Elementary Algebra

$n(n+5)(n+8)$
Each term has $n$ in it, so factor out $n$ to obtain: $=n(n^2+13n+40)$ The leading coefficient of the trinomial is 1. This means that it can be factored by looking for factors of the constant term (40) whose sum is equal to the numerical coefficient of the middle term (13). Note that $40=(5)(8)$ and that $5 + 8 = 13$, the numerical coefficient of the middle term. The factors of $40$ whose sum is $13$ are $5$ and $8$. Thus, the factored form of the trinomial is: $=(n+5)(n+8)$ Therefore $n(n^2+13n+40)$ can be factored as: $=n(n+5)(n+8)$