**Class 10 maths NCERT solution of the chapter 2 polynomial **

**10 class maths NCERT solutions of the chapter 2 Polynomial **are the** solutions of **unsolved questions of** Chapter 2 Polynomial of the NCERT maths** textbook of **class 10 CBSE** .** Chapter 2 Polynomial** contains the questions based on the **polynomial**. In the previous class, you would have studied the factors of the **polynomial**. If you would have understood the topics factorization of the **polynomial** of the previous** class** then it will be very easy for you to understand the problems in the **10 class maths NCERT chapter 2 polynomial. **If you are needed to clear your concept on polynomial then please go through the post-operation on** polynomial. **

The **NCERT solutions** of **c****hapter 2 polynomial** are solved by the expert teacher of **maths**. All questions of** chapter 2 polynomial** are **solved** as per the CBSE norms. Each answer of the unsolved questions of **chapter 2 polynomia**l are explained in such a way that every student can understand it. The purpose of the **NCERT solutions of chapter 2 polynomial** is to help you in your preparation of the unit test, half-yearly, preboard, and CBSE final exam of class 10 board.

Click for online shopping

**Future Study Point.Deal: Cloths, Laptops, Computers, Mobiles, Shoes etc**

At the end of this post please don’t forget to write a comment and if you want to update information regarding our future posts then you can subscribe to our website at free of cost. You can study in **Future Study Point**, **NCERT solutions of maths** and science, **maths** and science sample papers, **maths** and science assignments, guess papers, competitive examinations guide, the e-book of science and **maths** and the posts on online jobs and your carrier.

**NCERT Solutions Class 10 Science from chapter 1 to 16**

**Follow us on Pinterest**

**Click here to follow us on Pinterest**

**NCERT Solutions of class 10 maths**

**CBSE Class 10-Question paper of maths 2021 with solutions**

**CBSE Class 10-Half yearly question paper of maths 2020 with solutions**

**CBSE Class 10 -Question paper of maths 2020 with solutions**

**CBSE Class 10-Question paper of maths 2019 with solutions**

**See the video -10 class maths NCERT solutions of the chapter 2 :Polynomial**

**10 class maths NCERT solutions of chapter 2 :Polynomial**

**Exercise 2.1**

**Q1. The graphs of y = p(x) are given in the following figure for some polynomial p(x). Find the number of zeroes of p(x) in each case.**

Ans.(i) The number of zeroes is 0 because the graph of the polynomial p(x) does not intersect the x-axis.

(ii) The number of zeroes is 1 because the graph of the polynomial p(x) intersects at the x-axis at one point.

(iii) The number of zeroes are 3 because the graph of p(x) intersects the x-axis at 3 points.

(iv) The number of zeroes are 2 because the graph of p(x) intersects the x-axis at two points.

(v) The number of zeroes are 4 because the graph of p(x) intersects the x-axis at four points.

(vi) The number of zeroes are 24because the graph of p(x) intersects the x-axis at 4 points.

**See the video for Class 10 exercise 2.1 NCERT Maths Solutions Chapter 2 Polynomial**

**Exercise 2.2**

**Q1. Find the zeroes of the following quadratic polynomials and verify the relationship between zeroes and the coefficient.**

**(i) x² – 2x – 8 (ii) 4s² – 4s + 1 (iii) 6x² – 3 – 7x (iv) 4u² + 8u**

**(v) t² – 15 (vi) 3x² – x – 4**

Solutions.

(i) Factorizing the given polynomial

x² – 2x – 8

x² – 4x + 2x – 8

x( x – 4) + 2( x – 4)

( x – 4)(x +2)

Calculating the zeroes as following

x – 4 = 0, x +2 = 0

x = 4 and x = – 2

a = 1, b= – 2. c = – 8

The relationship between coefficient and zeroes is

2 = 2, – 8 = – 8

In both relationship LHS = RHS, Varified

(ii) The given polynomial is

4s² – 4s + 1

Factorizing it

4s² – 2s – 2s + 1

2s(2s – 1) – 1( 2s – 1)

(2s – 1)(2s – 1)

Calculating the zeroes as following

2s – 1 = 0, 2s – 1 = 0

a = 4, b = – 4, c = 1

The relationship between the coefficient and the zeroes are following

In both cases RHS = LHS, Varified

(iii) The given polynomial is as follows

6x² – 3 – 7x

Arranging it in the standard form ax² + bx + c

6x² – 7x – 3

Factorizing it

6x² – 9x + 2x – 3

3x ( 2x – 3) + 1(2x – 3)

( 2x – 3)(3x +1)

a = 6, b= – 7, c = – 3

The relationship between the coefficient and zeroes of the polynomial is as follows

In both relationship, RHS = LHS, Varified

(iv) We are given the polynomial

4u² + 8u

Factorizing it

4u( u + 2)

Determining its zeroes as following

4u = 0, u +2 = 0

u = 0, u = – 2

a = 4, b= 8, c =0

The relationship between coefficient and zeroes is as follows

In both cases LHS = RHS, Varified

(iv) The given polynomial is

t² – 15

Factorizing the polynomial

t² – (√15)²

(t + √15 )( t – √15)

t + √15 = 0, t – √15 = 0

t = – √15, t = √15

a = 1, b= 0, c = – 15

The relationship between coefficient and zeroes is as follows

0 = 0, –15 = – 15

in both cases ,RHS = LHS, verified

(vi) The given polynomial is as follows

3x² – x – 4

Factorizing the polynomial

3x² –4x + 3x – 4

x ( 3x –4 ) + 1(3x – 4)

( 3x –4 )(x + 1)

Determining the factor as follows

3x –4 = 0, x + 1 = 0

a = 3, b= – 1, c = – 4

The relationship between coefficient and zeroes is as follows

In both cases, it is verified RHS = LHS

**See the video for Class 10 exercise 2.2 NCERT Maths Solutions Chapter 2 Polynomial**

**Q2. Find a quadratic polynomial each with the given numbers as the sum and product of zeroes respectively.**

Ans. The sum of zeroes and their product is given to us

If zeroes are α and β, then the quadratic polynomial can be represented as follows

So, we have

Eliminating the denominator by multiplying with 4

4x² – x – 4

So, the required polynomial is 4x² – x – 4

(ii) The sum and product of the zeroes are given to us, so.

If zeroes are α and β, then the quadratic polynomial can be represented as follows

So, we have

Eliminating the denominator by multiplying with 3

3x² –√2x + 1

(iii) We are given the sum and product of zeroes of the polynomial as follows

α + β = 0, αβ = 1

³ – (α + β)x + αβ

x² – 0×x + 1

x² + 1

So, the required polynomial is x² + 1

(iv) The sum of the polynomial is = 1 and the product of zeroes is = 1

α + β = 1, αβ = 1

x² – (α + β)x + αβ

x² – x + 1

Therefore the required polynomial is x² – x + 1

(vi) We are given

4x² + x +1

So, the required polynomial is 4x² + x +1

(vi) We are given, The sum of the zeroes is = 4 and the product of zeroes is = 1

α + β = 4, αβ = 1

x² – (α + β)x + αβ

x² – 4x + 1

Therefore the required polynomial is x² – 4x + 1

**10 class maths NCERT solutions of chapter 2 :Polynomial**

**Exercise 2.3**

**Q1.Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in each of the following:**

(i) p(x)=x³ – 3x² + 5x – 3 , g(x)= x² – 2

(ii)p(x)= x^{4} – 3x^{2} + 4x + 5, g(x) = x^{2} + 1 – x

(iii)p(x)=x^{4}– 5x + 6, g(x) = 2 – x^{2}

Ans. (i) We are given p(x)=x³ – 3x² + 5x – 3 and g(x) = x² – 2

Dividing p(x) by g(x)

Quotient = x – 3 and remainder = 7x – 9

(ii) Dividing x^{4} – 3x^{2} + 4x + 5 by x^{2} + 1 – x

The quotient is x² + x – 3 and the remainder is 8

(iii) We are given p(x)= x^{4} – 3x^{2} + 4x + 5, g(x) = x^{2} + 1 – x

Dividing p(x) by g(x)

The quotient is –x²– 2 and the remainder is – 5x + 10

**Q2.Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial.**

**(i) t ^{2 }– 3 , 2t^{4 }+ 3t^{3 }− 2t^{2}− 9t −12**

**(ii) x ^{2} + 3x + 1, 3x^{4} + 5x^{3} – 7x^{2} + 2x + 2**

**(iii) x ^{2 }+ 3x + 1 , x^{5}− 4x^{3}+ x^{2}+ 3x + 1**

Ans.(i) Dividing 2t^{4 }+ 3t^{3 }− 2t^{2}− 9t −12 by t^{2 }– 3

The second polynomial is divided by the first polynomial

(ii) The first polynomial is x^{2} + 3x + 1

The second polynomial is 3x^{4} + 5x^{3} – 7x^{2} + 2x + 2

Dividing the second polynomial by the first polynomial

The second polynomial is divided by the first polynomial

(iii) The first polynomial is x^{2 }+ 3x + 1

The second polynomial is x^{5}− 4x^{3}+ x^{2}+ 3x + 1

Dividing the second polynomial by the first polynomial

The second polynomial is not divisible by the first polynomial

**Q3.Obtain all other zeroes of the polynomial 3x ^{4} + 6x^{3} – 2x^{2} – 10x – 5, if two of it’s zeroes are**

√(5/3), -√(5/3)

Ans. The given zeroes of the polynomial are

Therefore two of the factors of the polynomial will be

Their product will also be the factor of the given polynomial

Dividing 3x^{4} + 6x^{3} – 2x^{2} – 10x – 5 by

Factorizing the quotient further

3x² + 6x + 3

3x² + 3x + 3x + 3

3x(x + 1) + 3( x + 1)

(x + 1)(3x + 3) = 3(x + 1)(x + 1)

Hence the other two zeroes of the polynomial are –1 and –1

**See the video for question 3**

**Q4.On dividing x ^{3 }– 3x^{2} + x + 2 by a polynomial g(x), the quotient and remainder were x – 2 and -2x + 4 respectively. Find g(x).**

Ans. The dividend , p(x) =x^{3 }– 3x^{2} + x + 2, Quotient, q(x) = x – 2 and remainder, r(x) = -2x + 4 , Devisor, g(x) = ?

The relationship between p(x) ,q(x), g(x) and r(x) is given by

p(x) = g(x) .q(x) + r(x)

x^{3 }– 3x^{2} + x + 2 = g(x) (x – 2) + (-2x + 4)

g(x) (x – 2) = x^{3 }– 3x^{2} + x + 2 + 2x – 4 = x^{3 }– 3x^{2} + 3x – 2

Hence the value of g(x) is x² – x + 1

**Q5. Give examples of polynomial p(x), q(x), g(x) and r(x) which satisfy the division algorithm and**

**(i) deg.p(x) = deg.q(x)**

**(ii) deg.q(x) = deg.r(x)**

**(iii) deg.r(x) = 0**

Ans. (i) deg.p(x) = deg.q(x)

Choosing p(x) = 2x + 2 if it divided by 2 then the q(x) = x +1

So,example of the polynomial p(x), q(x), g(x) and r(x) which satisfy the given condition

p(x) = 2x + 2, q(x) = x +1, g(x) = 2, r(x) = 0

degree of 2x + 2 =degree of x + 1

deg.p(x) = deg.q(x) =1

(ii) deg.q(x) = deg.r(x)

Choosing p(x) = 3x² + 2, if it is divided by x + 2

So,example of the polynomial p(x), q(x), g(x) and r(x) which satisfy the given condition

P(x) = 3x² + 2, g(x) = x + 2, r(x) = –6x +2, q(x) = 3x

deg.q(x) = deg.r(x)

(iii) deg.r(x) = 0

Choosing p(x) and g(x) so that r(x) = constant(number)

If we suppose p(x) = 2x + 4 if we divide it by 2 the r(x) =4 and q(x) = x

So, p(x) = 2x + 4, g(x) = 2, r(x) = 4, q(x) = x

deg.r(x) = 0

**See the video of question number 4,5 of exercise 2.3 and question number 4 of exercise 2.4**

**10 class maths NCERT solutions of chapter 2 :Polynomial**

**Exercise. 2.4**

**Q1. Verify that the numbers given alongside the cubic polynomial below are their zeroes. Also, verify the relationship between the zeroes and the coefficient in each case.**

Ans.

If zeroes of a cubic polynomial are α, β and γ, then the polynomial is structured as follows

It has been proved that given numbers are the zeroes of the given polynomial

The relationship between the coefficient and zeroes of a cubic polynomial is given as

LHS = RHS, Varified

LHS = RHS, Varified

–1 = –1

LHS = RHS, Varified

If zeroes of a cubic polynomial are α, β and γ, then the polynomial is structured as follows

x³ – (2 + 1 + 1) x² + (2 × 1 + 1 × 1 + 1 ×2)x – 2× 1 ×1

x³ – 4x² + 5x – 2

It is verified that the numbers given are the zeroes of the given polynomial

4 = 4

LHS = RHS, Varified

5 = 5

LHS = RHS, Varified

2 = 2

LHS = RHS, Varified

**Q2. Find a cubic polynomial with the sum, sum of the product of its zeroes taken two at a time, and product of its zeroes as 2,–7, –14 respectively.**

Ans. If zeroes of a cubic polynomial are α, β, and γ, then the polynomial is structured as follows

We are given

α + β + γ = 2, αβ + βγ + γα = –7, αβγ = –14

Substituting all these values in the polynomial

x³ – 2x² –7 x + 14

Therefore the required polynomial is x³ – 2x² –7 x + 14

Q3. If the zeroes of the polynomial x³ – 3x² + x + 1 are a – b, a, a + b, find a and b.

Ans. The sum of zeroes of a cubic polynomial is given by

Where α, β, and γ are zeroes of the polynomial and a is the coefficient of x³, b is the coefficient of x², c is the coefficient of x and c is the constant.

Substituting the given values of zeroes in the above relation

3a = 3

a = 1

Applying another relationship between coefficient and the zeroes

Where a is the coefficient of x³ and d is the constant

Putting the values of a = 1

(1 – b)×1 × (1 + b) = –1

1 – b² = –1

b² = 2

b = ±√2

Therefore the values of a =1 and b = ±√2

**Q4. If two zeroes of the polynomial x ^{4} – 6x^{3} – 26x^{2 }+ 138x – 35 are 2 ± √3, find other zeroes.**

Ans. The zeroes of the given polynomial, x^{4} – 6x^{3} – 26x^{2 }+ 138x – 35 are as follows

2 + √3 and 2 – √3, therefore, two of the factors of the given polynomial are (x – 2 – √3) and (x – 2 + √3)

So, the product of both factor will also be the factor of the given polynomial

(x – 2 – √3)(x – 2 + √3) = (x – 2)² – √3² = x² – 4x + 4 – 3 = x² – 4x + 1

Dividing x^{4} – 6x^{3} – 26x^{2 }+ 138x – 35 by x² – 4x + 1

The quotient is x² – 2x – 35, factorizing it further

x² – 2x – 35

x² – 7x + 5x – 35

x ( x – 7) + 5( x – 7)

(x – 7)( x + 5)

x – 7 = 0 ⇒ x = 7

x + 5 = 0 ⇒ x = –5

Therefore the other zeroes of the polynomial are 7 and –5.

**Q5.If the polynomial X ^{4} −6x^{3}+ 16x^{2}− 25x + 10 is divided by another polynomial x² – 2x + k, the remainders comes out to be x + a,, find k and a.**

Ans. Dividing X^{4} −6x^{3}+ 16x^{2}− 25x + 10 by x² – 2x + k

According to question remainder is x + a

Comparing it with the remainder we got

x = (2k – 9) x, a = 10 – k( 8 – k)

2k – 9 = 1

k = 5

Putting the value of k =5 in a = 10 – k( 8 – k) = 10 – 5(8 – 5) = 10 – 15 = – 5

Hence the value of k =5 and value of a = – 5

**NCERT solutions of class 10 science**

**Solutions of class 10 last years Science question papers**

**CBSE Class 10 – Question paper of science 2020 with solutions**

**CBSE class 10 -sample paper of science**

**NCERT Solutions for class 9 maths**

**NCERT Solutions for class 9 science **

**NCERT Solutions for class 10 maths**

**CBSE Class 10-Question paper of maths 2021 with solutions**

**CBSE Class 10-Half yearly question paper of maths 2020 with solutions**

**CBSE Class 10 -Question paper of maths 2020 with solutions**

**CBSE Class 10-Question paper of maths 2019 with solutions**

**NCERT Solutions for Class 10 Science**

**NCERT Solutions for class 11 maths**

Chapter 1-Sets | Chapter 9-Sequences and Series |

Chapter 2- Relations and functions | Chapter 10- Straight Lines |

Chapter 3- Trigonometry | Chapter 11-Conic Sections |

Chapter 4-Principle of mathematical induction | Chapter 12-Introduction to three Dimensional Geometry |

Chapter 5-Complex numbers | Chapter 13- Limits and Derivatives |

Chapter 6- Linear Inequalities | Chapter 14-Mathematical Reasoning |

Chapter 7- Permutations and Combinations | Chapter 15- Statistics |

Chapter 8- Binomial Theorem | Chapter 16- Probability |

**CBSE Class 11-Question paper of maths 2015**

**CBSE Class 11 – Second unit test of maths 2021 with solutions**

**NCERT solutions for class 12 maths**

Chapter 1-Relations and Functions | Chapter 9-Differential Equations |

Chapter 2-Inverse Trigonometric Functions | Chapter 10-Vector Algebra |

Chapter 3-Matrices | Chapter 11 – Three Dimensional Geometry |

Chapter 4-Determinants | Chapter 12-Linear Programming |

Chapter 5- Continuity and Differentiability | Chapter 13-Probability |

Chapter 6- Application of Derivation | CBSE Class 12- Question paper of maths 2021 with solutions |

Chapter 7- Integrals | |

Chapter 8-Application of Integrals |

**Class 12 Solutions of Maths Latest Sample Paper Published by CBSE for 2021-22 Term 2**

**Class 12 Maths Important Questions-Application of Integrals**

**Solutions of Class 12 Maths Question Paper of Preboard -2 Exam Term-2 CBSE Board 2021-22**

**Solutions of class 12 maths question paper 2021 preboard exam CBSE Solution**