<s>[INST] 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?
 [/INST] 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.
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