Question 3. Verify Whether the Following are Zeroes of the Polyno Exercise 2.2 Chapter 2 Polynomial Maths Class 9

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 , 2/√3
(viii) p(x) = 2x + 1, x =1/2

Solution: (i) p(x) = 3x + 1, x = - 1/3
Plug x  = -1/3
=> p(x) = 3x + 1
=>p(-1/3) = 3(-1/3)  +1
=>p(-1/3) = -1 +1
=>p(-1/3) = 0
When P(a)  = 0 then a is always zero of polynomial
Hence  -1/3 is zero of polynomial p(x) = 3x + 1

(ii) p(x) = 5x – π, x = 4/5
Plug x = 4/5 we get
=> p(4/5) = 5x – π,
=> p(4/5) = 5(4/5) – π,
=> p(4/5) = 4 – π,
And pi = 22/7 so that
=> p(4/5) = 4 – 22/7 is not = 0
Hence 4/5 is not zero of polynomial p(x) = 5x – π

(iii) p(x) = x² – 1, x = 1, –1
Plug x = - 1
=> p(x) = x² – 1
=> p(-1) = (-1)² – 1
=> p(-1) =  1 – 1
=> p(-1) = 0

Plug x =  1
=> p(x) = x² – 1
=> p(1) = (1)² – 1
=> p(1) =  1 – 1
=> p(1) = 0
Hence both x = - 1 and 1 are zero of polynomial p(x) = x² – 1

(iv) p(x) = (x + 1) (x – 2), x = – 1, 2
Plug x = - 1
=>p(x) = (x + 1) (x – 2)
=>p(-1) = (-1 + 1) (-1 – 2)
=>p(-1) = (0) (-3)
=>p(-1) = 0

Now plug x = 2 we get
Plug x = 2
=>p(x) = (x + 1) (x – 2)
=>p(2) = (2 + 1) (2 – 2)
=>p(2) = (3) (0)
=>p(2) = 0
Hence -1 and  2 both are zero of the polynomial p(x) = (x + 1) (x – 2)

(v) p(x) = x², x = 0
Plug x = 0 we get
=>p(x) = x²
=>p(0) = (0)²
=>p(0) = 0
Hence 0 is the zero so polynomial p(x) = x²

(vi) p(x) = lx + m, x = –m/l
Plug x  = - m/l we get
=> p(x) = lx + m
=> p(-m/l) = l(-m/l) + m
=> p(-m/l) = -m + m
=> p(-m/l) = 0 
Hence  - m/l is the zero of polynomial p(x) = lx + m

(vii) p(x) = 3x² – 1, x = - 1/√3 , 2/√3

Plug x = - 1/ √3 we get
=>p(x) = 3x² – 1
=>p(-1/√3) = 3(-1/√3)² – 1
=>p(-1/√3) = 3(1/3) – 1
=>p(-1/√3) = 1 – 1
=>p(-1/√3) = 0

Plug x =  2/ √3 we get
=>p(x) = 3x² – 1
=>p(1/√3) = 3(2/√3)² – 1
=>p(1/√3) = 3(4/3) – 1
=>p(1/√3) = 4 – 1
=>p(1/√3) = 3

Hence x = - 1/√3 is zero of the polynomial p(x) = 3x² – 1
But x = 2/√3 is not a zero of the polynomial

 (viii) p(x) = 2x + 1, x =1/2
Plug x = ½ we get
=> p(x) = 2x + 1
=> p(1/2) = 2(1/2) + 1
=> p(1/2) = 2(1/2) + 1
=> p(1/2) = 1 + 1
=> p(1/2) = 2
Hence ½ is not a zero of polynomial p(x) = 2x + 1