NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.2 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.2.

- Polynomials Class 9 Ex 2.1
- Polynomials Class 9 Ex 2.2
- Polynomials Class 9 Ex 2.3
- Polynomials Class 9 Ex 2.4
- Polynomials Class 9 Ex 2.5

Board | CBSE |

Textbook | NCERT |

Class | Class 9 |

Subject | Maths |

Chapter | Chapter 2 |

Chapter Name | Polynomials |

Exercise | Ex 2.2 |

Number of Questions Solved | 4 |

Category | NCERT Solutions |

## NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.2

**Ex 2.2 Class 9 Maths Question 1.**

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

(i) x=0

(ii) x=-1

(iii) x=2**Solution:**

**Ex 2.2 Class 9 Maths Question 2.**

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)** We have,p(y) = y^{2}-y+1

∴ p(0)= (0)^{2}-0+1 = 0-0+1 =1,

p(1)=(1)^{2} -1+1 = 1-1+1 = 1,

and p(2) = (2)^{2} – 2 +1 = 4 – 2 +1 = 3

**(ii)** We have, p(t) = 2 +1 + 2t^{2} -t^{3}∴ p(0) = 2 + 0 + 2(0)^{2} – (0)^{3}= 2 + 0+ 0-0 = 2,

p(1) = 2 +1 + 2(1)^{2} – (l)^{3}= 2+l+2-l = 5-1 = 4

p(2) = 2 + 2 + 2(2)^{2} – (2)^{3}= 2 + 2 + 8-8 = 4

**(iii)** We have, p(x) = x^{3}p(o) = (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) = (2 -1) (1 +1) = (0)(2) = 0

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

**Ex 2.2 Class 9 Maths Question 3.**

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

(i) p(x) =3x+1,x=-

(ii) p(x)=5x-π,x=

(iii) p(x) =x^{2} -1,x=1,-1

(iv) p(x)=(x+1)(x-2),x = -1,2

(v) p(x) =x^{2 },x=0

(vi) p(x)=lx+m,x = –

(vii) p(x) =3x^{2 }-1,x= – ,

(viii) p(x) = 2x+1,x **Solution:****Ex 2.2 Class 9 Maths Question 4.****Find the zero of the polynomial in each of the following cases :**

(i) p(x) = x + 5

(ii) p(x) = x – 5

(iii) p(x) = 2x + 5

(vi) p(x) = 3x-2

(v) p(x) = 3x

(vi) p(x) = ax, a≠0

(vii) p(x) = cx + d, c≠0, c, d are real numbers.**Solution:**

(i) We have, p(x) = x + 5

Now, p(x) = 0 => x + 5 = 0 => x = -5

∴ 5 is a zero of the polynomial p(x).

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