The **NCERT Solutions Maths Class 10 for Chapter 2 Exercise 2.2 Polynomials** are provided here in pdf format. These solutions are created by Maths experts who have reviewed them from time to time. These NCERT chapter-wise solutions help the students to study and prepare well for their CBSE exams. The Solutions of NCERT for Exercise 2.2 Chapter 2- polynomials are given here in a step-wise and easy to understand method. You will surely be able to solve these easily, once you go through these **NCERT solutions for Class 10 Maths**.

All the important factors like paying attention to NCERT guidelines while preparing these solutions have been focused upon here. Exercise 2.2 explains the relationship between Zeroes and Coefficients of Polynomials.

### Download PDF of NCERT Solutions for Class 10 Maths Chapter 2- Polynomials Exercise 2.2

### Access Other Exercise Solutions of Class 10 Maths Chapter 2- Polynomials

Exercise 2.1 Solutions 1 Question

Exercise 2.3 Solutions 5 Questions (2 short, 3 long)

Exercise 2.4 Solutions 5 Questions (2 short, 3 long)

### Access Answers to NCERT Class 10 Maths Chapter 2 – Polynomials Exercise 2.2

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

**Solutions:**

**(i) x ^{2}–2x –8 **

**⇒**x^{2}– 4x+2x–8 = x(x–4)+2(x–4) = (x-4)(x+2)

Therefore, zeroes of polynomial equation x^{2}–2x–8 are (4, -2)

Sum of zeroes = 4–2 = 2 = -(-2)/1 = -(Coefficient of x)/(Coefficient of x^{2})

Product of zeroes = 4×(-2) = -8 =-(8)/1 = (Constant term)/(Coefficient of x^{2})

**(ii) 4s ^{2}–4s+1 **

⇒4s^{2}–2s–2s+1 = 2s(2s–1)–1(2s-1) = (2s–1)(2s–1)

Therefore, zeroes of polynomial equation 4s^{2}–4s+1 are (1/2, 1/2)

Sum of zeroes = (½)+(1/2) = 1 = -(-4)/4 = -(Coefficient of s)/(Coefficient of s^{2})

Product of zeros = (1/2)×(1/2) = 1/4 = (Constant term)/(Coefficient of s^{2 })

**(iii) 6x ^{2}–3–7x **

⇒6x^{2}–7x–3 = 6x^{2 }– 9x + 2x – 3 = 3x(2x – 3) +1(2x – 3) = (3x+1)(2x-3)

Therefore, zeroes of polynomial equation 6x^{2}–3–7x are (-1/3, 3/2)

Sum of zeroes = -(1/3)+(3/2) = (7/6) = -(Coefficient of x)/(Coefficient of x^{2})

Product of zeroes = -(1/3)×(3/2) = -(3/6) = (Constant term) /(Coefficient of x^{2 })

**(iv) 4u ^{2}+8u **

⇒ 4u(u+2)

Therefore, zeroes of polynomial equation 4u^{2} + 8u are (0, -2).

Sum of zeroes = 0+(-2) = -2 = -(8/4) = = -(Coefficient of u)/(Coefficient of u^{2})

Product of zeroes = 0×-2 = 0 = 0/4 = (Constant term)/(Coefficient of u^{2 })

**(v) t ^{2}–15 **

⇒ t^{2} = 15 or t = ±√15

Therefore, zeroes of polynomial equation t^{2} –15 are (√15, -√15)

Sum of zeroes =√15+(-√15) = 0= -(0/1)= -(Coefficient of t) / (Coefficient of t^{2})

Product of zeroes = √15×(-√15) = -15 = -15/1 = (Constant term) / (Coefficient of t^{2 })

**(vi) 3x ^{2}–x–4**

⇒ 3x^{2}–4x+3x–4 = x(3x-4)+1(3x-4) = (3x – 4)(x + 1)

Therefore, zeroes of polynomial equation3x^{2} – x – 4 are (4/3, -1)

Sum of zeroes = (4/3)+(-1) = (1/3)= -(-1/3) = -(Coefficient of x) / (Coefficient of x^{2})

Product of zeroes=(4/3)×(-1) = (-4/3) = (Constant term) /(Coefficient of x^{2 })

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

**(i) 1/4 , -1**

**Solution:**

From the formulas of sum and product of zeroes, we know,

Sum of zeroes = α+β

Product of zeroes = α β

Sum of zeroes = α+β = 1/4

Product of zeroes = α β = -1

∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly as:-

**x ^{2}–(α+β)x +αβ = 0**

**x ^{2}–(1/4)x +(-1) = 0**

**4x ^{2}–x-4 = 0**

**Thus,4x ^{2}–x–4 is the **quadratic polynomial.

**(ii)**√2, 1/3

**Solution:**

Sum of zeroes = α + β =√2

Product of zeroes = α β = 1/3

∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly as:-

**x ^{2}–(α+β)x +αβ = 0**

**x ^{2} –(**√2

**)x + (1/3) = 0**

**3x ^{2}-3**√2x+1 = 0

**Thus, 3x ^{2}-3**√2x+1

**is the**quadratic polynomial.

**(iii) 0, √5**

**Solution:**

Given,

Sum of zeroes = α+β = 0

Product of zeroes = α β = √5

∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly

as:-

**x ^{2}–(α+β)x +αβ = 0**

**x ^{2}–(0)x +**√5

**= 0**

**Thus, x ^{2}+**√5

**is the**quadratic polynomial.

**(iv) 1, 1 **

**Solution:**

Given,

Sum of zeroes = α+β = 1

Product of zeroes = α β = 1

∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly as:-

**x ^{2}–(α+β)x +αβ = 0**

**x ^{2}–x+1 = 0**

**Thus , x ^{2}–x+1is the **quadratic polynomial.

**(v) -1/4, 1/4 **

**Solution:**

Given,

Sum of zeroes = α+β = -1/4

Product of zeroes = α β = 1/4

**x ^{2}–(α+β)x +αβ = 0**

**x ^{2}–(-1/4)x +(1/4) = 0**

**4x ^{2}+x+1 = 0**

**Thus,4x ^{2}+x+1 is the **quadratic polynomial.

**(vi) 4, 1**

**Solution:**

Given,

Sum of zeroes = α+β =4

Product of zeroes = αβ = 1

**x ^{2}–(α+β)x+αβ = 0**

**x ^{2}–4x+1 = 0**

**Thus, x ^{2}–4x+1 is the **quadratic polynomial.

Exercise 2.2 of NCERT solutions for Class 10 Maths Chapter 2 is the second exercise of Polynomials of Class 10 Maths. Polynomials are introduced in Class 9 and is further discussed in detail in Class 10, by studying different cases of relationship between Zeroes and Coefficients of a Polynomial.

- Relationship between Zeroes and Coefficients of a Polynomial – It includes two questions with six different cases each.

### Key Features of NCERT Solutions for Class 10 Maths Chapter 2- Polynomials Exercise 2.2 Page number 33

- These NCERT Solutions helps you solve and revise all questions of Exercise 2.1.
- Solving these NCERT Solutions will help you score well in exams.
- These are the best study resources as they are prepared by Maths subject experts.
- It follows NCERT guidelines, which help in preparing the students accordingly.
- It contains all the important questions from the examination point of view.

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