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 – 4x2 + 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)=y2-y+1
(ii) p(t) =2+t+2t2-t3
(iii) p(x) =x3
(iv) p (x)=(x-1)(x+1)
Solution:
(i) We have,p(y) = y2-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 + 2t2 -t3
∴ 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) = x3
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=- \(\cfrac { 1 }{ 3 }\)
(ii) p(x)=5x-π,x=\(\cfrac { 4 }{ 5 }\)
(iii) p(x) =x2 -1,x=1,-1
(iv) p(x)=(x+1)(x-2),x = -1,2
(v) p(x) =x2 ,x=0
(vi) p(x)=lx+m,x = – \(\cfrac { m }{ l }\)
(vii) p(x) =3x2 -1,x= –\(\cfrac { 1 }{ \sqrt { 3 } } \) ,\(\cfrac { 2 }{ \sqrt { 3 } } \)
(viii) p(x) = 2x+1,x \(\cfrac { 1 }{ 2 }\)
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).
We hope the NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.2 help you. If you have any query regarding NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.2, drop a comment below and we will get back to you at the earliest.