(A) Let $f(t) = t^{3}-2 t^{2}-15 t$.
$= t(t^{2}-2 t-15)$
$= t(t^{2}-5 t+3 t-15)$ [by splitting the middle term]
$= t[t(t-5)+3(t-5)]$
$= t(t-5)(t+3)$
So,the value of $f(t)$ is zero when $t=0$,$t-5=0$,or $t+3=0$.
Thus,the zeroes are $t=0, t=5$,and $t=-3$.
Verification:
Sum of zeroes $= 0 + 5 + (-3) = 2 = -(-2)/1 = -(\text{Coefficient of } t^{2}) / (\text{Coefficient of } t^{3})$.
Sum of product of zeroes taken two at a time $= (0)(5) + (5)(-3) + (-3)(0) = 0 - 15 + 0 = -15 = (-15)/1 = (\text{Coefficient of } t) / (\text{Coefficient of } t^{3})$.
Product of zeroes $= (0)(5)(-3) = 0 = -(0)/1 = -(\text{Constant term}) / (\text{Coefficient of } t^{3})$.
Hence,the relationship is verified.