Sketch. Because the constant term is so large relative to the other terms, possible integer roots of such a polynomial lie in the relatively small interval $[160, 165]$ far to the right of the origin. The fact that $X_0$ is a root is equivalent to $4019680$ having a representation in base $X_0$ with particularly large coefficients, and by writing $4019680$ in the corresponding bases we see that it doesn't have such representations.
Proof. Reducibility is equivalent to the existence of an integer root. By inspection the polynomial can have at most one real root $X_0$, somewhere past $\sqrt[3]{4019680} = 159.0000131...$. By setting $a_0 = a_1 = a_2 = 20$ we check numerically that $X_0 < 166$, so the only possible roots are $X_0 = 160, 161, 162, 163, 164, 165$. Any integer root must divide the constant term, but: $4019680 \equiv 0 \bmod 160$ $4019680 \equiv 154 \bmod 161$ $4019680 \equiv 136 \bmod 162$ $4019680 \equiv 100 \bmod 163$ $4019680 \equiv 40 \bmod 164$ $4019680 \equiv 115 \bmod 165.$
Since $0 \le a_0 \le 20$, it is impossible for $X_0$ to divide the constant term unless $X_0 = 160, a_0 = 0$ or $X_0 = 161, a_0 = 7$. In either case, after dividing out by $X_0$, $X_0$ must divide the new constant terms, which are $-(25123+a_1), -(24967+a_1)$ respectively. But $25123 \equiv 3 \bmod 160$ $24967 \equiv 12 \bmod 161$
and again since $0 \le a_1 \le 20$, this is impossible.