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NCERT Solutions for Class 10 Maths Chapter 2 - Polynomials

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Complete Resource for NCERT Class 10 Maths Chapter 2 Polynomials - Free PDF Download

Mathematics is an important subject for Class 10 students. The syllabus is designed to help students gain knowledge and build a strong foundation for advanced classes. Chapter 2 of Class 10 Maths focuses on polynomials. NCERT Solutions for Class 10 Maths Chapter 2, prepared by top mentors of Vedantu, provide special assistance for this chapter. You can download the solutions Polynomial Class 10 PDF file for free and access it offline.

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Table of Content
1. Complete Resource for NCERT Class 10 Maths Chapter 2 Polynomials - Free PDF Download
2. Glance on Chapter 2 Maths Class 10 - Polynomial
3. Exercises under NCERT Solutions for Class 10 Maths Chapter 2 Polynomials
4. Access NCERT Solutions for Class 10 Maths Chapter 2 – Polynomials
5. Also You Can Find the Solutions of All the Class 10 Maths Chapters 2 Below.
6. Math Polynomials Mind Map for Class 10
7. Overview of Deleted Syllabus for CBSE Class 10 Maths Chapter 2 Polynomials
8. NCERT Solutions for Class 10 Maths Chapter 2 Exercises
9. Other Related Links for CBSE Class 10 Maths Chapter 2
10. Chapter-Specific NCERT Solutions for Class 10 Maths
FAQs


The chapter covers different types of equations and their components. By using the NCERT Solutions, you can easily learn new concepts and solve exercise questions. Vedantu offers solutions for all subjects and classes.


Glance on Chapter 2 Maths Class 10 - Polynomial

  • Polynomial are formed using addition, subtraction, and multiplication of terms.

  • Each term can have a coefficient, a variable raised to a whole number power (not fractions or decimals), or a constant.

  • Types of Polynomials (based on Degree):

  • Linear Polynomial (Degree 1): Has the highest exponent of 1 on the variable.

  • Quadratic Polynomial (Degree 2): Highest exponent is 2.

  • Cubic Polynomial (Degree 3): Highest exponent is 3. 

  • There are higher-degree polynomials as well (quartic, quintic, etc.).

  • Finding zeros helps solve equations and understand the polynomial's behavior.

  • Polynomials can be visualized as graphs. The shape of the graph depends on the degree of the polynomial.

  • There is a connection between the zeros of a polynomial and its coefficients.

  • There are two exercises (3 fully solved questions) in class 10th maths chapter 2 Polynomials.


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NCERT Solutions for Class 10 Maths Chapter 2 - Polynomials
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Exercises under NCERT Solutions for Class 10 Maths Chapter 2 Polynomials

NCERT Solutions for Class 10 Maths Chapter 2, "Polynomials," is based on the concept of polynomials and their applications. The chapter consists of the following exercises:


Exercise 2.1: This exercise discusses the concept of polynomials and their terms.

Exercise 2.2: This exercise covers the addition and subtraction of polynomials.


Access NCERT Solutions for Class 10 Maths Chapter 2 – Polynomials

Exercise - 2.1

1. The graphs of \[\text{y=p(x)}\] are given in following figure, for some Polynomials \[\text{p(x)}\]. Find the number of zeroes of \[\text{p(x)}\], in each case.


Straight line parallel to x-axis


Ans: The graph does not intersect the \[\text{x-axis}\] at any point. Therefore, it does not have any zeroes.


Graph intersect x-axis at one point


Ans: The graph intersects at the \[\text{x-axis}\] at only \[\text{1}\]point. Therefore, the number of zeroes is \[\text{1}\].


Graph intersect x-axis at three points


Ans: The graph intersects at the \[\text{x-axis}\] at \[\text{3}\] points. Therefore, the number of zeroes is \[\text{3}\].


Graph intersect x-axis at twice


Ans: The graph intersects at the \[\text{x-axis}\] at \[\text{2}\] points. Therefore, the number of zeroes is \[\text{2}\]. 


Graph intersect x-axis at one point and touch at two points


Ans: The graph intersects at the \[\text{x-axis}\] at \[\text{4}\] points. Therefore, the number of zeroes is \[\text{4}\].


Graph intersect x-axis at four points


Ans: The graph intersects at the \[\text{x-axis}\] at \[\text{3}\] points. Therefore, the number of zeroes is \[\text{3}\]. 

 

 Exercise - 2.2

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

(i) \[{{\text{x}}^{\text{2}}}-\text{2x}-\text{8}\]        

Ans:  Given: \[{{\text{x}}^{\text{2}}}-\text{2x}-\text{8}\].

Now factorize the given polynomial to get the roots.

\[\Rightarrow \text{(x}-\text{4)(x+2)}\]

The value of \[{{\text{x}}^{\text{2}}}-\text{2x}-\text{8}\] is zero. 

when \[\text{x}-\text{4=0}\] or \[\text{x+2=0}\]. i.e., \[\text{x = 4}\] or \[\text{x = }-\text{2}\]

Therefore, the zeroes of \[{{\text{x}}^{\text{2}}}-\text{2x}-\text{8}\] are \[\text{4}\] and \[-2\].

Now, Sum of zeroes\[\text{=4}-\text{2= 2 =}-\dfrac{\text{2}}{\text{1}}\text{=}-\dfrac{\text{Coefficient of x}}{\text{Coefficient of }{{\text{x}}^{\text{2}}}}\] $ \therefore Sum\,of\,zeroes=-\dfrac{Coefficient\,of\,x}{Coefficient\,of\,{{x}^{2}}} $ 

Product of zeroes \[\text{=4 }\times\text{ (}-\text{2)=}-\text{8=}\dfrac{\text{(}-\text{8)}}{\text{1}}\text{=}\dfrac{\text{Constant term}}{\text{Coefficient of }{{\text{x}}^{\text{2}}}}\]

 $ \therefore \text{Product of zeroes}=\dfrac{\text{Constant term}}{\text{Coefficient of }{{\text{x}}^{\text{2}}}} $ .

(ii) \[\mathbf{4{{s}^{2}}-4s+1}\]

Ans: Given: \[\text{4}{{\text{s}}^{\text{2}}}-\text{4s+1}\]

Now factorize the given polynomial to get the roots.

\[\Rightarrow {{\text{(2s}-\text{1)}}^{\text{2}}}\]

The value of \[\text{4}{{\text{s}}^{\text{2}}}-\text{4s+1}\] is zero. 

when \[\text{2s}-\text{1=0}\], \[\text{2s}-\text{1=0}\]. i.e., \[\text{s =}\dfrac{\text{1}}{\text{2}}\] and \[\text{s =}\dfrac{\text{1}}{\text{2}}\]

Therefore, the zeroes of \[\text{4}{{\text{s}}^{\text{2}}}-\text{4s+1}\] are \[\dfrac{\text{1}}{\text{2}}\] and \[\dfrac{\text{1}}{\text{2}}\].

Now, Sum of zeroes\[\text{=}\dfrac{\text{1}}{\text{2}}\text{+}\dfrac{\text{1}}{\text{2}}\text{=1=}\dfrac{\text{(}-\text{4)}}{\text{4}}\text{=}\dfrac{-\text{(Coefficient of s)}}{\text{Coefficient of }{{\text{s}}^{\text{2}}}}\]

\[\therefore \text{Sum of zeroes =}\dfrac{-\text{(Coefficient of s)}}{\text{Coefficient of }{{\text{s}}^{\text{2}}}}\]

Product of zeroes\[=\dfrac{\text{1}}{\text{2}}\text{ }\times\text{ }\dfrac{\text{1}}{\text{2}}=\dfrac{\text{1}}{\text{4}}=\dfrac{\text{Constant term}}{\text{Coefficient of }{{\text{s}}^{\text{2}}}}\]

 $ \therefore Product of zeroes=\dfrac{\text{Constant term}}{\text{Coefficient of }{{\text{s}}^{\text{2}}}} $ .

(iii) \[\mathbf{6{{x}^{2}}-3-7x}\]

Ans: Given: \[\text{6}{{\text{x}}^{\text{2}}}-\text{3}-\text{7x }\]

\[\Rightarrow \text{6}{{\text{x}}^{\text{2}}}-\text{7x}-\text{3}\]

Now factorize the given polynomial to get the roots.

\[\Rightarrow \text{(3x+1)(2x}-\text{3)}\]

The value of  \[\text{6}{{\text{x}}^{\text{2}}}-\text{3}-\text{7x }\] is zero. 

when \[\text{3x+1=0}\] or \[\text{2x}-\text{3=0}\]. i.e., \[\text{x =}\dfrac{-\text{1}}{\text{3}}\] or \[\text{x =}\dfrac{\text{3}}{\text{2}}\].

Therefore, the zeroes of  \[\text{6}{{\text{x}}^{\text{2}}}-\text{3}-\text{7x }\] are \[\dfrac{\text{-1}}{\text{3}}\] and \[\dfrac{\text{3}}{\text{2}}\].

Now, Sum of zeroes\[\text{=}\dfrac{-\text{1}}{\text{3}}\text{+}\dfrac{\text{3}}{\text{2}}\text{=}\dfrac{\text{7}}{\text{6}}\text{=}\dfrac{-\text{(}-\text{7)}}{\text{6}}\text{=}\dfrac{-\text{(Coefficient of x)}}{\text{Coefficient of }{{\text{x}}^{\text{2}}}}\]

 $ \therefore Sum of zeroes\text{=}\dfrac{-\text{(Coefficient of x)}}{\text{Coefficient of }{{\text{x}}^{\text{2}}}} $ 

Product of zeroes\[\text{=}\dfrac{-\text{1}}{\text{3}}\text{ }\times\text{ }\dfrac{\text{3}}{\text{2}}\text{=}\dfrac{-\text{3}}{\text{6}}\text{=}\dfrac{\text{Constant term}}{\text{Coefficient of }{{\text{x}}^{\text{2}}}}\]

 $ \therefore \text{Product of zeroes=}\dfrac{\text{Constant term}}{\text{Coefficient of }{{\text{x}}^{\text{2}}}} $ 

(iv) \[\mathbf{4{{u}^{2}}+8u}\]                          

Ans: Given: \[\text{4}{{\text{u}}^{\text{2}}}\text{+8u}\]

\[\Rightarrow \text{4}{{\text{u}}^{\text{2}}}\text{+8u+0}\] 

\[\Rightarrow \text{4u(u+2)}\]

The value of \[\text{4}{{\text{u}}^{\text{2}}}\text{+8u}\] is zero. 

when \[\text{4u=0}\] or \[\text{u+2=0}\]. i.e., \[\text{u = 0}\] or \[\text{u =}-\text{2}\]

Therefore, the zeroes of \[\text{4}{{\text{u}}^{\text{2}}}\text{+8u}\] are \[\text{0}\] and \[\text{-2}\].

Now, Sum of zeroes\[\text{=0+(}-\text{2)=}-\text{2=}\dfrac{-\text{8}}{\text{4}}\text{=}\dfrac{-\text{(Coefficient of u)}}{\text{Coefficient of }{{\text{u}}^{\text{2}}}}\]

 $ \therefore \text{Sum of zeroes=}\dfrac{-\text{(Coefficient of u)}}{\text{Coefficient of }{{\text{u}}^{\text{2}}}} $ 

Product of zeroes\[\text{=0 }\times\text{ (}-\text{2)= 0 =}\dfrac{\text{0}}{\text{4}}\text{=}\dfrac{\text{Constant term}}{\text{Coefficient of }{{\text{u}}^{\text{2}}}}\]

 $ \therefore \text{Product of zeroes=}\dfrac{\text{Constant term}}{\text{Coefficient of }{{\text{u}}^{\text{2}}}} $ 

(v) \[\mathbf{{{t}^{2}}-15}\]

Ans: Given: \[{{\text{t}}^{\text{2}}}-\text{15}\]

\[\Rightarrow {{\text{t}}^{\text{2}}}-\text{0t}-\text{15}\] 

Now factorize the given polynomial to get the roots.

\[\Rightarrow \text{(t}-\sqrt{\text{15}}\text{)(t +}\sqrt{\text{15}}\text{)}\] 

The value of \[{{\text{t}}^{\text{2}}}-\text{15}\] is zero. 

when \[\text{t}-\sqrt{\text{15}}\text{=0}\] or \[\text{t+}\sqrt{\text{15}}\text{=0}\], i.e., \[\text{t=}\sqrt{\text{15}}\] or \[\text{t=}-\sqrt{\text{15}}\]

Therefore, the zeroes of \[{{\text{t}}^{\text{2}}}-\text{15}\] are \[\sqrt{\text{15}}\] and \[-\sqrt{\text{15}}\].

Now, Sum of zeroes\[\text{=}\sqrt{\text{15}}\text{+(}-\sqrt{\text{15}}\text{)=0=}\dfrac{-\text{0}}{\text{1}}\text{=}\dfrac{-\text{(Coefficient of t)}}{\text{Coefficient of }{{\text{t}}^{\text{2}}}}\]

 $ \therefore \text{Sum of zeroes =}\dfrac{-\text{(Coefficient of t)}}{\text{Coefficient of }{{\text{t}}^{\text{2}}}} $ 

Product of zeroes\[\text{=}\left( \sqrt{\text{15}} \right)\times \left( -\sqrt{\text{15}} \right)\text{=}-\text{15=}\dfrac{-\text{15}}{\text{1}}\text{=}\dfrac{\text{Constant term}}{\text{Coefficient of }{{\text{t}}^{\text{2}}}}\]

 $ \therefore \text{Product of zeroes =}\dfrac{\text{Constant term}}{\text{Coefficient of }{{\text{t}}^{\text{2}}}} $ .

(vi) \[\mathbf{3{{x}^{2}}-x-4}\]

Ans:  Given: \[\text{3}{{\text{x}}^{\text{2}}}-\text{x}-\text{4}\]

Now factorize the given polynomial to get the roots.

 $ \Rightarrow \left( 3x-4 \right)(x+1) $ 

The value of \[\text{3}{{\text{x}}^{\text{2}}}-\text{x}-\text{4}\] is zero.

when \[\text{3x}-\text{4=0}\] or \[\text{x+1=0}\], i.e., \[\text{x=}\dfrac{\text{4}}{\text{3}}\] or \[\text{x=}-\text{1}\]

Therefore, the zeroes of \[\text{3}{{\text{x}}^{\text{2}}}-\text{x}-\text{4}\] are \[\dfrac{\text{4}}{\text{3}}\] and \[\text{-1}\].

Now, Sum of zeroes\[\text{=}\dfrac{\text{4}}{\text{3}}\text{+(}-\text{1)=}\dfrac{\text{1}}{\text{3}}\text{=}\dfrac{-\text{(}-\text{1)}}{\text{3}}\text{=}\dfrac{-\text{(Coefficient of x)}}{\text{Coefficient of }{{\text{x}}^{\text{2}}}}\]

 $ \therefore \text{Sum of zeroes=}\dfrac{-\text{(Coefficient of x)}}{\text{Coefficient of }{{\text{x}}^{\text{2}}}} $ 

Product of zeroes\[\text{=}\dfrac{\text{4}}{\text{3}}\times \text{(}-\text{1)=}\dfrac{-\text{4}}{\text{3}}\text{=}\dfrac{\text{Constant term}}{\text{Coefficient of }{{\text{x}}^{\text{2}}}}\]

 $ \therefore \text{Product of zeroes =}\dfrac{\text{Constant term}}{\text{Coefficient of }{{\text{x}}^{\text{2}}}} $ .


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

(i)  \[\mathbf{\dfrac{1}{4},-1}\]  

Ans: Given: \[\dfrac{\text{1}}{\text{4}}\text{,-1}\]

Let the zeroes of polynomial be \[\text{ }\alpha\text{ }\] and \[\text{ }\beta\text{ }\].

Then, 

\[\text{ }\alpha\text{ + }\beta\text{  =}\dfrac{\text{1}}{\text{4}}\]

\[\text{ }\alpha\text{  }\beta\text{ =}-\text{1}\] 

Hence, the required polynomial is \[{{\text{x}}^{\text{2}}}-\left( \alpha +\beta  \right)\text{x+}\alpha \beta \]. 

 $ \Rightarrow {{x}^{2}}-\dfrac{1}{4}x-1 $ 

 $ \Rightarrow 4{{x}^{2}}-x-4 $ 

Therefore, the quadratic polynomial is \[\text{4}{{\text{x}}^{\text{2}}}-\text{x}-\text{4}\].

(ii) \[\mathbf{\sqrt{2},\dfrac{1}{3}}\]      

Ans: Given: \[\sqrt{\text{2}}\text{,}\dfrac{\text{1}}{\text{3}}\]

Let the zeroes of polynomial be \[\text{ }\alpha \text{ }\] and \[\text{ }\beta\text{ }\].

Then, \[\text{ }\alpha\text{ + }\beta\text{  =}\sqrt{\text{2}}\]

\[\text{ }\alpha\text{  }\beta\text{ =}\dfrac{\text{1}}{\text{3}}\] 

Hence, the required polynomial is \[{{\text{x}}^{\text{2}}}-\left( \alpha +\beta  \right)\text{x+}\alpha \beta \]. 

\[\Rightarrow {{\text{x}}^{\text{2}}}-\sqrt{2}\text{x+}\dfrac{1}{3}\]

\[\Rightarrow 3{{\text{x}}^{\text{2}}}-3\sqrt{2}\text{x+1}\]

Therefore, the quadratic polynomial is \[3{{\text{x}}^{\text{2}}}-3\sqrt{2}\text{x+1}\].

(iii) \[\mathbf{0,\sqrt{5}}\]  

(here, root is missing)

Ans: Given: \[\text{0,}\sqrt{\text{5}}\]

Let the zeroes of polynomial be \[\text{ } \alpha \text{ }\] and \[\text{ } \beta \text{ }\].

Then,

\[\text{ } \alpha \text{ + } \beta \text{  = 0}\]

\[\text{ } \alpha \text{  } \beta \text{ =}\sqrt{\text{5}}\] 

Hence, the required polynomial is \[{{\text{x}}^{\text{2}}}-\left( \alpha +\beta  \right)\text{x+}\alpha \beta \]. 

\[\Rightarrow {{\text{x}}^{\text{2}}}-0\text{x+}\sqrt{5}\]

\[\Rightarrow {{\text{x}}^{\text{2}}}\text{+}\sqrt{5}\]

Therefore, the quadratic polynomial is \[{{\text{x}}^{\text{2}}}\text{+}\sqrt{\text{5}}\].

(iv) \[\mathbf{1,1}\]     

Ans: Given: \[\text{1,1}\]

Let the zeroes of polynomial be \[\text{ } \alpha \text{ }\] and \[\text{ } \beta \text{ }\].

Then,

\[\text{ } \alpha \text{ + } \beta \text{  =1}\]

\[\text{ } \alpha \text{  } \beta \text{  =1}\] 

Hence, the required polynomial is \[{{\text{x}}^{\text{2}}}-\left( \alpha +\beta  \right)\text{x+}\alpha \beta \]. 

\[\Rightarrow {{\text{x}}^{\text{2}}}-\text{1x+1}\]

Therefore, the quadratic polynomial is \[{{\text{x}}^{\text{2}}}-\text{x+1}\].

(v) \[\mathbf{-\dfrac{1}{4},\dfrac{1}{4}}\]      

Ans: Given: \[-\dfrac{\text{1}}{\text{4}}\text{,}\dfrac{\text{1}}{\text{4}}\]

Let the zeroes of polynomial be \[\text{ } \alpha \text{ }\] and \[\text{ } \beta \text{ }\].

Then,

\[\text{ } \alpha \text{ + } \beta\text{ =}-\dfrac{\text{1}}{\text{4}}\]

\[\text{ } \alpha \text{  } \beta \text{  =}\dfrac{\text{1}}{\text{4}}\] 

Hence, the required polynomial is \[{{\text{x}}^{\text{2}}}-\left( \alpha +\beta  \right)\text{x+}\alpha \beta \]. 

 $ \Rightarrow {{\text{x}}^{\text{2}}}-\left( -\dfrac{1}{4} \right)\text{x+}\dfrac{1}{4} $ 

 $ \Rightarrow \text{4}{{\text{x}}^{\text{2}}}\text{+ x +1} $ 

Therefore, the quadratic polynomial is  $ \text{4}{{\text{x}}^{\text{2}}}\text{+ x +1} $ .

(vi) \[\mathbf{4,1}\]

Ans: Given: \[\text{4,1}\]

Let the zeroes of polynomial be \[\text{ } \alpha \text{ }\] and \[\text{ }\beta \text{ }\].

Then,

\[\text{ } \alpha \text{ + } \beta \text{  = 4}\]

\[\text{ } \alpha \text{  } \beta \text{  =1}\] 

Hence, the required polynomial is \[{{\text{x}}^{\text{2}}}-\left( \alpha +\beta  \right)\text{x+}\alpha \beta \]. 

 $ \Rightarrow {{\text{x}}^{\text{2}}}-\text{4x+1} $ 

Therefore, the quadratic polynomial is  $ {{\text{x}}^{\text{2}}}-\text{4x+1} $ .


NCERT Solutions for Class 10 Maths Chapter 2 PDF for Upcoming Exams and Also You Can Find the Solutions of All the Maths Chapters Below.

2.1 Introduction

Given the importance of this subject, it falls under the unit of algebra, which is worth 20 marks on the Class 10 CBSE maths exams. One question is often asked about this chapter on average. This chapter covers the following topics:


  • Overview of Polynomials

  • Geometric Interpretation of Polynomial Zeros 

  • The Connection between Zeros and Coefficients in a Polynomial 

  • Division Algorithm


Section 2.1 of the chapter begins with an introduction to polynomials, and parts 2.2 and 2.3 cover two crucial subjects.


  • Geometric Interpretation of a Polynomial's Zeroes - There is one question with six possible answers.

  • Relationship between a Polynomial's Zeroes and Coefficients Examine the relationship between a quadratic polynomial's zeroes and coefficients by working through the answers to the two problems (each consisting of six pieces) in Exercise 2.2.

  • Division Algorithm for Polynomials: This provides answers to the five issues (three lengthy questions) in Exercise 2.3.


2.2 Important Topics Under NCERT Solutions for Class 10 Maths Chapter 2 Polynomials

Chapter 2 Polynomials is an important chapter in the mathematics syllabus for Class 10. This chapter on Polynomials includes a number of topics, and in order to internalize the ideas properly, students are advised to go through the important topics under Polynomials, thoroughly. We have provided the following list of important topics covered in this chapter for a better understanding of the concept of Polynomials.


  • Definition

  • Degree of a Polynomial

  • Types of Polynomial

  1. Constant Polynomial

  2. Linear Polynomial

  3. Quadratic Polynomial

  4. Cubic Polynomial

  • Value of a Polynomial

  • Zero of a Polynomial

  • Graph of a Polynomial

  • Meaning of the Zeroes of a Quadratic Polynomial


2.3 Importance of Polynomials

Polynomials are expressions that have more than two algebraic terms. They can also be defined as the sum of several terms where the same variable/variables has/have different powers.


The topics in ch 2 class 10 maths are important because they have applications in most mathematical expressions. They are used to represent appropriate relations between different variables or numbers. We encourage students to learn from this chapter to be able to solve tricky problems easily in exams.


2.4 Polynomials: NCERT Solutions for Class 10 Maths Chapter 2 Summary

Mathematics is an important subject for Class 10 students. The syllabus is designed to help students gain knowledge and build a strong foundation for advanced classes. Chapter 2 of Class 10 Maths focuses on polynomials. NCERT Solutions for Chapter 2 Maths Class 10, prepared by top mentors of Vedantu, provide special assistance for this chapter. You can download the solutions PDF file for free and access it offline. The chapter covers different types of equations and their components. By using the NCERT Solutions, you can easily learn new concepts and solve exercise questions. Vedantu offers solutions for all subjects and classes, including NCERT Solutions for Class 10 Science.


2.5 Benefits of Using NCERT Solutions for Class 10 Chapter 2 Maths - Polynomials

The Polynomials Class 10 Solutions have been designed to deliver the following benefits to the students.


  • Answers to Exercise Questions

All the answers provided in the NCERT Solutions of Class 10 Maths Chapter 2 Solutions are framed by the expert teachers of Vedantu. You can rely on the quality of the answers in Class 10 Maths Ch 2 Solutions. Maths NCERT Solutions of Chapter 2 Maths Class 10  is formulated following the CBSE guidelines.


  • Understanding the Concepts of Chapter 2 Polynomials Class 10

As mentioned earlier, new concepts will be taught in ch 2 maths class 10 (Polynomials). These concepts will then have to be used to solve the problems in the exercises. You can easily grasp the concepts by using the NCERT Solutions Class 10 Maths Ch 2. The simplification and utilization of the concepts in the answers will help you solve the problems in the future.


  • Developing a Strategy

There is no easiest way to develop a strategy other than using ch 2 maths class 10 NCERT Solutions to practice solving the exercise problems. Learn efficient approaches and develop a strong strategy to answer questions and save time during an exam.


  • Doubt Clarification

The doubts arising during studying Maths Class 10th Chapter 2 can be resolved using the simplest NCERT Class 10 Maths Chapter 2 Solutions prepared by the teachers. You can now resolve your doubts on your own and complete preparing the chapter efficiently.


  • Quick Revision

Revise Class 10th Maths Chapter 2 before an exam by referring to the Class 10th Maths Chapter 2 Solutions and save time. Use your time to complete other chapters and their NCERT solutions too


Math Polynomials Mind Map for Class 10

2.6.1 Polynomial

A polynomial in variable x is an algebraic statement of the form f(x) = a0 + a1x + a2x2 +…. + anxn, where a0, a1,…., and an are real numbers and all variable indices are non-negative integers.


(i) The polynomial's degree is its largest power of x.

(ii) The polynomial's terms are a0, a1x,…, and anxn.

iii) The polynomial's coefficients are a0, a1,... an.


2.6.2 Common Formats for Cubic, Quadratic, and Linear Polynomials

(i) Linear polynomial: axe + b, where a ≠ 0 and a, b are real values.

(ii) A quadratic polynomial, where a, b, and c are real values and a ≠ 0.

(iii) Cubic polynomials, where a, b, c, and d are real integers and a ≠ 0, are represented as: x3 + bx2 + cx + d.


2.6.3 What a Polynomial is Worth

By replacing x = an in the provided polynomial, the value of a polynomial f(x) at x = an is found and is represented by/(a).


If p(r) = 0, then a zero of a polynomial p(x) equals Zero(es)/Root(s) of Polynomial x = r.


2.6.4 Geometric Interpretation of a Polynomial's Zeroes

The x-coordinate of the point or points where the graph y = fix) intersects the x-axis is the zero(es) of a polynomial.


(i) Polynomial that is Linear: A linear polynomial graph has exactly one zero and is a straight line.

(ii) Quadratic Polynomial: This type of polynomial can have a maximum of two zeros and its graph is invariably a parabola.

(iii) A cubic polynomial can have a maximum of three zeros.


2.6.5 Examples of Polynomial Cases

  • Case-I: The graph of a quadratic polynomial P(x) = ax2 + bx + c will intersect the x-axis at two different positions, A and B, as indicated in the image, if the polynomial has two zeros.

  • Case-II: The graph of a quadratic polynomial P(x) = ax2 + bx + c will touch the x-axis at only one point A, as indicated in the picture, if it contains only one zero.

  • Case-III: A quadratic polynomial P(x) = ax2 + bx + c will not intersect or touch the x-axis at any point, as illustrated in the image, if it lacks a zero.


2.6.6 Relationship between a Polynomial's Zeroes and Coefficients 

(i) A linear polynomial axe + b has zero at x = −𝑏𝑎 .

(ii) The quadratic polynomial ax2 + bx + c has zeroes at α and β. Therefore, α + β = −𝑏𝑎 .

(iii) The cubic polynomial ax3 + bx2 + cx + d has zeroes at α, β, and γ.


2.6.7 Algorithm for Division

We can find polynomials q(x) and r(x) such that p(x) = g(x) × q(x) + r(x) where either r(x) = 0 or degree of r(x) < degree of g(x) if p(x) and g(x) are any two polynomials with g(x) ≠ 0.


Overview of Deleted Syllabus for CBSE Class 10 Maths Chapter 2 Polynomials

Chapter

Dropped Topics

Polynomials

Page Number 33 - 37

2.4 Division algorithm for polynomials


NCERT Solutions for Class 10 Maths Chapter 2 Exercises

Exercise

Number of Questions

Exercise 2.1

1 Questions & Solutions

Exercise 2.2

2 Questions & Solutions


Conclusion

To sum up, Chapter 2 of Class 10 Maths on Polynomials establishes the framework for comprehending algebraic expressions and equations. Understanding fundamental ideas such as degrees, polynomial types, polynomial operations, and factorization is essential. Higher level maths courses and a variety of real-life situations can be solved with an understanding of these foundational concepts. Practice a variety of question types, particularly those pertaining to remainder theorem and factorization. Exams from prior years usually have 3-5 questions from this chapter, with an emphasis on conceptual knowledge and using polynomial operations. To master this chapter, consistent practice and a firm comprehension of the topics are essential.


Other Related Links for CBSE Class 10 Maths Chapter 2


Chapter-Specific NCERT Solutions for Class 10 Maths

Given below are the chapter-wise NCERT Solutions for Class 10 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.


FAQs on NCERT Solutions for Class 10 Maths Chapter 2 - Polynomials

1. How can You Practice Polynomials Questions for Class 10?

To practice Polynomials questions for Class 10, solve a variety of exercises from textbooks or practice guides. Understand the concepts, formulas, and methods of polynomial manipulation. Attempt sample papers or previous year's question papers to familiarize yourself with different types of polynomial problems and enhance problem-solving skills.

2. Why use Class 10 Maths Chapter 2 NCERT Solutions?

Using Class 10 Maths Chapter 2 NCERT solutions is beneficial as they provide accurate and comprehensive answers to all the questions in the textbook. These solutions are prepared by subject-matter experts and are aligned with the latest CBSE curriculum. By using NCERT solutions, students can clarify their doubts, understand the step-by-step solution approach, and improve their problem-solving skills. It helps in building a strong foundation in mathematics and ensures thorough preparation for exams.

3. What are the real-life applications of Class 10th Mathematics Chapter 2?

Class 10 Mathematics Chapter 2, "Polynomials," has several real-life applications. Some include:

  • Finance: Polynomials are used in financial calculations, such as compound interest, loan amortization and investment analysis.

  • Engineering: Polynomial equations are utilized in various engineering fields, including electrical circuit analysis, control systems, and structural design.

  • Data Analysis: Polynomial regression is used to fit curves to data points, helping in trend analysis and predicting future values.

  • Physics: Polynomial equations are applied in physics for modeling motion, calculating forces, and understanding phenomena like projectile motion.

  • Computer Graphics: Polynomials play a crucial role in computer graphics, enabling the creation of smooth curves and surfaces in animations and 3D rendering.

4. Which questions and examples are important in Class 10 Maths Chapter 2?

Students can find a list of the most important questions and examples from Chapter 2 of Class 10 Maths on Vedantu (vedantu.com). Subject experts at Vedantu prepare the important questions, revision notes, and other study materials for the students. Hence, to find out the most important questions that may be asked from the Polynomials chapter in Class 10 Maths, students can refer to the important questions and revision notes on Vedantu (vedantu.com).

5. What is a polynomial in Class 10 NCERT?

Polynomial is an algebraic expression that has variables and coefficients in it. The meaning of a polynomial is “many terms”. An example of a polynomial is $4x^2 + 3x + 7$. They are usually a sum or difference of exponents and variables. For an expression to be a polynomial, it should not contain the square root, negative powers, or fractional powers on the variables. It should also not have variables in the denominator of any fractions.

6. How do you identify a polynomial?

An elementary mathematical expression is called a polynomial. Here's how to identify one:

  • Important characteristics: Exponents and Variables: Polynomials employ variables (such as x, y) with exponents that are whole numbers (0, 1, 2, 3...).

  • Operations: Addition, subtraction, and multiplication are used to make them. Square roots, division (apart from by a constant), and other intricate operations on the variable are not permitted.

  • Consider it this way: Consider a system of building blocks where each block stands for a word (a variable, a number, or a variable times a number).

  • Illustrations of polynomials:

3x^2 + 2x - 1 where x^2 is the exponent of 2.

5y - 2 (y is equivalent to only y when raised to the power of 1).

-7 (This term is regarded as a polynomial in addition to being constant.)

7. What is zero in a polynomial?

The particular values of a variable that cause a polynomial to equal zero are known as its zeros. They are also referred to as the polynomial's "roots" on occasion.


Take the polynomial x^2 - 4x + 4, for example. Among other solutions, x = 2 can be found by solving the equation x^2 - 4x + 4 = 0. This indicates that you get 0 when you substitute 2 for x in the polynomial: (2)^2 - 4(2) + 4 = 0. Thus, the zero of this polynomial is 2.

8. How many types of polynomials are there?

Polynomials can be categorised in two primary ways: according to their degree (highest exponent) and phrase count.


Zero Polynomial: This is only the constant term zero (0x + 0), and it has a degree of zero.

Linear Polynomial: In thisthe highest exponent is 1. For instance, 3x plus 5.

Quadratic Polynomial:The highest exponent is 2. Let's say x^2 + 2x - 1.

Cubic Polynomial: Variable cubed has the maximum exponent of 3. Take x^3 + 4x^2 - 2x + 1 as an example.

9. What are the facts about polynomials Class 10?

The following are some important polynomial facts that you will learn in maths class 10:

  • The fundamentals: Expressions consisting of variables and constants joined by addition, subtraction, and multiplication are called polynomials.
  • Degrees and Types: A polynomial's degree is its highest exponent for the variable it contains in the expression.
  • Operations: Polynomials can be multiplied, subtracted, and added using the standard algebraic formulas.
  • Importance: The basis of algebra is polynomials. Comprehending them is essential while attempting to solve equations, inequalities, and other algebraic expressions.

10. Is zero a polynomial or not?

It is true that zero, or the number 0 is a polynomial. In reality, there are two interpretations of "zero" in relation to polynomials:

  • The Constant Zero Polynomial: The zero sign itself can be expressed as a polynomial. When a polynomial's coefficients are all 0, something occurs. As an illustration, consider the polynomial expression 0x + 0, which always has a value of zero when x is entered. The degree of this kind of polynomial is zero.

  • A polynomial's zeros: are the particular numbers that bring the polynomial to zero. Take the polynomial x^2 - 4x + 4, for example. In this polynomial, the number 2 is a zero since, when you substitute 2 for x, you obtain 0: (2)^2 - 4(2) + 4 = 0. Thus, zero in this instance refers to a particular value that causes the polynomial to evaluate to zero rather than the polynomial as a whole.

11. What is the root of a polynomial?

The value that reduces a polynomial to zero is known as its root, or simply a zero. For instance, the roots of the polynomial x^2 + 2x - 3 are x = 1 and x = -3. Multiple roots are possible for polynomials, and not all of them are rational numbers. The polynomial's complexity determines the root-finding strategy.

12. Is one a polynomial?

One digit on its own is not a polynomial. This is so because polynomials contain both constants, or numbers, and variables, such as x, y, and z. They show a broad link between the value of the expression and the variable. "One" does not meet the definition of a polynomial because it is only a single number without a variable.


Additionally, "one" can be thought of as a shorthand for 1^1 (one to the power of one), because polynomials usually involve variables raised to whole number exponents bigger than one (0, 1, 2, 3, etc.). A polynomial's constant term is merely one term in the polynomial formulation and can be any integer, even 1.

13. What are 2 characteristics of polynomials?

Two essential properties of polynomials are as follows:

  • Variable Exponents: Whole number exponents (0, 1, 2, 3, and so forth) are associated with variables (such as x, y, and z) in polynomials.

  • Basic Operations: When adding, subtracting, and multiplying polynomials, whole number exponents and variables are used. Polynomials cannot be divided by complex functions or variables.

14. Where can I find the precise answer to Chapter 2 of the NCERT Solutions for Maths in Class 10?

You may obtain the precise NCERT Solutions for Class 10 Maths Chapter 2 answer in PDF format at Vedantu. An authentic design of the NCERT Textbook Solutions for the chapter Polynomials has been created by Vedantu's mathematics experts. All of these solutions are offered while taking into account the updated CBSE pattern, giving pupils comprehensive test preparation.

15. Does each issue in the NCERT Solutions for Chapter 2 Maths Class 10 have to be solved?

Indeed. Thus these questions are significant in terms of exams. Experts have answered these questions to make it easier for students to complete each assignment. These answers aid in pupils' familiarisation with polynomials. On the Vedantu's website, solutions are provided in PDF format.