Answer: The correct option is
(B) 5, − one third , one half.
Step-by-step explanation: We are given to select the correct option that represents the zeroes of the following cubic function :

We know that
if for any number a, the function f(x) is zero at x =a , then x = a is a zero of the function f(x).
We have, for the given function, at x = 5,

So, x = 5 is a zero of the function f(x).
Now, we have

The zeroes of f(x) are given by

Thus, the zeroes of f(x) are 5, -one-third, one half.
Option (B) is CORRECT.