# NCERT Solutions for Class 10 Maths (Chapter 2, Exercise 2.4)

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 2.4, Chapter 2 of Class 10 Mathematics designed in the easiest possible way. There is a total of 5 questions in the Ex. 2.4 and all the questions are covered in this section.

### NCERT Solutions for Class 10 Maths, Chapter 2 – Polynomials (Ex. 2.4)

NCERT Solutions for Class 10 Maths, Chapter 2 (Polynomials), Exercise 2.4 are described below. The Exercise 2.4 is based on Division Algorithm for Polynomials and each and every question of Ex. 2.4 are solved (step-by-step) below.

## Question 1: Verify that the numbers given alongside of the cubic polynomials below are their zeroes. Also, verify the relationship between the zeroes and coefficients in each case.

### i. $2x^3+x^2-5x+2$; $\frac{1}{2},\,1,\,-2$

Solution:

Let, $p(x)=2x^3+x^2-5x+2$

Now,

$p\left(\frac{1}{2}\right)=2\left(\frac{1}{2}\right)^3+\left(\frac{1}{2}\right)^2-5\left(\frac{1}{2}\right)+2$

$p\left(\frac{1}{2}\right)=\frac{1}{4}+\frac{1}{4}-\frac{5}{2}+2=0$

Similarly,

$p(1)=2(1)^3+(1)^2-5(1)+2$

$p(1)=2+1-5+2=0$

Again,

$p(-2)=2(-2)^3+(-2)^2-5(-2)+2$

$p(-2)=-16+4+10+2=0$

Therefore, $\frac{1}{2},\,1$ and $-2$ are the zeroes of the given polynomial.

Verifying:

Let,

$\alpha=\frac{1}{2}$

$\beta=1$ and

$\gamma=-2$

Now,

$\alpha+\beta+\gamma=\frac{1}{2}+1-2$

$\Rightarrow\alpha+\beta+\gamma=\frac{1+2-4}{2}=-\frac{1}{2}=-\frac{Coefficient\,of\,x^2}{Coefficient \,of\,x^3}$

And

$\alpha \beta+\beta \gamma+\gamma \alpha=\left(\frac{1}{2}\right)(1)+(1)(-2)+(-2)\left(\frac{1}{2}\right)$

$\Rightarrow \alpha \beta+\beta \gamma+\gamma \alpha=\frac{1}{2}-2-1=\frac{1-4-2}{2}=\frac{-5}{2}=\frac{Coefficient\,of\,x}{Coefficient\,of\,x^3}$

Again

$\alpha \beta \gamma=\frac{1}{2}\times 1\times (-2)$

$\alpha \beta \gamma=-\frac{2}{2}=-\frac{Constant\,term}{Coefficient\,of\,x^3}$

Therefore, the relationship between the zeroes and the coefficients of the polynomial is verified.

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

Solution:

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

Now,

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

$p(2)=8-16+10-2=0$

Similarly,

$p(1)=(1)^3-4(1)^2+5(1)-2$

$p(1)=1-4+5-2=0$

Therefore, $2,\,1$ and $1$ are the zeroes of the given polynomial.

Verifying:

Let,

$\alpha=2$

$\beta=1$ and

$\gamma=1$

Now,

$\alpha+\beta+\gamma=2+1+1=4=-\frac{-4}{1}=-\frac{Coefficient\,of\,x^2}{Coefficient\,of\,x^3}$

And,

$\alpha \beta+\beta \gamma+\gamma \alpha=2\times 1+1\times 1+1\times 2=2+1+2=5=\frac{5}{1}=\frac{Coefficient\,of\,x}{Coefficient\,of\,x^3}$

Again,

$\alpha \beta \gamma=2\times 1\times 1=2=-\frac{-2}{1}=-\frac{Constant\,term}{Coefficient\,of\,x^3}$

Therefore, the relationship between the zeroes and the coefficients of the polynomial is verified.

## Question 2: Find a cubic polynomial with the sum, sum of the product of its zeroes taken two at a time, and the product of its zeroes as $2,\,-7,\,-14$ respectively.

Solution:

Let us consider the zeroes of the cubic polynomial be $\alpha,\,\beta$ and $\gamma$.

Then, according to the question,

$\alpha+\beta+\gamma=2$

$\alpha \beta+\beta \gamma+\gamma \alpha=-7$ and

$\alpha \beta \gamma=-14$

Therefore, the required polynomial will be,

$x^3-(\alpha+\beta+\gamma)x^2+(\alpha \beta+\beta \gamma+\gamma \alpha)x-(\alpha \beta \gamma)=x^3-2x^2-7x+14$

## Question 3: If the zeroes of the polynomial $x^3-3x^2+x+1$ are $a-b,\,a$ and $a+b$, then find $a$ and $b$.

Solution:

Given, $(a-b),\,a$ and $(a+b)$ are the zeros of the polynomial $x^3-3x^2+x+1$.

Comparing the given polynomial with $Ax^3+Bx^2+Cx+D$, we get

$A=1,\,B=-3,\,C=1$ and $D=1$

Now, sum of zeros$=(a-b)+a+(a+b)=-\frac{B}{A}$

Putting the values of $B$ and $A$,

$\Rightarrow 3a=\frac{-(-3)}{1}$

$\Rightarrow 3a=3$

$\Rightarrow a=1$

Again, the sum of the product of zeroes taken two at a time,

$\Rightarrow a(a-b)+a(a+b)+(a+b)(a-b)=\frac{C}{A}$

Putting the values of $C$ and $A$ we get

$\Rightarrow a^2-ab+a^2+ab+a^2-b^2=\frac{1}{1}$

$\Rightarrow 3a^2-b^2=1$

$\Rightarrow 3(1)^2-b^2=1$

$\Rightarrow -b^2=1-3$

$\Rightarrow b^2=2$

$\Rightarrow b=\pm \sqrt{2}$

Therefore, $a=1$ and $b=\pm \sqrt{2}$

### NCERT Solutions for Class 10 Maths Chapter 2, Exercise 2.4 PDF Download

The PDF of NCERT solutions for Class 10 Maths Chapter 2, Exercise 2.4 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 2.4 have helped you. The solutions are designed keeping in mind that everyone can understand very easily if previous class concepts are clear.

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