NCERT Solutions Class 10 Maths Chapter 2 Polynomials Exercise 2.2

NCERT Solutions Class 10 Maths Chapter 2 Exercise 2.2 will help you understand the coefficients and zeroes of a polynomial and the relation of both of these entities. It facilitates an understanding of the basic ideas of polynomials, which play a critical role in understanding advanced mathematics concepts. The step-by-step explanation of the exercise below adheres to the newest CBSE syllabus and examination protocols. Hence, by understanding these relations, students can approach polynomial problems confidently in exams.

1.0Download NCERT Solutions for Class 10 Maths Chapter 2 Exercise 2.2 : Free PDF

2.0Introduction to Polynomial: Coefficients and Zeroes of a Polynomial

A polynomial is a mathematical expression involving variables, constants, and variable exponents combined through addition, subtraction, and multiplication operations. The standard form of a polynomial in one variable can be expressed as: 

$p(x)=a{x}^n+b{x}^{n-1}+c{x}^{n-2}+....+k$

Here, a, b, c,… are constants, and n is a non-negative integer. The degree of a polynomial depends upon the value of n, in other words, the degree of a polynomial is the highest power of the variable x. For example, a polynomial with degree 1 is known as a Linear polynomial, with degree 2 it is a quadratic one, while with degree 3 the polynomial is a cubic polynomial.

Coefficients of a Polynomial

The coefficients of a polynomial are the constant factors multiplied by the variable terms. In a polynomial expression, the coefficients are the numbers appearing before every power of the variable. For instance, in the polynomial, 3x² + 5x – 2, 3 is the coefficient of x², 5 is the coefficient of x, and –2 is the constant term. 

Zeroes of a Polynomial

The zeroes or roots of a polynomial are the values of the variable for which the polynomial equals zero. Or, in simple terms, the zeroes are solutions to the equation that is formed by setting the polynomial equal to zero. For Example, take the polynomial p(x) = 2x² – 8x + 6, the zeroes for this equation are 1 and 3, as when put in the equation the equation becomes zeroes for both the values. 

3.0Exercise 2.2 Overview: Key Concepts

Relationship Between Zeroes and Coefficients of Polynomials

The connection between the zeroes (roots) of a polynomial and the coefficients is one of the most basic ideas in algebra, particularly when handling quadratic and cubic polynomials. These connections enable us to determine significant characteristics of a polynomial without necessarily solving it fully. Since we know the coefficients, we can conclude facts about the zeroes, and the same applies in reverse. Let's see this connection for quadratic and cubic polynomials: 

Relationship for Quadratic Polynomials: 

To understand the relationship between Quadratic Polynomials, let’s take a quadratic polynomial in standard form: 

$p(x)=a{x}^2+bx+c$

Here, a and b are the coefficients of the variable x, and c is the constant, where a0. The zeroes (also known as roots) of the quadratic polynomials are denoted by and . 

• The sum of the zeroes: The sum of the zeroes of a quadratic polynomial is expressed as:   

$\alpha +\beta =-\frac{Coefficientofx}{Coefficientof{x}^2}=-\frac{b}{a}$

• The product of the zeroes: The product of zeroes of a quadratic polynomial is expressed as:  

$\alpha \beta =-\frac{Constantterm}{Coefficientof{x}^2}=\frac{c}{a}$

Relationship for Cubic Polynomials: 

Take a cubic polynomial in its standard form to understand the relationship for cubic polynomials: 

$p(x)=a{x}^3+b{x}^2+cx+d$

Here, a, b, and c are the coefficients of the variable. In this case, x and d are the constant terms, where $a\ne 0$. The zeroes or the roots of a cubic polynomial are denoted by $\alpha ,\beta and\gamma$ . These zeroes are the values of x, which makes the polynomial equal to zero. 

• The sum of the zeroes: The sum of the zeroes of a cubic polynomial is equal to: 

$\alpha +\beta +\gamma =-\frac{coefficientof{x}^2}{coefficientof{x}^3}=-\frac{b}{a}$

• The sum of the products of the zeroes taken two at a time: It is equal to: 

$\alpha \beta +\beta \gamma +\gamma \alpha =\frac{coefficientofx}{coefficientof{x}^3}=\frac{c}{a}$

• Product of the Zeroes: The product of the zeroes is equal to:  

$\alpha \beta \gamma =\frac{constantterm}{coefficientof{x}^3}=-\frac{d}{a}$

Formation of Polynomials from Zeroes

1. Formula of Quadratic Polynomial: The equation of quadratic polynomial is formed by the formula: 

$p(x)=x^2-(\alpha +\beta )x+\alpha \beta$

2. The formula for Cubic Polynomial: The equation of cubic polynomial is formed by the formula:

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

Also Read: CBSE Class 10 Maths Chapter 2 Polynomials

4.0NCERT Class 10 Maths Chapter 2 Exercise 2.2 : Detailed Solution

1. Find the zeros of the following quadratic polynomials and verify the relationship between the zeros and the coefficients.

(i) x² - 2x - 8

x² - 2x - 8 = x² - 4x + 2x - 8 = x(x - 4) + 2(x - 4) = (x + 2)(x - 4)

Zeros are -2 and 4.

Sum of zeros = (-2) + (4) = 2 = -(-2)/1 = -(Coefficient of x) / (Coefficient of x²)

Product of zeros = (-2)(4) = -8 = -8/1 = (Constant term) / (Coefficient of x²)

(ii) 4s² - 4s + 1

4s² - 4s + 1 = (2s - 1)²

Zeros are 1/2, 1/2.

Sum of zeros = 1/2 + 1/2 = 1 = -(-4)/4 = -(Coefficient of s) / (Coefficient of s²)

Product of zeros = (1/2)(1/2) = 1/4 = (Constant term) / (Coefficient of s²)

(iii) 6x² - 3 - 7x

6x² - 7x - 3 = 6x² - 9x + 2x - 3 = 3x(2x - 3) + 1(2x - 3) = (2x - 3)(3x + 1)

Zeros are 3/2, -1/3.

Sum of zeros = 3/2 + (-1/3) = 7/6 = -(-7)/6 = -(Coefficient of x) / (Coefficient of x²)

Product of zeros = (3/2)(-1/3) = -1/2 = -3/6 = (Constant term) / (Coefficient of x²)

(iv) 4u² + 8u

4u² + 8u = 4u(u + 2)

Zeros are 0, -2.

Sum of zeros = 0 + (-2) = -2 = -8/4 = -(Coefficient of u) / (Coefficient of u²)

Product of zeros = 0(-2) = 0 = 0/4 = (Constant term) / (Coefficient of u²)

(v) t² - 15

t² - 15 = (t - √15)(t + √15)

Zeros are √15, -√15.

Sum of zeros = √15 + (-√15) = 0 = 0/1 = -(Coefficient of t) / (Coefficient of t²)

Product of zeros = (√15)(-√15) = -15 = -15/1 = (Constant term) / (Coefficient of t²)

(vi) 3x² - x - 4

3x² - x - 4 = 3x² - 4x + 3x - 4 = x(3x - 4) + 1(3x - 4) = (3x - 4)(x + 1)

Zeros are 4/3, -1.

Sum of zeros = 4/3 - 1 = 1/3 = -(-1)/3 = -(Coefficient of x) / (Coefficient of x²)

Product of zeros = (4/3)(-1) = -4/3 = (Constant term) / (Coefficient of x²)

2. Find a quadratic polynomial each with the given numbers as the sum and product of its zeros respectively.

(i) 1/4, -1

Required polynomial = x² - (sum of zeros)x + (product of zeros) = x² - (1/4)x - 1 = (1/4)(4x² - x - 4)

(ii) √2, 1/3

Required polynomial = x² - (sum of zeros)x + (product of zeros) = x² - √2x + 1/3 = (1/3)(3x² - 3√2x + 1)

(iii) 0, √5

Required polynomial = x² - (sum of zeros)x + (product of zeros) = x² - 0x + √5 = x² + √5

(iv) 1, 1

Required polynomial = x² - (sum of zeros)x + (product of zeros) = x² - 1x + 1 = x² - x + 1

(v) -1/4, 1/4

Required polynomial = x² - (sum of zeros)x + (product of zeros) = x² - (-1/4)x + 1/4 = x² + (1/4)x + 1/4 = (1/4)(4x² + x + 1)

(vi) 4, 1

Required polynomial = x² - (sum of zeros)x + (product of zeros) = x² - 4x + 1

5.0Benefits of Studying NCERT Solutions Class 10 Maths Chapter 2 Exercise 2.2

• Fundamentals strengthened for better learning
• Learn smart strategies for problem-solving
• Confident exam preparation
• Easy breakdown of key mathematical concepts