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.

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

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

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

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

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

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

NCERT Solutions for Class 10 Maths Chapter 2 Exercises

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.