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

Exercise 2.2

1. Find the value of the polynomial $\text{5x - 4}{{\text{x}}^{\text{2}}}\text{ + 3}$  at 

i) $\text{x = 0}$

Ans: Let $\text{p}\left( \text{x} \right)\text{ = 5x - 4}{{\text{x}}^{\text{2}}}\text{ + 3}$

We will simply put the value of $\text{x}$ in the given polynomial.

$\Rightarrow \text{p}\left( \text{0} \right)\text{ = 5}\left( \text{0} \right)\text{ - 4}\left( {{\text{0}}^{\text{2}}} \right)\text{ + 3}$

$\Rightarrow \text{p}\left( \text{0} \right)\text{ = 3}$

Hence, the value of the polynomial at $\text{x = 0}$ is $\text{3}$.

ii) $\text{x = -1}$ 

Ans: Let $\text{p}\left( \text{x} \right)\text{ = 5x - 4}{{\text{x}}^{\text{2}}}\text{ + 3}$

We will simply put the value of $\text{x}$ in the given polynomial.

$\Rightarrow \text{p}\left( \text{-1} \right)\text{ = 5}\left( \text{-1} \right)\text{ - 4}\left[ {{\left( \text{-1} \right)}^{\text{2}}} \right]\text{ + 3}$

$\Rightarrow \text{p}\left( \text{-1} \right)\text{ = - 5 - 4 + 3}$

$\Rightarrow \text{p}\left( \text{-1} \right)\text{ = - 6}$

Hence, the value of the polynomial at $\text{x = -1}$ is $\text{-6}$.

iii) $\text{x = 2}$ 

Ans: Let $\text{p}\left( \text{x} \right)\text{ = 5x - 4}{{\text{x}}^{\text{2}}}\text{ + 3}$

We will simply put the value of $\text{x}$ in the given polynomial.

$\Rightarrow \text{p}\left( \text{2} \right)\text{ = 5}\left( \text{2} \right)\text{ - 4}\left( {{\text{2}}^{\text{2}}} \right)\text{ + 3}$

$\Rightarrow \text{p}\left( \text{2} \right)\text{ = 10 - 16 + 3}$

$\Rightarrow \text{p}\left( \text{2} \right)\text{ = - 3}$

Hence, the value of the polynomial at $\text{x = 2}$ is $\text{-3}$.

2. Find $\text{p}\left( \text{0} \right)$, $\text{p}\left( \text{1} \right)$  and $\text{p}\left( \text{2} \right)$  for each of the following polynomials:

i) $\text{p}\left( \text{y} \right)\text{ = }{{\text{y}}^{\text{2}}}\text{ - y + 1}$ 

Ans:

• $\text{p}\left( \text{0} \right)\text{ = }{{\text{0}}^{\text{2}}}\text{ - 0 + 1}$.

So, we get the value of $\text{p}\left( \text{0} \right)\text{ = 1}$.

• $\text{p}\left( \text{1} \right)\text{ = }{{\text{1}}^{\text{2}}}\text{ - 1 + 1 = 1}$

So, we get the value of $\text{p}\left( \text{1} \right)\text{ = 1}$.

• $\text{p}\left( \text{2} \right)\text{ = }{{\text{2}}^{\text{2}}}\text{ - 2 + 1}$

$\Rightarrow \text{p}\left( \text{2} \right)\text{ = 4 - 2 + 1}$

$\Rightarrow \text{p}\left( \text{2} \right)\text{ = 3}$

So, we get the value of $\text{p}\left( \text{2} \right)\text{ = 3}$.

ii) $\text{p}\left( \text{t} \right)\text{ = 2 + t + 2}{{\text{t}}^{\text{2}}}\text{ - }{{\text{t}}^{\text{3}}}$

Ans:

• $\text{p}\left( \text{0} \right)\text{ = 2 + 0 + 2}\left( {{\text{0}}^{\text{2}}} \right)\text{ - }\left( {{\text{0}}^{\text{3}}} \right)\text{ = 2}$

So, we get the value of $\text{p}\left( \text{0} \right)\text{ = 2}$.

• $\text{p}\left( \text{1} \right)\text{ = 2 + 1 + 2}\left( {{\text{1}}^{\text{2}}} \right)\text{ - }\left( {{\text{1}}^{\text{3}}} \right)\text{ = 4}$

So, we get the value of $\text{p}\left( \text{1} \right)\text{ = 4}$.

• $\text{p}\left( \text{2} \right)\text{ = 2 + 2 + 2}\left( {{\text{2}}^{\text{2}}} \right)\text{ - }\left( {{\text{2}}^{\text{3}}} \right)$

$\Rightarrow \text{p}\left( \text{2} \right)\text{ = 4 + 8 - 8}$

$\Rightarrow \text{p}\left( \text{2} \right)\text{ = 4}$

So, we get the value of $\text{p}\left( \text{2} \right)\text{ = 4}$.

iii) $\text{p}\left( \text{x} \right)\text{ = }{{\text{x}}^{\text{3}}}$

Ans:

• $\text{p}\left( \text{0} \right)\text{ = }{{\text{0}}^{\text{3}}}\text{ = 0}$

So, we get the value of $\text{p}\left( \text{0} \right)\text{ = 0}$.

• $\text{p}\left( \text{1} \right)\text{ = }{{\text{1}}^{\text{3}}}\text{ = 1}$

So, we get the value of $\text{p}\left( \text{1} \right)\text{ = 1}$.

• $\text{p}\left( \text{2} \right)\text{ = }{{\text{2}}^{\text{3}}}\text{ = 8}$

So, we get the value of $\text{p}\left( \text{2} \right)\text{ = 8}$.

iv) $\text{p}\left( \text{x} \right)\text{ =}\left( \text{x - 1} \right)\left( \text{x + 1} \right)$

Ans:

• $\text{p}\left( \text{0} \right)\text{ =}\left( \text{0 - 1} \right)\left( \text{0 + 1} \right)\text{ = -1}$

So, we get the value of $\text{p}\left( \text{0} \right)\text{ = -1}$.

• $\text{p}\left( \text{1} \right)\text{ =}\left( \text{1 - 1} \right)\left( \text{1 + 1} \right)\text{ = 0}$

So, we get the value of $\text{p}\left( \text{1} \right)\text{ = 0}$.

• $\text{p}\left( \text{2} \right)\text{ =}\left( \text{2 - 1} \right)\left( \text{2 + 1} \right)$

$\Rightarrow \text{p}\left( \text{2} \right)\text{ = }\left( \text{1} \right)\left( \text{3} \right)\text{ = 3}$

So, we get the value of $\text{p}\left( \text{2} \right)\text{ = 3}$.

3. Verify whether the following are zeroes of the polynomial, indicated against them.

i) $\text{p}\left( \text{x} \right)\text{ = 3x + 1}$,$\text{x = -}\frac{\text{1}}{\text{3}}$

Ans: We are given: $\text{x = -}\frac{\text{1}}{\text{3}}$. If it is the zero of the polynomial $\text{p}\left( \text{x} \right)\text{ = 3x + 1}$, then $\text{p}\left( \text{-}\frac{\text{1}}{\text{3}} \right)$ should be  $\text{0}$.

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

Hence, we can say that $\text{x = -}\frac{\text{1}}{\text{3}}$ is a zero of the given polynomial.

ii) $\text{p}\left( \text{x} \right)\text{= 5x - }\!\!\pi\!\!\text{ }$ ,$\text{x = }\frac{\text{4}}{\text{5}}$

Ans: We are given: $\text{x = }\frac{\text{4}}{\text{5}}$. If it is the zero of the polynomial $\text{p}\left( \text{x} \right)\text{ = 5x -  }\text{ }\!\!\pi\!\!\text{ }$, then $\text{p}\left( \frac{\text{4}}{\text{5}} \right)$ should be  $\text{0}$.

$\text{p}\left( \frac{\text{4}}{\text{5}} \right)\text{ = 5}\left( \frac{\text{4}}{\text{5}} \right)\text{ - 3}\text{.14 = 4 - 3}\text{.14 }\ne \text{ 0}$

Hence, we can say that $\text{x = }\frac{\text{4}}{\text{5}}$ is not a zero of the given polynomial.

iii) $\text{p(x) = }{{\text{x}}^{\text{2}}}\text{-1, x = 1, -1}$

Ans: We are given: $\text{x = 1}$  and  $\text{x = -1}$ . 

If they are zeros of polynomial  $\text{p(x) = }{{\text{x}}^{\text{2}}}\text{-1}$ , then $\text{p}\left( \text{1} \right)$ and $\text{p}\left( \text{-1} \right)$ should both be $\text{0}$.

$\text{p}\left( \text{1} \right)\text{ = }{{\left( \text{1} \right)}^{\text{2}}}\text{-1 = 0}$

$\text{p}\left( \text{-1} \right)\text{ = }{{\left( \text{-1} \right)}^{\text{2}}}\text{-1 = 0}$

Hence, we can say that $\text{x = 1}$ and $\text{x = -1}$  are zeroes of the given polynomial.

iv) $\text{p(x) = (x+1)(x-2), x = -1, 2}$

Ans: We are given: $\text{x = -1}$ and $\text{x = 2}$. 

If they are zeroes of the polynomial $\text{p}\left( \text{x} \right)\text{ = }\left( \text{x+1} \right)\left( \text{x-2} \right)$ , then $\text{p}\left( \text{-1} \right)$ and $\text{p}\left( \text{2} \right)$ should be $\text{0}$.

$\text{p}\left( \text{-1} \right)\text{ = }\left( \text{-1+1} \right)\left( \text{-1-2} \right)\text{ = }\left( \text{0} \right)\left( \text{-3} \right)\text{ = 0}$

$\text{p}\left( \text{2} \right)\text{ = }\left( \text{2+1} \right)\left( \text{2-2} \right)\text{ = }\left( \text{3} \right)\left( \text{0} \right)\text{ = 0}$

Hence, we can say that $\text{x = -1}$ and $\text{x = 2}$  are zeroes of the given polynomial.

v) $\text{p(x) = }{{\text{x}}^{\text{2}}}\text{, x = 0}$

Ans: We are given $\text{x = 0}$. 

If it is a zero of the polynomial $\text{p}\left( \text{x} \right)\text{ = }{{\text{x}}^{\text{2}}}$  , then $\text{p}\left( \text{0} \right)$ should be $\text{0}$.

$\text{ p}\left( \text{0} \right)\text{ = }{{\left( \text{0} \right)}^{\text{2}}}\text{ = 0}$

Hence, we can say that $\text{x = 0}$ is a zero of the given polynomial.

vi) $\text{p(x) = lx + m, x = -}\frac{\text{m}}{\text{l}}$

Ans: We are given: $\text{x = -}\frac{\text{m}}{\text{l}}$. 

If it is a zero of the polynomial  $\text{p}\left( \text{x} \right)\text{ = lx + m}$ , then $\text{p}\left( \text{-}\frac{\text{m}}{\text{l}} \right)$ should be  $\text{0}$ .

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

Hence, we can say that $\text{x = -}\frac{\text{m}}{\text{l}}$  is a zero of the given polynomial.

vii)$\text{p(x)=3}{{\text{x}}^{\text{2}}}\text{-1,x=-}\frac{\text{1}}{\sqrt{\text{3}}}\text{,}\frac{\text{2}}{\sqrt{\text{3}}}$

Ans: We are given:$\text{x = }\frac{\text{-1}}{\sqrt{\text{3}}}$  and  $\text{x = }\frac{\text{2}}{\sqrt{\text{3}}}$. 

If they are zeroes of the polynomial $\text{p}\left( \text{x} \right)\text{ = 3}{{\text{x}}^{\text{2}}}\text{-1}$, then $\text{p}\left( \frac{\text{-1}}{\sqrt{\text{3}}} \right)$and $\text{p}\left( \frac{\text{2}}{\sqrt{\text{3}}} \right)$   should be $\text{0}$ .

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

$\text{p}\left( \frac{\text{2}}{\sqrt{\text{3}}} \right)\text{ = 3}{{\left( \frac{\text{2}}{\sqrt{\text{3}}} \right)}^{\text{2}}}\text{-1 = 3}\left( \frac{\text{4}}{\text{3}} \right)\text{-1 = 4-1 = 3}$

Hence, we can say that $\text{x = }\frac{\text{-1}}{\sqrt{\text{3}}}$ is a zero of the given polynomial.

However, the value of $\text{x = }\frac{\text{2}}{\sqrt{\text{3}}}$ is not a zero of the given polynomial.

viii) $\text{p(x) = 2x+1, x = }\frac{\text{1}}{\text{2}}$

Ans: We are given: $\text{x = }\frac{\text{1}}{\text{2}}$. 

If it is a zero of polynomial $\text{p}\left( \text{x} \right)\text{ = 2x+1}$ , then $\text{p}\left( \frac{\text{1}}{\text{2}} \right)$ should be $\text{0}$

Here, $\text{p}\left( \frac{\text{1}}{\text{2}} \right)\text{ = 2}\left( \frac{\text{1}}{\text{2}} \right)\text{+1 = 1+1 = 2}$.

So, we get the value $\text{p}\left( \frac{\text{1}}{\text{2}} \right)\ne \text{0}$  .

Hence, we can say that $\text{x = }\frac{\text{1}}{\text{2}}$ is not a zero of the given polynomial.

4. Find the zero of the polynomial in each of the following cases:

i) $\text{p}\left( \text{x} \right)\text{ = x+5}$

Ans:  If $\text{x}$ is zero of the polynomial, then we can say that $\text{p}\left( \text{x} \right)\text{ = 0}$.

Let $\text{p}\left( \text{x} \right)\text{ = 0}$

$\Rightarrow \text{x+5 = 0}$

$\Rightarrow \text{x = -5}$

Therefore, $\text{x = -5}$ is a zero of the given polynomial.

ii) $\text{p}\left( \text{x} \right)\text{ = x-5}$ 

Ans:  If $\text{x}$ is zero of the polynomial, then we can say that $\text{p}\left( \text{x} \right)\text{ = 0}$.

Let $\text{p}\left( \text{x} \right)\text{ = 0}$

$\Rightarrow \text{x-5 = 0}$

$\Rightarrow \text{x = 5}$

Therefore, $\text{x = 5}$ is a zero of the given polynomial.

iii) $\text{p}\left( \text{x} \right)\text{ = 2x+5}$

Ans:  If $\text{x}$ is zero of the polynomial, then we can say that $\text{p}\left( \text{x} \right)\text{ = 0}$.

Let $\text{p}\left( \text{x} \right)\text{ = 0}$

$\Rightarrow \text{2x+5 = 0}$

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

Therefore, $\text{x = -}\frac{\text{5}}{\text{2}}$ is a zero of the given polynomial.

iv) $\text{p}\left( \text{x} \right)\text{ = 3x-2}$

Ans:  If $\text{x}$ is zero of the polynomial, then we can say that $\text{p}\left( \text{x} \right)\text{ = 0}$.

Let $\text{p}\left( \text{x} \right)\text{ = 0}$

$\Rightarrow \text{3x-2 = 0}$

$\Rightarrow \text{x = }\frac{\text{2}}{\text{3}}$

Therefore, $\text{x = }\frac{\text{2}}{\text{3}}$  is a zero of the given polynomial.

v) $\text{p}\left( \text{x} \right)\text{ = 3x}$

Ans:  If $\text{x}$ is zero of the polynomial, then we can say that $\text{p}\left( \text{x} \right)\text{ = 0}$.

Let $\text{p}\left( \text{x} \right)\text{ = 0}$

$\Rightarrow \text{3x = 0}$

$\Rightarrow \text{x = 0}$

Therefore, $\text{x = 0}$ is a zero of the given polynomial.

vi) $\text{p}\left( \text{x} \right)\text{ = ax, a}\ne \text{0}$

Ans:  If $\text{x}$ is zero of the polynomial, then we can say that $\text{p}\left( \text{x} \right)\text{ = 0}$.

Let $\text{p}\left( \text{x} \right)\text{ = 0}$

$\Rightarrow \text{ax = 0}$

It is also given that $\text{a}$ is non-zero. 

$\Rightarrow \text{x = 0}$

Therefore, $\text{x = 0}$ is a zero of the given polynomial.

vii) $\text{p}\left( \text{x} \right)\text{ = cx+d, c}\ne \text{0, c, d}$ are real numbers.

Ans:  If $\text{x}$ is zero of the polynomial, then we can say that $\text{p}\left( \text{x} \right)\text{ = 0}$.

Let $\text{p}\left( \text{x} \right)\text{ = 0}$

$\Rightarrow \text{cx+d = 0}$

It is also given that $\text{c}$ is non-zero.

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

Therefore, $\text{x = -}\frac{\text{d}}{\text{c}}$  is a zero of the given polynomial.

Class 9 Maths Chapter 2 - Exercise 2.2

Vedantu’s Class 9 Maths Chapter 2 Exercise 2.2 includes the following problems from polynomial-

1. Finding the value of polynomial when the value of the variable is randomly fixed by the teacher,For example, if I ask you to find the value of 2x + 3x when I have randomly set the value of X as 5, you will put the value - 5 in place of X and then multiply 2 and 3 with 5.

2. Finding whether a particular value of the variable will result in the polynomial to be equated as  For example, if X equals to 5 in the equation 2X - 10, then the X = 5 is a zero of the given polynomial.

3. Finding which value of the variable will result in the polynomial to be equated as 0. For example, if I want to make the value of 3X+1-7 then I have to assign the value of X as 2.

What are Polynomials? Constants and Variables

Before discussing polynomials, it is necessary that you have a clear idea about variables and exponents. Consider the following equation -

2x + 3x = 10

Here we don't know at first what X is. When we solve the equation, we come to the conclusion that here X is 2. So, what X does here is it answers the ‘what’ of a question. In this equation, the question that is implied is - what should we multiply 2 and 3 with so that the addition of the product results in 10?

Again when we are confronted with the following equation - 

2x + 3x = 20

the result of X changes. So, the value of X varies the situation.

This is what we call variable. This variable is generally denoted by x,y,z etc.

Again, if you look at the above two examples, the values of 2 and 3 never change. In both situations, the values of 2 and 3 remain fixed. These are called constants.

Now, look at this,

 2³

Here the value of 2 increases exponentially ( 2*2*2). The power 3 is known as an exponent.

Terms

If you look at the algebraic expression - 2x + 3x - written above, you will see there are two parts in it - 2x and 3x. These parts are called terms.

Polynomial

Polynomial is a mathematical expression that has multiple terms and consists of constants, variables and non-negative or non-fractionated exponents. The polynomial will only involve addition, subtraction or multiplication.

So 2x + 3x is a polynomial in the variable X.

Again, 2³ + X³ + 3² is a polynomial.

But X + 1∕X is not a polynomial since 1/x can be written as . As said earlier, a polynomial does not allow negative exponents.

Expressions that have √X are not polynomials too since √X can be written as and we know that this is not accepted by polynomials.

If you look at the expression 2x + 3x, the numbers before the variables are called coefficients. 

Class 9 Maths Chapter 2 – Other Exercises (2.1 and 2.3)

Here are a few other exercises that you will find in the NCERT Solutions for Class 9 Maths Chapter 2 PDF – 

Exercise 2.1  

In this exercise of Class 9 Maths Chapter 2, students will be introduced to the concept of polynomials. They will obtain an indent knowledge regarding polynomials in one variable and how to solve the associated sums through shortcut techniques. Here are some important questions for exams that are provided in the exercise.

Question 1: Finding the polynomials among the expressions in one variable and stating reasons for the answers.

Question 2: Finding the coefficients of x2 in the provided expressions.

Question 3: Examples of a binomial of degree 35 and monomial of degree 100.

Question 4: Determining the degree of polynomials.

Question 5: Classifying linear, quadratic and cubic polynomial expressions.

Exercise 2.3

This chapter is a continuation of Exercise 2.2 Class 9 of NCERT maths book. Students will get in-depth knowledge about the Factor Theorem by solving all questions provided in this exercise step by step. 

Question 1: Finding the remainder when divided by expressions provided.

Question 2: Determining remainders. 

Question 3: Checking if provided expressions are factors.

How Can Vedantu Help You?

Vedantu's Polynomials Class 9 solutions will help you understand all the topics included in the Polynomials chapter. NCERT guidelines specifically advise teachers to build the fundamental knowledge of the subjects in the students. Vedantu follows this NCERT advice thoroughly. That is why you will find not only solutions to the questions on polynomials but also the explanations on the logic behind our solutions. For example - in question number 2 of Ex 2.1 of our Polynomials Class 9 NCERT Solutions when we are asked to find the coefficients of given polynomials we showed how has an invisible 1 before it. So the coefficient of X² is 1.

Our answers in Class 9th Maths NCERT Solutions Chapter 2 include every step. No step has been skipped. You will not struggle to understand how we reached the result from a given equation. So in question 1(i) of Ex 2.2, we showed how we arrived at the answer 3 by solving the equation step by step. This will not only help the students to understand the granularities in the chapter but will also help them to get good marks.

Every Polynomial formula Class 9 has been used in our solution. The students will get to learn all the aspects of the chapter. Reading our PDF, a student can answer any question from this chapter with confidence.

The Polynomials Class 9 PDF is written in a lucid language. Apart from the Maths expressions, we have used simple, easy to understand words so that students can understand the solutions. Even the math expressions are adequately explained wherever deemed fit.

The Class 9 Maths Chapter 2 PDF has accurate solutions to every polynomials problem. The answers are written by expert teachers who know how to solve a Maths problem so that students understand the process and even the examiners get impressed by the detailed process. These teachers are well aware of NCERT guidelines and follow them to write their answers. There is no unnecessary information in our solutions.

Vedantu’s Polynomial Class 9 PDF is neatly organised. The solutions are written in an uncluttered, easy to understand way. That is why in question 1 of Ex2.3, you will find that we have used red ink to indicate the cutting off of -2x² and +2x². This visually pleasurable way of reading the solutions will help the students to maintain their focus on the solutions.

You can use our NCERT Solutions for Class 9 Maths Chapter 2 Exercise 2.2 to learn all the intricacies of polynomials. These solutions can provide you with everything that you need to excel in the chapter. The answers in our PDF are not written by part-time bloggers, but they are written by professional teachers who took care to make the solutions as helpful as possible.

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Our online teachers are experts in their subjects and come from respectable institutions like IITs. These teachers know how to teach Maths to students in such a way that students do not get bored. The teachers are aided by simultaneous images, videos and slides. This visual way of teaching ensures easy retention in students.

With our Master Class, you can experience a whole new learning experience. We are offering free seats in our master class for a limited period of time. These masterclasses can enhance your knowledge in ways you cannot fathom. We do not employ cheap tricks and tips. Our main focus is on building the basic knowledge of the students. Our Master Class is not just for advanced students, intermediate students can get benefit from the Master Classes as they will shed fear for Maths and grow a love for the subject.

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NCERT Solutions for Class 9 Maths

• Chapter 1 - Number System

• Chapter 3 - Coordinate Geometry

• Chapter 4 - Linear Equations in Two Variables

• Chapter 5 - Introductions to Euclid's Geometry

• Chapter 6 - Lines and Angles

• Chapter 7 - Triangles

• Chapter 8 - Quadrilaterals

• Chapter 9 - Areas of Parallelogram and Triangles

• Chapter 10 - Circles

• Chapter 11 - Constructions

• Chapter 12 - Heron's formula

• Chapter 13 - Surface area and Volumes

• Chapter 14 - Statistics

• Chapter 15 - Probability