# Class 9 Maths NCERT Solution for Polynomial part 2

In this page we have NCERT book Solutions for Class 9th Maths:Polynomial for EXERCISE 2 . 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-4x2 +3 at
(i) x = 0           (ii)       x = −1 (iii)      x = 2
Solution:
(i) P(x)= 5x-4x2 +3
P(0)= 0-0+3=3
(ii) P(x)= 5x-4x2 +3
P(-1) =-5-4+3=-6
(iii) P(x)= 5x-4x2 +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) = y2 − y + 1     (ii) p(t) = 2 + t + 2t2 − t3
(iii) p(x) = x3 (iv)         p(x) = (x − 1) (x + 1)
Solution:
(i) p(y) = y2 − 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 + 2t2 − t3
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) = x3
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 – π, x =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 = – m/l
(vii) p(x) = 3x2  – 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−π, 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 − π.
(iii) p(x) = x2−1,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− 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) = x2,x = 0
p(0) = 02 = 0
p(0) = 0 which means that 0 is the zero of the polynomial p(x) = x2.
(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) = 3x2 − 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)−1 = 3 × (4/3)−1= 4−1 = 3
p(-1/√3) = 0 which means that -1/√3 is zero of the polynomial p(x) = 3x− 1.
p(2/√3) ≠ 0 which means that 2/√3 is not zero of the polynomial p(x)=3x− 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 = - 5/2  is 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