Prove that $2x^3+3x^2+x$ is always divisible by 6 if x is an integer. Prove that $2x^3+3x^2+x$ is always divisible by $6$ if $x$ is an integer.
I started by factoring the expression:
$x(2x^2+3x+1)=x(2x+1)(x+1)$
However I wasn’t sure how to progress from here to prove that it is always divisible by 6. Any ideas?
 A: To be divisible by $6$, you need to prove, that it's divisible by $2$ and $3$.
The first one is easy: $x(x+1)$ is always divisible by 2, because either of consecutive natural numbers is even.
With three if $x$ or $x + 1$ is divisible by 3, then it's completed. Let's consider the third case, when neither of the former is divisible by $3$: $x \equiv 1 \pmod3$, then $2x + 1=3 \equiv 0 \pmod3$.
A: You can use induction. Let $f(x)=2x^3+3x^2+x$. Then
$$
f(x+1)-f(x)=6 (x + 1)^2
$$
implies that $f(x)$ is a multiple of $6$ for all $x \in \mathbb Z$ because $f(0)=0$ is a multiple of $6$.
A: You are almost done
$$x(2x^2+3x+1)=x(2x+1)(x+1)=x(2x-2+3)(x+1)
\\=2(x-1)x(x+1)+3x(x+1)$$
Now,  among any three consecutive integers one is a multiple of 3, and among two consecutive integers one is even.
A: Newton's interpolation formula
gives
$$
2x^3+3x^2+x = 6 \binom{x}{1} + 18 \binom{x}{2} + 12 \binom{x}{3}
$$
A: Put $\,p,q = 2,3\,$ in $\,pq\mid p(\color{#c00}{x^{\large q}\!-\!x}) + q(\color{#c00}{x^{\large p}\!-\!x}) + pq\,x\,$ by little $\rm\color{#c00}{Fermat}$.
A: You can easily prove by induction that$$6+6\times2^2+6\times3^2+\cdots+6\times n^2=2n^3+3n^2+1.$$It follows immediately from this fact that $6\mid2n^3+3n^2+1$.
