Proof using concept of polynomials. This question is from Challenge and Thrill of Pre-College Mathematics.
Prove that $2n^3-3n^2+n$ is divisible by 6 for any natural number $n$.
I know there are many proofs of this using induction and many other ways. Is there any proof using concept of polynomials particularly using remainder theorem and factorization of polynomials? These concepts were introduced just before the exercise and the whole exercise was based on divisibility of a polynomial by another polynomial. Any hint/suggestions are really appreciated. Thank you! 
I tried to prove that $2n^3-3n^2+n$ is divisible by $2b_1(n)$ and $3b_2(n)$ where $b_1(n)$ and $b_2(n)$ are polynomials over $\mathbb Z$ but could not found such polynomials.
 A: \begin{align}2n^3-3n^2+n &= n(2n^2-3n+1)\\&=n(2n-1)(n-1)\\
&= (n-1)n(n+1+n-2)\\
&=(n-1)n(n+1)+(n-2)(n-1)n \end{align}
Try to interpret the meaning of the last expression.
A: The polynomial factors to $$2n^3-3n^2+n=(n-1)n(2n-1)$$
One of the numbers $n-1$ and $n$ must be even and modulo $3$, the factors are $n+2$ , $n$ , $2n+2=2(n+1)$ , so one of them must be divisible by $3$.
A: Others have already given a factorization as $(n-1)n(2n-1)$
Now look at what $n$ is modulo 6, then calculate what $(n-1) $ and $2n-1$ are modulo 6 and multiply all three.
n=0 mod 6. Clearly true
n=1 mod 6, then $n-1$ is a multiple of 6 and hence the product is.
n= 2mod 6 then the three numbers are 1,2, and 3 mod 6 and product is 0 mod6.
n=3mod 6, then n-1 is 2 mod6 hence  n(n-1) is multiple of 6
n=4mod 6. Then n-1 is 3 mod 6 and so n(n-1) is also a multipe of 6.
n=5mod 6 Then n-1 is 4mod6 and (2n-1) is 3 mod 6, hence the product is 0 mod6
n=3 mod 6 then three numbers are 2,3 and some x, product is 0 mod6
ALTERNATIVE: (a corollary of the formula for sum of squares)
Prove the formula $1^2+2^2+\cdots +n^2= n(n+1)(2n+1)/6$
S0 the sum of first $n-1$ squares is  $(n-1)n(2n-1)/6$, as this quantity is an integer (sum of integers!) the numerator should be a multiple of 6, the denominator.
