(i) f(x) = 3x + 1, x = \(\frac{-1}{3}\) 

f(x) = 3x + 1 

Substitute x =  \(\frac{-1}{3}\)  in f(x) 

\(f (\frac{-1}{3})\)= \(3(\frac{-1}{3}) + 1\)

= -1 + 1

= 0 

Since, the result is 0, so x =  \(\frac{-1}{3}\)  is the root of 3x + 1 

(ii) f(x) = x² – 1, x = 1,−1 

f(x) = x² – 1 

Given that x = (1 , -1) 

Substitute x = 1 in f(x) 

f(1) = 1² – 1 

= 1 – 1 

= 0 

Now, substitute x = (-1) in f(x) 

f(-1) = (−1)² – 1 

= 1 – 1 

= 0 

Since , the results when x = 1 and x = -1 are 0, so (1 , -1) are the roots of the polynomial f(x) = x² – 1 

(iii) g(x) = 3x² – 2 , \(x = \frac{2}{\sqrt3}\) ,\(\frac{-2}{\sqrt3}\) 

g(x) = 3x² – 2 

Substitute \(x = \frac{2}{\sqrt3}\) in g(x) 

g\((\frac{2}{\sqrt3})\) = \(3(\frac{2}{\sqrt3})^2 - 2\)

= \(3 (\frac{4}{3}) - 2\)

= 4 – 2 

= 2 ≠ 0 

Now, Substitute x =  \(\frac{-2}{\sqrt3}\) in g(x) 

\(g(\frac{2}{\sqrt3}) = 3 (\frac{-2}{\sqrt3})^2 - 2\)

= \(3 (\frac{4}{3}) - 2\)

= 4 – 2 

= 2 ≠ 0 

Since, the results when x = \(\frac{2}{\sqrt3}\) and x = \(\frac{-2}{\sqrt3}\) are not 0. Therefore (\(\frac{2}{\sqrt3}\) ,\(\frac{-2}{\sqrt3}\) ) are not zeros of 3x² – 2. 

(iv) p(x) = x³ – 6x² + 11x – 6 , x = 1, 2, 3 

p(1) = 1³ – 6(1)² + 11 x 1 – 6 

= 1 – 6 + 11 – 6 

= 0 

p(2) = 2³ – 6(2)² + 11 x 2 – 6 

= 8 – 24 – 22 – 6

= 0 

p(3) = 3³ – 6(3)² + 11 x 3 – 6 

= 27 – 54 + 33 – 6 

= 0 

Therefore, x = 1, 2, 3 are zeros of p(x). 

(v) f(x) = 5x – π, x = \(\frac{4}{5}\) 

f( \(\frac{4}{5}\) ) = 5 x  \(\frac{4}{5}\)  – π 

= 4 – π ≠ 0 

Therefore, x =  \(\frac{4}{5}\)  is not a zeros of f(x). 

(vi) f(x) = x² , x = 0 

f(0) = 0² = 0 

Therefore, x = 0 is a zero of f(x). 

(vii) f(x) = lx + m, x = \(\frac{-m}{l}\) 

f( \(\frac{-m}{l}\) ) = l x  \(\frac{-m}{l}\)  + m 

= -m + m 

= 0 

Therefore, x =  \(\frac{-m}{l}\)  is a zero of f(x). 

(viii) f(x) = 2x + 1, x = \(\frac{1}{2}\) 

f(\(\frac{1}{2}\)) = 2 x \(\frac{1}{2}\) + 1 

= 1 + 1 

= 2 ≠ 0 

Therefore, x = \(\frac{1}{2}\) is not a zero of f(x).