Answer was given to this problem by plotting different possible integers for "a", and finding the crossing of the x-axis (roots of the polynomial)
After trying 1, -1, 2, -2, 3, -3, 4, -4, 5.-5, 6, -6, 7, -7, 8, =8, 9, =9, 10, -10, 11, -11, 12, -12
we found that with x = 2, and a=5, the polynomial gave zero.
The other value we didn't try was a = 0,
which we could try now:, and see that we don't get a rational root for it.
So the only answer we found was for a = 5, which makes the polynomial:
f(x) = 5 x^3 + 4 x^2 - 24 x - 8
and x = 2 is a root, therefore the polynomial is factorable by the binomial (x - 2)