In this article we provide the **NCERT Solutions for class 10 Maths Chapter 2 (New Syllabus)**. Here you can find the **NCERT Solutions for class 10 Maths Chapter 2 **. We at Study Circle have created step-by-step solutions with thorough explanations for those times when students feel overwhelmed about looking for the most complete and detailed **NCERT Solutions for Class 10 Maths Chapter- 2 (New Syllabus)**.

We suggest that students who wish to do well in maths read over these solutions and brush up on their skills.

There are 2 exercises in Chapter 2 Polynomials , and the NCERT Solutions provided on this page answer the exercises’ questions. Let’s now take a closer look at some of the ideas covered in this chapter.

**NCERT Solutions for Class 10 Maths Chapter 2 **

** EXERCISE 2.1**

**Question 1. The graphs of y = p(x) are given in Fig. 2.10 below, for some polynomials p(x). Find the number of zeroes of p(x), in each case.**

**Solutions:–**

**Graphical method to find zeroes:-**

Total number of zeroes in any polynomial equation = total number of times the curve intersects x-axis.

(i) In the given graph, the number of zeroes of p(x) is 0 because the graph is parallel to x-axis does not cut it at any point.

(ii) In the given graph, the number of zeroes of p(x) is 1 because the graph intersects the x-axis at only one point.

(iii) In the given graph, the number of zeroes of p(x) is 3 because the graph intersects the x-axis at any three points.

(iv) In the given graph, the number of zeroes of p(x) is 2 because the graph intersects the x-axis at two points.

(v) In the given graph, the number of zeroes of p(x) is 4 because the graph intersects the x-axis at four points.

(vi) In the given graph, the number of zeroes of p(x) is 3 because the graph intersects the x-axis at three points.

**NCERT Solutions for Class 10 Maths Chapter 2**

** EXERCISE 2.2**

**Question 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

^{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+1 is 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.