proof that $n^4+22n^3+71n^2+218n+384$ is divisible by $24$ Let $n$ be a positive integer number.

How do we show that the irreducible polynomial $n^4+22n^3+71n^2+218n+384$ is divisible by $24$ for all $n$? 


Another way to do it is to note:
$n^4+22n^3+71n^2+218n+384 \equiv $
$n^4 - 2n^3 - n^2 + 2n \mod 8$.
If $n \equiv 0, \pm 1, \pm 2, \pm3, 4 \mod 8$ then $n^4 - 2n^3 - n^2 + 2n\equiv$
$0, 1\mp 2 - 1 \pm 1, 16 \mp 16 -4 \pm 4, 81 \mp 54 - 9 \pm 6, 4^4 \mp 2*4^3 - 16 + 8 \equiv$
$0, 0, 72 \mp 48, 0 \equiv 0 \mod 8$.
So $8$ divides $n^4+22n^3+71n^2+218n+384$
and $n^4+22n^3+71n^2+218n+384 \equiv $
$n^4 + n^3 - n^2 -n \mod 3$
If $n \equiv 0, \pm 1 \mod 3$ then $n^4 + n^3 - n^2 - n \equiv 0, 1 \pm 1 -1 \mp 1 \equiv 0 \mod 3$.
So $3$ divides $n^4+22n^3+71n^2+218n+384$.
So $3*8 =24$ divides $n^4+22n^3+71n^2+218n+384$.
