Factoring $3n^3 - 39n^2 + 360n + 20$ I am wondering how to factor 
$$f(n) = 3n^3 - 39n^2 + 360n + 20$$ the right way. I think the factors are equal to 
$$(n - 39.9762)(n - 12.0791)(n + 0.055248)$$
 A: There is only one real root which is exactly
$$n_1=\frac{13}{3}-\frac{2}{3} \sqrt{191} \sinh \left(\frac{1}{3} \sinh
   ^{-1}\left(\frac{4913}{191 \sqrt{191}}\right)\right)\approx -0.055223771734378147887$$
So, numerically
$$f(n) = 3n^3 - 39n^2 + 360n + 20$$ $$f(n)=3(n-n_1)(n^2-13.055223771734378148 n+120.72095869751148663)$$
A: Set
$$
P(x)=3x^3-39x^2+360 x+20. 
$$
Then set $x=y+\frac{13}{3}$, then
$$
\frac{1}{3}P(x)=P_1(y):=y^3+\frac{191}{3}y+\frac{9826}{27}
$$
If $\rho_0=\sqrt[3]{A}+\sqrt[3]{B}$ is the real root of $P_1(y)=0$, then 
$$
\rho_0^3=A+B+3(AB)^{1/3}(\sqrt[3]{A}+\sqrt[3]{B})=s+3p^{1/3}\rho_0.
$$
Hence $s=A+B=-\frac{9826}{27}$, $p=AB=-\left(\frac{191}{9}\right)^3$ and the equation
$$
X^2+\frac{9826}{27}X-\left(\frac{191}{9}\right)^3=0,
$$
have roots
$$
A=\frac{1}{27}(-4913-12\sqrt{216010})\textrm{ , }B=\frac{1}{27}(-4913+12\sqrt{216010}).
$$
Hence we find 
$$
\rho_0=\sqrt[3]{\frac{1}{27}(-4913+12\sqrt{216010})}-\sqrt[3]{\frac{1}{27}(4913+12\sqrt{216010})}.
$$
Now $P_1(y)$ have the other two roots such (Vieta)
$$
\rho_1+\rho_2=-\rho_0\textrm{ and }\rho_1\rho_2=\frac{9826}{27}\rho_0^{-1}
$$
Solving
$$
X^2+\rho_0X+\frac{9826}{27}\rho_0^{-1}=0
$$
we get the other two roots of $P_1(y)=0$. 
By this way every third degree polynomial equation reduced solving only two degree equations.
NOTE. We have used $\sqrt[3]{-|a|^3}=-|a|$, since the equation $x^3+|a|^3=0$, have solution $x=-|a|$. 
