**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.

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**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.

**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 exercise 2.2

**See the video , solutions of question 1 of exercise 1 and question 1 of exercise 2.2-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

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

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Chapter 1- Number System | Chapter 9-Areas of parallelogram and triangles |

Chapter 2-Polynomial | Chapter 10-Circles |

Chapter 3- Coordinate Geometry | Chapter 11-Construction |

Chapter 4- Linear equations in two variables | Chapter 12-Heron’s Formula |

Chapter 5- Introduction to Euclid’s Geometry | Chapter 13-Surface Areas and Volumes |

Chapter 6-Lines and Angles | Chapter 14-Statistics |

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Chapter 8- Quadrilateral |

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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 |

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Chapter 8- Binomial Theorem | Chapter 16- Probability |

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

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Chapter 4-Determinants | Chapter 12-Linear Programming |

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

Chapter 7- Integrals | |

Chapter 8-Application of Integrals |

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