Our ncert solutions for Exercise 2.4 Class 10 maths 2. Polynomials - ncert solutions - Toppers Study is the best material for English Medium students cbse board and other state boards students.

# Exercise 2.4 Class 10 maths 2. Polynomials - ncert solutions - Toppers Study

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## Exercise 2.4 Class 10 maths 2. Polynomials - ncert solutions - Toppers Study

CBSE board students who preparing for **class 10 ncert solutions maths and Mathematics** solved exercise **chapter 2. Polynomials** available and this helps in upcoming exams
2023-2024.

### You can Find Mathematics solution Class 10 Chapter 2. Polynomials

- All Chapter review quick revision notes for chapter 2. Polynomials Class 10
- NCERT Solutions And Textual questions Answers Class 10 Mathematics
- Extra NCERT Book questions Answers Class 10 Mathematics
- Importatnt key points with additional Assignment and questions bank solved.

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### Exercise 2.4 class 10 Mathematics Chapter 2. Polynomials

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- Exercise 2.4 Class 10 Maths 2. Polynomials - Ncert Solutions - Toppers Study
- Class 10 Ncert Solutions
- Solution Chapter 2. Polynomials Class 10
- Solutions Class 10
- Chapter 2. Polynomials Exercise 2.4 Class 10

## 2. Polynomials

### | Exercise 2.4 |

## Exercise 2.4 Class 10 maths 2. Polynomials - ncert solutions - Toppers Study

### Exercise 2.4 Class 10 maths chapter 2. Polynomials

**Q1. Verify that the numbers given alongside of the cubic polynomials below are their zeroes. Also verify the relationship between the zeroes and the coefficients in each case:**

(i) 2x^{3} + x^{2} - 5x + 2; α = ½, β = 1 and γ = – 2;

(ii) x^{3} – 4x^{2} + 5x – 2; α = 2, β = 1, and γ = 1

**Solution:** (i) 2x^{3} + x^{2} - 5x + 2; α = ½, β = 1 and γ = – 2;

Given: p(x) = 2x^{3} + x^{2} - 5x + 2

p(1) = 2(1)^{3} + (1)^{2} - 5(1) + 2

= 2 + 1 - 5 + 2

= 5 - 5 = 0

p(x) = 0

Therefore, 1 is the zero of p(x)

अब, p(-2) = 2(-2)^{3} + (-2)^{2} - 5(-2) + 2

= -16 + 4 + 10 + 2

= 16 - 16 = 0

p(x) = 0

Therefore, -2 is the zero of p(x)

So, α = ½, β = 1 and γ = – 2 are zeroes.

And coefficients a = 2, b = 1, c = - 5 and d = 2

Verification of relation between zeroes and coefficients:

Verified by equation (1) (2) and (3)

**Solution:** (ii) x^{3} – 4x^{2} + 5x – 2; α = 2, β = 1, and γ = 1

Given that: p(x) = x^{3} – 4x^{2} + 5x – 2

p(2) = (2)^{3} – 4(2)^{2} + 5(2) – 2

= 8 – 4.4 + 10 – 2

= 8 – 16 + 10 – 2

= 18 – 18 = 0

Hence, α = 2 is the zero of p(x)

Now for β = 1

p(x) = x^{3} – 4x^{2} + 5x – 2

p(1) = (1)^{3} – 4(1)^{2} + 5(1) – 2

= 1 – 4.1 + 5 – 2

= 1 – 4 + 5 – 2

= 6 – 6 = 0

Hence, β = 1 is the zero of p(x)

Now, for γ = 1

p(x) = x^{3} – 4x^{2} + 5x – 2

p(1) = (1)^{3} – 4(1)^{2} + 5(1) – 2

= 1 – 4.1 + 5 – 2

= 1 – 4 + 5 – 2

= 6 – 6 = 0

Hence, γ = 1 is the zero of p(x)

So, α = 2, β = 1 and γ = 1 are zeroes.

And coefficients a = 1, b = – 4, c = 5 and d = – 2

Verification of relation between zeroes and coefficients:

##### Other Pages of this Chapter: 2. Polynomials

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