How to know that $2n^3+9n^2+13n+6$ factors into $(n+1)(n+2)(2n+3)$?

The Rational Root Theorem tells you that any rational root of the polynomial $2 n^3 + 9 n^2 + 13 n + 6$ has the form $\frac{p}{q}$ for integers $q \mid 2$, $p \mid 6$, leaving 12 possible rational roots. Since all of the coefficients are positive, all of the (real) roots are negative, leaving only 6 possible rational roots, $-6, -3, -2, -\frac{3}{2}, -1, -\frac{1}{2}$.

Substituting, for example, $-1$ (because it's fast to evaluate a polynomial at $-1$---just take the alternating sum of the coefficients) gives $-2 + 9 - 13 + 6 = 0$, so $n = -1$ is a root of the polynomial, and equivalently the polynomial has factor $n + 1$. Dividing the polynomial by $n + 1$ and factoring gives $2 n^2 + 7 n + 6 = (n + 2)(2 n + 3)$, and putting this all together gives the desired factorization: $$\boxed{2 n^3 + 9 n^2 + 13 n + 6 = (n + 1) (n + 2) (2 n + 3)}.$$

Remark As you probably know, not all polynomials with integer coefficients factor into products of linear terms with integer coefficients as they did in this case. For example, the general formula for $\sum i^4$ analogous to the one in this question contains a factor of $3 n^2 + 3 n - 1$, which is not factorable over the integers.

Recall (in your question) the line ... $$=\frac{(n^2+n)(2n+1)+6(n+1)^2}{6}$$

What followed, however was a missed opportunity, for $n^2+n$ factors as $n(n+1)$, so your expression becomes $$\begin{align} \frac{(n^2+n)(2n+1)+6(n+1)^2}{6} &= \frac{n(n+1)(2n+1)+6(n+1)^2}{6}\\ &= \frac{(n+1)\bigg(n(2n+1)+6(n+1)\bigg)}{6}\\ &= \frac{(n+1)(2n^2+n+6n+6)}{6}\\ &= \frac{(n+1)(2n^2+7n+6)}{6}\\ \end{align} $$ and all that remains to do is to factor the quadratic $2n^2+7n+6$, and you're done. :-)

Find a single root of the polynomial through trial and error and then apply long division. Dividing $(2n^3+9n^2+13n+6)$ by $(n+1)$ yields $(n+2)(2n+3)=2n^2+7n+6$. A root of $-1$ yields a factor of $(n+1)$.