CLASS 9 MATH NCERT SOLUTION FOR CHAPTER – 2 POLYNOMIALS EX – 2.2

Polynomials

Question 1.
Find the value of the polynomial 5x – 4x2 + 3 at
(i) x=0
(ii) x=-1
(iii) x=2

Solution:

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.2.1

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

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) =x,x=0
(vi) p(x)=lx+m,x = – \cfrac { m }{ l }
(vii) p(x) =3x-1,x= –\cfrac { 1 }{ \sqrt { 3 } }  ,\cfrac { 2 }{ \sqrt { 3 } }
(viii) p(x) = 2x+1,x \cfrac { 1 }{ 2 }

Solution:

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.2.2



NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.2.3
NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.2.4
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).

Lorem ipsum dolor sit amet, consectetur adipiscing elit. Phasellus cursus rutrum est nec suscipit. Ut et ultrices nisi. Vivamus id nisl ligula. Nulla sed iaculis ipsum.