$f(n) = 3n^3 - 39n^2 + 360n + 20$. We need constants $c, n_0 > 0$ such that $f(n) \geq$ c$n^3$ $\,$ $\forall n\geq n_0$. Fix $c=2.25$. The smallest integer value that works can be found using calculus to find the range where the functions have $0$'s. The factors are equal to (n - 39.9762)(n - 12.0791)(n + 0.055248). Upon further inspection that for $-0.055248 \leq n \leq 12.0791$ and $n\geq 39.9762$ we have that $f(n) \geq 2.25n^3$. In order to satisfy $\forall n\geq n_0$ $f(n) \geq 2.25n^3$ we can't take $n_0$ in the range -0.55248 to 12.0791 since, for instance $n=13$ (greater than any of those numbers) would not satisfy the inequality. Thus, we can set $n_0 = 40$, the smallest integer value that satisfies are constraints.
My question is how to get the factors equal to (n - 39.9762)(n - 12.0791)(n + 0.055248)? Also, how would I use calculus to find the range where the functions have $0$'s and the smallest integer value that works? What about the roots?