Methods for Factoring Cubics I am looking for some advice and tips/help about something. I am in calculus now and have been doing well but I recently realized to a bit of my own embarrassment that I am still not fully comfortable factoring cubics. I can do it usually, it just takes me a while and consists of mostly guess and checking, and not a certain process.
Ill give an example,
suppose I want to factor to solve for the roots $$-x^3+9x^2-15x+2=0$$
Now; I know there are a few general methods.
I know one way is to  factor by grouping, but this cannot be done in the example here. I know if possible you can also factor in such a manner that you have one root and can use the quadratic formula on the other.
I think I am looking for other possibilities, mainly a nice explanation of what I believe is called the "Rational Root Theorem".
It involves something along the lines of first looking at the constant term and finding at least one root by plugging in factors of that number, then using this root and the coefficient of the highest x, you use synthetic division to find the other roots. 
Anyways, I am basically looking for any explanations on methods, and if someone could show me how to solve this using synthetic division that would be great.
Update: And I am also wondering for the rational root test, if the coefficient on the $x^3$ is not a 1 or -1, would not the amount of possible factors potentially be very large? how do you deal with this?
Update 2: I solved the above question as follows; I noted x=2 was a factor and did synthetic division as 
x+2 | -1 9 -15 2 

    | -1 7 -1 0 

then did the quadratic on $-x^2+7x-1=0$
Thank you all,
 A: In a first step set $$x = y + 3$$
and you will get something like this:
$${y^3} - 12y - 11 = 0$$
Then set:
$$\begin{gathered}
  y = u + v \hfill \\
  u \cdot v = 4 \hfill \\ 
\end{gathered}$$
should be:
$${u^3} + {v^3} - 11 + (3 \cdot u \cdot v - 12) \cdot (u + v) = 0$$
Now follows:
$$\begin{gathered}
  {u^3} + {v^3} = 11 \hfill \\
  {u^3} \cdot {v^3} = {4^3} \hfill \\
  {u^3}{u^3} + {u^3}{v^3} = 11{u^3} \hfill \\
  {u^6} - 11{u^3} + 64 = 0 \hfill \\
  {\left( {{u^3} - \frac{{11}}{2}} \right)^2} - {\left( {\frac{{\sqrt {135} }}{2}i} \right)^2} = 0 \hfill \\
  \left( {{u^3} - \frac{1}{2}\left( {11 + \sqrt {135}i } \right)} \right) \cdot \left( {{u^3} - \frac{1}{2}\left( {11 - \sqrt {135}i } \right)} \right) = 0 \hfill \\
  {u^3} = \frac{1}{2}\left( {11 + \sqrt {135}i } \right) \hfill \\
  {u^3} \cdot {v^3} = 64 \hfill \\
  \frac{1}{2}\left( {11 + \sqrt {135}i } \right){v^3} = {u^3}{v^3} = 64 \hfill \\
  {v^3} =  - \frac{{128}}{{11 + \sqrt {135}i }} = \frac{1}{2}\left( {11 - \sqrt {135}i } \right) \hfill \\ 
\end{gathered}$$
Right combinations of $u$ and $v$ and re-substitutions will give factorization.
In this case over field $\mathbb{C}$
Key I used: Cardano-Formula.
A: You should look up the Rational Root Theorem. Basically it says that any rational root of a polynomial with integer coefficients is of the form $\frac{a}{b}$ (in lowest terms) where $a$ divides the constant term and $b$ divides the coefficient of the highest-order term.
It's also sometimes possible to find small integer roots just "by inspection". For example, $x=1$ and $x=-1$ are easy to test just by looking at the coefficients (though neither is a root of the cubic you gave).
