# NCERT Solution for Polynomial Chapter 2 Exercise 2.1

In this page we have Class 9 Maths NCERT Solution for Polynomial Chapter-2 for Exercise 2.1 . Hope you like them and do not forget to like , social share and comment at the end of the page.
Question 1:
Find the value of the polynomial $P(x)= 5x-4x^2 +3$ at
(i) x = 0 (ii) x = -1 (iii) x = 2
Solution:
(i)$P(x)= 5x-4x^2 +3$
P(0)= 0-0+3=3
(ii)$P(x)= 5x-4x^2 +3$
P(-1) =-5-4+3=-6
(iii)$P(x)= 5x-4x^2 +3$
P(2)= 10-16+3=-3
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) $p(y) = y^2 - y + 1$
p(0) = (0)2- (0) + 1 = 1
p(1) = (1)2- (1) + 1 = 1
p(2) = (2)2- (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
= 2 + 1 + 2 - 1 = 4
p(2) = 2 + 2 + 2(2)2- (2)3
= 2 + 2 + 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 (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 =- 1/3
(ii) $p(x) = 5x - \pi$, x =4/5
(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 = - m/l
(vii) $p(x) = 3x^2 - 1$, x =-1/√3 and 2/√3
(viii) $p(x) = 2x + 1$, x =1/2

Solution:
(i) $p(x) = 3x + 1$,x=-1/3
p(-1/3) = 3 (-1/3)+1=-1+1=0
p(-1/3) = 0 which means that-1 /3is zero of the polynomial p(x) = 3x+1.

(ii) $p(x) = 5x - \pi$, x =4/5
p(4/5) = 5(4/5)-π = 4-π
p(4/5) ≠ 0 which means that 4/5 is not zero of the polynomial $p(x) = 5x - \pi$.

(iii) $p(x) = x^2 - 1$,x = 1,- 1
p(1)=12-1=1-1=0
p(-1)=(-1)2-1=1-1=0
Both p(1) and p(-1) are equal to 0. It means that 1 and -1 are zeroes of the polynomial $p(x) = x^2 - 1$.

(iv) $p(x) = (x + 1) (x - 2)$,x=-1,2
p(-1) = (-1+1)(-1-2) = 0×-3 = 0
p(2) = (2+1)(2-2) = 3×0 = 0
Both p(-1) and p(2) are equal to 0. It means that -1 and 2 are zeroes of the polynomial $p(x) = (x + 1) (x - 2)$.

(v) $p(x) = x^2$,x = 0
p(0) = 02= 0
p(0) = 0 which means that 0 is the zero of the polynomial $p(x) = x^2$.

(vi) $p(x) = lx + m$,x =-m/
p(-m/l) =l(-m/l) + m = -m + m = 0
p(-m/l) = 0 which means that(-m/l)is zero of the polynomial $p(x) = lx + m$

(vii) $p(x) = 3x^2 - 1$, x =-1/√3 and 2/√3
p(-1/√3) = 3(-1/√3)2-1=3(1/3)-1=1-1=0
p(2/√3) = 3(2/√3)2-1 = 3 ×(4/3)-1= 4-1 = 3
p(-1/√3) = 0 which means that-1/√3is zero of the polynomial p(x) = 3x2- 1.
p(2/√3) ≠ 0 which means that2/√3is not zero of the polynomial $p(x) = 3x^2 - 1$.

(viii) p(x) = 2x + 1 ,x =1/2
p(1/2)=2 ×(1/2)+ 1=1+1=2
p(1/2) ≠ 0. It means that 1/2 is not zero of the polynomial p(x)=2x+1.

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) $p(x) = 2x + 5$
(iv) $p(x) = 3x - 2$
(v) $p(x) = 3x$
(vi) $p(x) = ax$, a ≠ 0
(vii) $p(x) = cx + d$, c ≠ 0, c, d are real numbers.
Solution:
Zero of a polynomial is that value of the variable at which the value of the polynomial is obtained as 0.
(i) $p(x) = x + 5$
p(x) = 0
x + 5 = 0
x = - 5
Therefore, for x = -5, the value of the polynomial is 0 and hence, x = -5 is a zero of the given polynomial.

(ii) $p(x) = x - 5$
p(x) = 0
x - 5 = 0
x = 5
Therefore, for x = 5, the value of the polynomial is0 and hence, x = 5 is a zero of the given polynomial.

(iii) $p(x) = 2x + 5$
p(x) = 0
2x + 5 = 0
2x = - 5
x = -5/2
Therefore, for x = -5/2 , the value of the polynomial is 0 and hence, x = -5/2is a zero of the given polynomial.

(iv) $p(x) = 3x - 2$
p(x) = 0
3x - 2 = 0
x=2/3, so 2/3 is the zero of the polynomial

(v) $p(x) = 3x$
p(x) = 0
3x = 0
x = 0
So x=0 is the zero of the polynomial

(vi) $p(x) = ax$
p(x) = 0
ax = 0
x = 0
Therefore, for x = 0, the value of the polynomial is 0 and hence, x = 0 is a zero of the given polynomial.

(vii) $p(x) = cx + d$
p(x) = 0
cx+ d = 0
x=-d/c