NCERT Solutions for Exercise 2.2 Class 9 Maths Chapter 2

NCERT Solutions for Exercise 2.2 Class 9 Maths Chapter 2 - Polynomials

Edited By Vishal kumar | Updated on Oct 03, 2023 01:23 PM IST

NCERT Solutions for Class 9 Maths Chapter 2: Polynomials Exercise 2.2- Download Free PDF

NCERT Solutions for Class 9 Maths Chapter 2: Polynomials Exercise 2.2- NCERT Solutions for Exercise 2.2 Class 9 Maths is the subpart of NCERT Solutions for Class 9 Mathematics. So the solutions given are relatively helpful while studying and pursuing homework. Here In this NCERT book Class 9 Maths Exercise 2.2, we will be studying Polynomials. In Chapter 2 of Class 9 NCERT Mathematics, the concept of the relation between polynomials and non-polynomials is discussed. The Class 9 Maths chapter 2 exercise 2.2 covers the topic verification of zeroes of a particular equation. This main lesson of NCERT solutions for Class 9 Maths chapter 2 exercise 2.2. is to find the functional value of the equation.

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NCERT Solutions for Exercise 2.2 Class 9 Maths Chapter 2 - Polynomials

Class 9 Maths Chapter 2 exercise 2.2 consists of four questions, each with multiple parts. These solutions have been expertly crafted by Careers360 subject experts. Students can download the PDF of these 9th class maths exercise 2.2 answers for offline use, and they are provided free of charge. Apart from Class 9 Maths Chapter 1 exercise 2.2, the following exercises are also present.

NCERT Solutions for Class 9 Maths Chapter 2 – Polynomials Exercise 2.2

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Access Polynomials Class 9 Chapter 2 Exercise: 2.2

Q1 (i) Find the value of the polynomial $5x - 4x^2 +3$ at $x = 0$

Answer:

Given polynomial is $5x - 4x^2 +3$

Now, at $x = 0$ value is

$\Rightarrow 5(0)-4(0)^2+3 = 0 - 0 + 3 = 3$

Therefore, value of polynomial $5x - 4x^2 +3$ at x = 0 is 3

Q1 (ii) Find the value of the polynomial $5x - 4x^2 +3$ at $x = -1$

Answer:

Given polynomial is $5x - 4x^2 +3$

Now, at $x = -1$ value is

$\Rightarrow 5(-1)-4(-1)^2+3 = -5 - 4 + 3 = -6$

Therefore, value of polynomial $5x - 4x^2 +3$ at x = -1 is -6

Q1 (iii) Find the value of the polynomial $5x - 4x^2 +3$ at $x = 2$

Answer:

Given polynomial is $5x - 4x^2 +3$

Now, at $x = 2$ value is

$\Rightarrow 5(2)-4(2)^2+3 = 10 - 16 + 3 = -3$

Therefore, value of polynomial $5x - 4x^2 +3$ at x = 2 is -3

Q2 (i) Find p(0) , p(1) and p(2) for each of the following polynomials: $p(y)= y^2 - y + 1$

Answer:

Given polynomial is

$p(y)= y^2 - y + 1$

Now,

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

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

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

Therefore, values of p(0) , p(1) and p(2) are 1 , 1 and 3 respectively .

Q2 (ii) Find p(0) , p(1) and p(2) for each of the following polynomials: $p(t) = 2 + t + 2t^2 - t^3$

Answer:

Given polynomial is

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

Now,

$p(0) = 2 + 0 + 2(0)^2 - (0)^3 = 2$

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

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

Therefore, values of p(0) , p(1) and p(2) are 2 , 4 and 4 respectively

Q2 (iii) Find p(0), p(1) and p(2) for each of the following polynomials: $p(x) = x^3$

Answer:

Given polynomial is

$p(x) = x^3$

Now,

$p(0) = (0)^3 =0$

$p(1) = (1)^3=1$

$p(2) = (2)^3=8$

Therefore, values of p(0) , p(1) and p(2) are 0 , 1 and 8 respectively

Q2 (iv) Find p(0), p(1) and p(2) for each of the following polynomials: $p(x)= (x-1)(x+ 1)$

Answer:

Given polynomial is

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

Now,

$p(0) = (0)^2-1 = -1$

$p(1) = (1)^2-1 = 0$

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

Therefore, values of p(0) , p(1) and p(2) are -1 , 0 and 3 respectively

Q3 (i) Verify whether the following are zeroes of the polynomial, indicated against it. $p(x) = 3x + 1, x = -\frac{1}{3}$

Answer:

Given polynomial is $p(x) = 3x + 1$

Now, at $x = -\frac{1}{3}$ it's value is

$p\left ( -\frac{1}{3} \right )=3\times \left ( -\frac{1}{3} \right )+1 = -1+1=0$

Therefore, yes $x = -\frac{1}{3}$ is a zero of polynomial $p(x) = 3x + 1$

Q3 (ii) Verify whether the following are zeroes of the polynomial, indicated against it. $p(x) = 5x - \pi, x = \frac{4}{5}$

Answer:

Given polynomial is $p(x) = 5x - \pi$

Now, at $x =\frac{4}{5}$ it's value is

$p\left ( \frac{4}{5} \right )=5\times \left ( \frac{4}{5} \right ) -\pi = 4-\pi \neq 0$
Therefore, no $x =\frac{4}{5}$ is not a zero of a polynomial $p(x) = 5x - \pi$

Q3 (iii) Verify whether the following are zeroes of the polynomial, indicated against it. $p(x) = x^2 -1, x = 1,-1$

Answer:

Given polynomial is $p(x) = x^2-1$

Now, at x = 1 it's value is

$p(1) = (1)^2-1 = 1 -1 =0$

And at x = -1

$p(-1) = (-1)^2-1 = 1 -1 =0$
Therefore, yes x = 1 , -1 are zeros of polynomial $p(x) = x^2-1$

Q3 (iv) Verify whether the following are zeroes of the polynomial, indicated against it. $p(x) = (x + 1)(x-2), x = -1,2$

Answer:

Given polynomial is $p(x) = (x+1)(x-2)$

Now, at x = 2 it's value is

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

And at x = -1

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

Therefore, yes x = 2 , -1 are zeros of polynomial $p(x) = (x+1)(x-2)$

Q3 (v) Verify whether the following are zeroes of the polynomial, indicated against it. $p(x) = x^2. x =0$

Answer:

Given polynomial is $p(x) = x^2$

Now, at x = 0 it's value is

$p(0) = (0)^2=0$

Therefore, yes x = 0 is a zeros of polynomial $p(x) = (x+1)(x-2)$

Q3 (vi) Verify whether the following are zeroes of the polynomial, indicated against it. $p(x) = lx + m, \ x =- \frac{m}{l}$

Answer:

Given polynomial is $p(x) = lx+m$

Now, at $x = -\frac{m}{l}$ it's value is

$p\left ( -\frac{m}{l} \right )= l \times \left ( -\frac{m}{l} \right )+m = -m+m =0$

Therefore, yes $x = -\frac{m}{l}$ is a zeros of polynomial $p(x) = lx+m$

Q3 (vii) Verify whether the following are zeroes of the polynomial, indicated against it. $p(x) = 3x^2 - 1, \ x = -\frac{1}{\sqrt3}, \frac{2}{\sqrt3}$

Answer:

Given polynomial is $p(x) = 3x^2-1$

Now, at $x = -\frac{1}{\sqrt3}$ it's value is

$p\left ( -\frac{1}{\sqrt3} \right )= 3 \times \left ( -\frac{1}{\sqrt3} \right )^2-1 = 1-1 =0$

And at $x = \frac{2}{\sqrt3}$

$p\left ( \frac{2}{\sqrt3} \right )= 3 \times \left ( \frac{2}{\sqrt3} \right )^2-1 = 4-1 =3\neq 0$

Therefore, $x = -\frac{1}{\sqrt3}$ is a zeros of polynomial $p(x) = 3x^2-1$ .

whereas $x = \frac{2}{\sqrt3}$ is not a zeros of polynomial $p(x) = 3x^2-1$

Q3 (viii) Verify whether the following are zeroes of the polynomial, indicated against it. $p(x) = 2x +1,\ x = \frac{1}{2}$

Answer:

Given polynomial is $p(x) = 2x+1$

Now, at $x = \frac{1}{2}$ it's value is

$p\left ( \frac{1}{2} \right )= 2 \times \left ( \frac{1}{2} \right )+1 = 1+1=2 \neq 0$

Therefore, $x = \frac{1}{2}$ is not a zeros of polynomial $p(x) = 2x+1$

Q4 (i) Find the zero of the polynomial in each of the following cases: $p(x)= x + 5$

Answer:

Given polynomial is $p(x)= x + 5$

Zero of a polynomial is that value of the variable at which the value of the polynomial is obtained as 0.

Now,

$p(x)=0$

$\Rightarrow x+5 = 0$

$\Rightarrow x=-5$

Therefore, x = -5 is the zero of polynomial $p(x)= x + 5$

Q4 (ii) Find the zero of the polynomial in each of the following cases: $p(x) = x - 5$

Answer:

Given polynomial is $p(x)= x - 5$

Zero of a polynomial is that value of the variable at which the value of the polynomial is obtained as 0.

Now,

$p(x)=0$

$\Rightarrow x-5 = 0$

$\Rightarrow x=5$

Therefore, x = 5 is a zero of polynomial $p(x)= x - 5$

Q4 (iii) Find the zero of the polynomial in each of the following cases: $p(x)= 2x + 5$

Answer:

Given polynomial is $p(x)= 2x + 5$

Zero of a polynomial is that value of the variable at which the value of the polynomial is obtained as 0.

Now,

$p(x)=0$

$\Rightarrow 2x+5 = 0$

$\Rightarrow x=-\frac{5}{2}$

Therefore, $x=-\frac{5}{2}$ is a zero of polynomial $p(x)= 2x + 5$

Q4 (iv) Find the zero of the polynomial in each of the following cases: $p(x) = 3x - 2$

Answer:

Given polynomial is $p(x) = 3x - 2$

Zero of a polynomial is that value of the variable at which the value of the polynomial is obtained as 0.

Now,

$p(x)=0$

$\Rightarrow 3x-2 = 0$

$\Rightarrow x=\frac{2}{3}$
Therefore, $x=\frac{2}{3}$ is a zero of polynomial $p(x) = 3x - 2$

Q4 (v) Find the zero of the polynomial in each of the following cases: $p(x) = 3x$

Answer:

Given polynomial is $p(x) = 3x$

Zero of a polynomial is that value of the variable at which the value of the polynomial is obtained as 0.

Now,

$p(x)=0$

$\Rightarrow 3x = 0$

$\Rightarrow x=0$

Therefore, $x=0$ is a zero of polynomial $p(x) = 3x$

Q4 (vi) Find the zero of the polynomial in each of the following cases: $p(x) = ax, \ a\neq 0$

Answer:

Given polynomial is $p(x) = ax$

Zero of a polynomial is that value of the variable at which the value of the polynomial is obtained as 0.

Now,

$p(x)=0$

$\Rightarrow ax = 0$

$\Rightarrow x=0$

Therefore, $x=0$ is a zero of polynomial $p(x) = ax$

Q4 (vii) Find the zero of the polynomial in each of the following cases: $p(x) = cx + d, c\neq 0, c,d$ are real numbers

Answer:

Given polynomial is $p(x) = cx+d$

Zero of a polynomial is that value of the variable at which the value of the polynomial is obtained as 0.

Now,

$p(x)=0$

$\Rightarrow cx+d = 0$

$\Rightarrow x=-\frac{d}{c}$

Therefore, $x=-\frac{d}{c}$ is a zero of polynomial $p(x) = cx+d$

Benefits of NCERT Solutions for Class 9 Maths Chapter 2 Exercise 2.2

The NCERT solutions for class 9 maths ex 2.2 and the solved example before exercise 2.2 Class 9 Maths are crucial since they feature problems from the basic Polynomials analysis, which will provide students with an understanding of numbers other than whole numbers.
Students would be able to grasp the genuine concept of polynomials supplied in 9th class maths exercise 2.2 answers if they can solve and understand each question in this exercise 2.2 Class 9 Maths.
Students may receive short-answer or long answer questions from the types presented in Class 9 Maths chapter 1 Exercise 2.2 for final exams.

Key features of Class 9 Maths Chapter 2 Exercise 2.2

Comprehensive Coverage: Ex 2.2 class 9 provides a comprehensive set of questions that cover various aspects of polynomials.
Polynomial Basics: This class 9 maths ex 2.2 reinforces fundamental concepts related to polynomials, including their definition, degree, leading coefficient, and classification.
Identification of Polynomials: Students learn how to identify polynomials among different types of expressions and equations.
Expert-Created Solutions: exercise 2.2 class 9 mathsto the problems are typically provided in the exercise. These solutions are crafted by subject matter experts to ensure accuracy and clarity.
PDF Availability: Students can often download a PDF version of the solutions, allowing them to access and use them offline. This resource is provided at no cost.

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Frequently Asked Questions (FAQs)

1. How do you define polynomials?

The name "polynomial" derives from the words "poly" (which means numerous) and "nomial" (which means phrase).

2. In chapter 2, how many exercises are there?

Ans: 5 exercises in chapter 2 there are,. Ex 2.2, 2.2, 2.3, 2.4, 2.5.

3. Give an example of a monomial?

A monomial is a form of the polynomial with only one term, such as ax^2.

4. Can you give a binomial example?

A binomial polynomial is one with only two terms, such as = ax^2 + b.

5. How many degrees does the polynomial x^2 +x have?

Since the highest power of x^2 +x is 2, the degree equals 2 and the polynomial is quadratic.

6. What is the name of the polynomial with degree = 4?

A biquadratic polynomial is a polynomial with degree = 4, which is not covered in Chapter 2 class 9.

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