<s>[INST] Solving this 3-degree polynomial I'm trying to factor the following polynomial by hand:
$-x^3 + 9x^2 - 24x + 20 = 0$
The simplest I could get is:
$-x^2(x-9) - 4(5x+5) = 0$
Any ideas on how I could go ahead and solve this by hand? This seems pretty tough.
 [/INST] You have not a polynomial, rather an equation, whose LHS is the cubic polynomial $$-x^{3}+9x^{2}-24x+20.\tag{0}$$
To get rid of the negative coefficient of $x^3$ multiply $(0)$ by $-1$
$$
\begin{equation*}
-\left( -x^{3}+9x^{2}-24x+20\right) =x^{3}-9x^{2}+24x-20.\tag{1}
\end{equation*}
$$
By trial and error$^1$, inspection, or per DonAntonio's hint (Rational root test), we find that $x=5$ is a root: $5^{3}-9\left( 5\right)
^{2}+24\left( 5\right) -20=0$. As a consequence we can divide $(1)$ by $x-5$, using polynomial long division or Ruffini's rule to obtain a quadratic polynomial, since the remainder must be zero
$$
\begin{equation*}
\frac{x^{3}-9x^{2}+24x-20}{x-5}= x^{2}-4x+4 =\left( x-2\right)\tag{2}
^{2}.
\end{equation*}
$$
So
$$
\begin{equation*}
-x^{3}+9x^{2}-24x+20=-\left( x-5\right) \left( x-2\right) ^{2},\tag{3}
\end{equation*}
$$
whose roots are $x=5$ (single root) and $x=2$ (double root).
If you start with the root $x=2$, you get the same result:
$$\frac{x^{3}-9x^{2}+24x-20}{x-2}=x^{2}-7x+10=\left( x-2\right) \left(
x-5\right).\tag{2'} $$
$^1$ Trial and error is a fundamental method of solving problems.
</s>