How to know that $2n^3+9n^2+13n+6$ factors into $(n+1)(n+2)(2n+3)$? Apologies in advance, my math is very rusty.
I'm slowly working my way through Schaum's Outline of Discrete Math for some self-study, occasionally filling in large knowledge gaps in my grasp on algebra. In one of the supplementary questions I'm asked to prove (by induction) that:
$$
\sum^n_{i=1}i^2 = \frac{n(n+1)(2n+1)}{6}
$$
In the inductive step I add $(n+1)^2$ to either side and then try to work my way through, multiplying out the factors:
$$
\frac{n(n+1)(2n+1)}{6} + (n+1)^2 \\
= \frac{n(n+1)(2n+1)+6(n+1)^2}{6} \\
= \frac{(n^2+n)(2n+1)+6(n+1)^2}{6} \\
= \frac{2n^3+n^2+2n^2+n+6(n^2+2n+1)}{6} \\
= \frac{2n^3+n^2+2n^2+n+6n^2+12n+6}{6} \\
= \frac{2n^3+9n^2+13n+6}{6}
$$
I think everything up to this point is pretty trivial, but it's around this point that I get a bit lost. I know I need to factor the terms in the 3-degree polynomial in the numerator somehow, but I'm having a hard time with the actual mechanics given the lack of obvious common factors & the fact that 13 is prime.
I'm pretty sure the final result should look something like:
$$
\frac{(n+1)((n+1)+1)(2(n+1)+1)}{6} \\
= \frac{(n+1)(n+2)(2n+3)}{6}
$$
Multiplying the expected result out I can see the two statements are equal and I'm confident that
$$
\frac{2n^3+9n^2+13n+6}{6} = \frac{(n+1)(n+2)(2n+3)}{6}
$$
but I'm just not quite sure how to factor the polynomial myself to arrive at the final result.
Any hints on how to proceed from here, or what I need to be reading up on to get my head around this?

Step-by-step simplification
$$2n^3 + 9n^2+13n+6$$ $$= 2n(n^2 + 4n + 4) + (n^2+5n+6)$$$$=2n(n+2)^2+(n+2)(n+3)$$$$=(n+2)(n+3+2n(n+2))$$$$=(n+2)(n+3+2n^2+4n)$$$$=(n+2)(2n^2+5n+3)$$$$=(n+2)(2n(n+1) + 3(n+1))$$$$ = (n+2)(n+1)(2n+3)$$
