polynomial with positive integer coefficients divisible by 24? I have to show that  $n^4+ 6n^3 + 11n^2+6n$ is divisible by 24 for every natural number, n, so I decided to show that this polynomial is divisible by 8 and 3, but I'm having trouble showing that it is divisible by 8.
Divisible by 3: $$n^4+ 6n^3 + 11n^2+6n \equiv n^4+2n^2 \pmod3 $$$$n^4+2n^2 = n^2(n^2+2)=n^2(n^2-1)=n^2(n+1)(n-1)\equiv 0 \pmod3$$
Divisble by 8: $$n^4+ 6n^3 + 11n^2+6n \equiv n(n^3-2n^2-5n-2) \pmod8 $$$$n(n^3-2n^2-5n-2) = n(n+1)(n^2-3n-2) = n(n+1)(n^2+5n+6) = n(n+1)(n+2)(n+3) = ??? \pmod8$$
So it seems that the polynomial has to be divisible by four, but I'm not sure how to show that it has to be divisible by two one more time to show that it's divisible by 8.
Any other approaches to solving the problem are welcome, as I'm preparing for an exam and would appreciate having multiple ways to tackle the problem. I think this problem was meant to be an exercise in proof by induction, but the other approach seemed more approachable.
 A: $$n^4+6n^3+11n^2+6n=n(n^3+6n^2+11n+6)=n(n+1)(n^2+5n+6)$$
$$=n(n+1)(n+2)(n+3)$$ which is a Product of $4$ consecutive integers 
Now see The product of n consecutive integers is divisible by n factorial
A: We have
$$
n^4+6n^3+11n^2+6n=n(n+1)(n+2)(n+3)=24\binom{n+3}4.
$$
This settles the claim when $n\ge1$. To get it for all $n$ observe that all integers are congruent to one $>1$ modulo $24$.
A: If we write the polynomial as a linear combination of combinatorial polynomials, then the polynomial is divisible by $24$ for all integer arguments if and only if all of the coefficients are divisible by $24$.
$$
n^4+6n^3+11n^2+6n=24\binom{n}{4}+72\binom{n}{3}+72\binom{n}{2}+24\binom{n}{1}
$$
and so the polynomial is divisible by $24$ for all integer arguments.
A: $n(n+1)(n+2)(n+3)$ is divisible by $8$ because there are two even numbers and one of them is divisible by $4$.
A: You can always do a brute force method. Check n=0,1,2,3,4,5,6,7
$$
n=0, 
n(n+1)(n+2)(n+3)=0(1)(2)(3)=0 (mod 8)
$$
$$
n=1,
1(2)(3)(4)=24=0 (mod8)
$$
$$....$$
$$n=7,
7(8)(9)(10)=7(0)(1)(2)=0 (mod8)
$$
This is sufficient since if $$n=8m+r, r=0,1,2,3,4,5,6,7, m\in Z $$ then
$$
n(n+1)(n+2)(n+3)=(8m+r)(8m+r+1)(8m+r+2)(8m+r+3)=r(r+1)(r+2)(r+3)(mod8)$$
