Verify whether the indicated numbers are zeros of the polynomials corresponding to them in the following cases:
(i) f(x) = 3x+1; x = 
(ii) f(x) = x2-1; x = 1, -1
(iii) g(x) = 3x2-2; x = 
, -
(iv) p(x) = x3-6x2+11x-6, x = 1,2,3
(v) f(x) = 5x-π, x= 
(vi) f(x) = x2, x=0
(vii) f(x) =1x+m, x= 
(viii) f(x) = 2x+1, x =
(i) f(x) = 3x + 1
Put x = -1/3
f (-1/3) = 3 * (-1/3) + 1
= -1 + 1
= 0
Therefore, x = -1/3 is a root of f (x) = 3x + 1
(ii) We have,
f (x) = x2 – 1
Put x = 1 and x = -1
f (1) = (1)2 – 1 and f (-1) = (-1)2 – 1
= 1 – 1 = 1- 1
= 0 = 0
Therefore, x = -1 and x = 1 are the roots of f(x) = x2 – 1
(iii) g (x) = 3x2 – 2
Put x = 
and x = 
g (
) = 3 (
)2 – 2 and g (
) = 3 (
)2 – 2
= 3 * 
– 2 = 3 * 
– 2
= 2 
0 = 2 
0
Therefore, x = 
and x = 
are not the roots of g (x) = 3x2 – 2
(iv) p (x) = x3 – 6x2 + 11x – 6
Put x = 1
p (1) = (1)3 – 6 (1)2 + 11 (1) – 6
= 1 – 6 + 11 – 6
= 0
Put x = 2
p (2) = (2)3 – 6 (2)2 + 11 (2) – 6
= 8 – 24 + 22 – 6
= 0
Put x = 3
p (3) = (3)3 – 6 (3)2 + 11 (3) – 6
= 27 – 54 + 33 – 6
= 0
Therefore, x = 1, 2, 3 are roots of p (x) = x3 – 6x2 + 11x – 6
(v) f (x) = 5x – 
Put x = 
f (
) = 5 * 
– 
= 4 – 
0
Therefore, x = 
is not a root of f (x) = 5x – 
(vi) f (x) = x2
Put x = 0
f (0) = (0)2
= 0
Therefore, x = 0 is not a root of f (x) = x2
(vii) f (x) = lx + m
Put x = 
f (
) = l * (
) + m
= -m + m
= 0
Therefore, x = 
is a root of f (x) = lx + m
(viii) f (x) = 2x + 1
Put x = 
f (
) = 2 * 
+ 1
= 1 + 1
= 2 
0
Therefore, x = 
is not a root of f (x) = 2x + 1