Our ncert solutions for Exercise 2.2 Class 9 maths 2. Polynomials - ncert solutions - Toppers Study is the best material for English Medium students cbse board and other state boards students.

# Exercise 2.2 Class 9 maths 2. Polynomials - ncert solutions - Toppers Study

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## Exercise 2.2 Class 9 maths 2. Polynomials - ncert solutions - Toppers Study

CBSE board students who preparing for **class 9 ncert solutions maths and Mathematics** solved exercise **chapter 2. Polynomials** available and this helps in upcoming exams
2024-2025.

### You can Find Mathematics solution Class 9 Chapter 2. Polynomials

- All Chapter review quick revision notes for chapter 2. Polynomials Class 9
- NCERT Solutions And Textual questions Answers Class 9 Mathematics
- Extra NCERT Book questions Answers Class 9 Mathematics
- Importatnt key points with additional Assignment and questions bank solved.

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### Exercise 2.2 class 9 Mathematics Chapter 2. Polynomials

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- Exercise 2.2 Class 9 Maths 2. Polynomials - Ncert Solutions - Toppers Study
- Class 9 Ncert Solutions
- Solution Chapter 2. Polynomials Class 9
- Solutions Class 9
- Chapter 2. Polynomials Exercise 2.2 Class 9

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## 2. Polynomials

### | Exercise 2.2 |

## Exercise 2.2 Class 9 maths 2. Polynomials - ncert solutions - Toppers Study

**2. Polynomials**

**Exercise 2.2 **

Q1. Find the value of the polynomial 5x – 4x^{2} + 3 at

(i) x = 0 (ii) x = –1 (iii) x = 2

**Solution:**

(i) p(x) = 5x - 4x^{2 }+ 3

The value of the polynomial p(x) at x = 0 is given by

P(0) = 5(0) - 4(0)^{2 }+ 3

= 0 - 0 + 3

= 3

(ii) p(x) = 5x - 4x^{2 }+ 3

The value of the polynomial p(x) at x = 1 is given by

P(1) = 5(1) - 4(1)^{2 }+ 3

= 5 - 4 + 3

= 4

(iii) p(x) = 5x - 4x^{2 }+ 3

The value of the polynomial p(x) at x = 2 is given by

P(2) = 5(2) - 4(2)^{2 }+ 3

= 10 -16 + 3

= - 9

Q2.Find p(0), p(1) and p(2) for each of the following polynomials:

(i) p(y) = y^{2} – y + 1

(ii) p(t) = 2 + t + 2t^{ 2} – t ^{3}

(iii) p(x) = x^{3}

(iv) p(x) = (x – 1) (x + 1)

**Solution:**

(i) p(y) = y^{2 }- y + 1

P(0) = (0)^{2}- 0 + 1

=1

P(1) = (1)^{2}- 1 + 1

= 1 - 1 + 1

= 1

P(2) = (2)^{2}- 2 + 1 = 4 - 2 + 1 = 3

(ii) p(t) = 2 + t + 2t^{2}- t^{3}

P(0) = 2 + 0 + 2(0)^{2}- (0)^{3}

=2

P(1) = 2 + 1 + 2(1)^{2 }- (1)^{3}

= 4

P(2) = 2 + 2 + 2(2)^{2}- (2)^{3}

= 4 + 8 - 8

= 4

(iii) p(x) = x^{3}

P(0)=(0)^{3} =0

P(1)=(1)^{3 }=1

P(2)=(2)^{3}=8

(iv)p(x) = (x – 1) (x + 1)

P(0)= (0-1) (0+1)=(-1) (1) =-1

P(1)= (1-1) (1+1) =0(1) =0

P(2)= (2-1) (2+1)=1(3) =3

Q3. Verify whether the following are zeroes of the polynomial, indicated against them.

**Solution: **

Q4. Find the zero of the polynomial in each of the following cases:

(i) P(x) = x + 5

(ii) P(x) = x **–** 5

(iii) Px) = 2x + 5

(iv) P(x) = 3x **–** 2

(v) P(x) = 3x

(vi) P(x) = ax, a ≠ 0

**Solution (i) :**

(i) P(x) = x + 5

⇒ x + 5 = 0

⇒ x = - 5

The zero of the polynomial is - 5.

** Solution (ii) :**

(ii) P(x) = x **–** 5

⇒ x **–** 5 = 0

⇒ x = 5

The zero of the polynomial is 5.

The zero of the polynomial is - 5/2.

(iv) P(x) = 3x - 2

3x - 2 = 0

The zero of the polynomial is 2/3.

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

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