### ii. $x^2+3x+1$, $3x^4+5x^3-7x^2+2x+2$

Solution:

Dividing the second polynomial by the first polynomial,

Here, the remainder is $0$. Therefore, the first polynomial is a factor of the second polynomial.

### iii. $x^3-3x+1$, $x^5-4x^3+x^2+3x+1$

Solution:

Dividing the second polynomial by the first polynomial,

Here, the remainder is $2 (\ne 0)$. Therefore, the first polynomial is not a factor of the second polynomial.

## Question 3: Obtain all the zeros of $3x^4+6x^3-2x^2-10x-5$, if two of its zeros are $\sqrt{\frac{5}{3}}$ and $-\sqrt{\frac{5}{3}}$.

Solution:

Given, two zeros of the polynomial are, $\sqrt{\frac{5}{3}}$ and $-\sqrt{\frac{5}{3}}$.

$\therefore$ the two factors of the polynomial will be,

$\left(x-\sqrt{\frac{5}{3}}\right)$ and $\left(x+\sqrt{\frac{5}{3}}\right)$

Multiplying the two factors we get,

$\left(x-\sqrt{\frac{5}{3}}\right) \left(x+\sqrt{\frac{5}{3}}\right)=x^2-\frac{5}{3}=\frac{3x^2-5}{3}$

Now, $\frac{3x^2-5}{3}$ is a factor of the polynomial.

$\therefore\,3x^2-5$ is also a factor of the given polynomial.

Dividing $3x^4+6x^3-2x^2-10x-5$ by $3x^2-5$ we get

Here, the quotient is $x^2+2x+1$ and the remainder is $0$.

The quotient can also be written as,

$x^2+2x+1=(x+1)^2$

The zeros of the quotient will be, $-1$ and $-1$.

Therefore, all the zeros of the given polynomial are, $\sqrt{\frac{5}{3}}$, $-\sqrt{\frac{5}{3}}$, $-1$ and $-1$.

## Question 4: On dividing $x^3-3x^2+x+2$ by a polynomial $g(x)$, the quotient and the remainder were $x-2$ and $-2x+4$, respectively. Find $g(x)$.

Solution:

Let us consider the polynomial as $p(x)=x^3-3x^2+x+2$, the quotient as $q(x)=x-2$ and the remainder as $r(x)=-2x+4$.

According to the division algorithm for polynomials,

$Dividend=(Divisor\times Quotient)+Remainder$

$\Rightarrow p(x)=g(x)\times q(x)+r(x)\,\,\,\,\,\,\,\,\,\,……(i)$

Putting the values of $p(x),\,q(x)$ and $r(x)$ in $equation(i)$ we get,

$x^3-3x^2+x+2=g(x)\times (x-2)+(-2x+4)$

$\Rightarrow x^3-3x^2+x+2=g(x)\times (x-2)-2x+4$

$\Rightarrow x^3-3x^2+x+2+2x-4=g(x)\times (x-2)$

$\Rightarrow x^3-3x^2+3x-2=g(x)\times (x-2)$

$\Rightarrow g(x)=\frac{x^3-3x^2+3x-2}{x-2}$

Now,

Therefore, $g(x)=x^2-x+1$.

## Question 5: Give examples of polynomial $p(x),\,g(x),\,q(x)$ and $r(x)$, which satisfy the division algorithm and

### i. deg $p(x)=$deg $q(x)$

Solution:

Let,

$p(x)=10x^2+15x+20$

$g(x)=5$

$q(x)=2x^2+3x+4$ and

$r(x)=0$

According to the division algorithm for polynomials,

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

Now,

$g(x)\times q(x)+r(x)=5\times (2x^2+3x+4)+0$

$\Rightarrow g(x)\times q(x)+r(x)=10x^2+15x+20=p(x)$

Therefore, the above polynomials satisfy the division algorithm for polynomials and also deg $p(x)=$deg $q(x)=2$.

### ii. deg $q(x)=$deg $r(x)$

Solution:

Let,

$p(x)=x^5+2x^4++3x^3+5x^2+2$

$g(x)=x^3+x^2+x+1$

$q(x)=x^2+x+1$ and

$r(x)=2x^2-2x+1$

According to the division algorithm for polynomials,

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

Now,

$g(x)\times q(x)+r(x)=(x^3+x^2+x+1)(x^2+x+1)+2x^2-2x+1$

$\Rightarrow g(x)\times q(x)+r(x)= x^5+2x^4++3x^3+5x^2+2 =p(x)$

Therefore, the above polynomials satisfy the division algorithm for polynomials and also deg $q(x)=$deg $r(x)=2$.

### iii. deg $r(x)=0$

Solution:

Let,

$p(x)=2x^4+8x^3+6x^2+4x+12$

$g(x)=x^4+4x^3+3x^2+2x+1$

$q(x)=2$ and

$r(x)=10$

According to the division algorithm for polynomials,

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

Now,

$g(x)\times q(x)+r(x)=(x^4+4x^3+3x^2+2x+1\times 2+10$

$\Rightarrow g(x)\times q(x)+r(x)= 2x^4+8x^3+6x^2+4x+12=p(x)$

Therefore, the above polynomials satisfy the division algorithm for polynomials and also deg $r(x)=0$.

### NCERT Solutions for Class 10 Maths Chapter 2, Exercise 2.3 PDF Download

The PDF of NCERT solutions for Class 10 Maths Chapter 2, Exercise 2.3 is provided below.

### List of Exercises in Polynomials, Chapter 2, Class 10 Maths

I hope the NCERT Solutions for Class 10 Maths Chapter 2, Exercise 2.3 have helped you. The solutions are designed keeping in mind that everyone can understand very easily if previous class concepts are clear.

If you have any issues/queries related to NCERT solutions for Class 10 Maths, Chapter 2, Ex 2.3, feel free to contact me at [email protected] or fill the form here.

### What are the main topics covered in the NCERT Solutions for Class 10 Maths Chapter 2?

The main topics covered in the NCERT Solutions for Class 10 Maths, Chapter 2 are: Geometrical Meaning of the Zeros of a Polynomial, Relationship between Zeros and Coefficients of a Polynomial and the Division Algorithm for Polynomials.

### How many exercises are there in the NCERT Solutions for Class 10 Maths Chapter 2?

There are 4 exercises in the NCERT Solutions for Class 10 Maths Chapter 2.

### Is the NCERT solutions for Class 10 Maths Chapter 2 important for exam point of view?

Yes, the NCERT solutions for Class 10 Maths Chapter 2 are very much important for exam point of view as it covers very basic concepts on Polynomials.

### Is Exercise 2.3 is of NCERT solutions for Class 10 Maths important for examination point of view?

Yes, Exercise 2.3 of NCERT solutions for Class 10 Maths is very important for examination point of view as covers the topic Division Algorithm for Polynomials which is very important for understanding the Polynomials.

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