Is Rufinni's rule the quickest hand-method to find roots in high order polynomials? I'm wondering if there's another method where I do not have to "trial-error" with every guess neither using numerical methods. When the searched root is 3*π/5 with Ruffini's rule I cannot find it quickly, and numerical methods are a dead end because I need the analytic response.
 A: One way to solve cubic equations is to reduce it and use Vietes formula. 
Here is an example from Kent on how to do this
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=296&t=422006&


Kent Merryfield: 
Some mathematical history, from the 16th century. This was the century in which cubic equations were solved. First on the scene was the method we know as the Cardano-Tartaglia method. This expresses the solution of the cubic as $u-v$, where $u^3$ and $v^3$ are solutions of a quadratic equation. So if you can take cube roots, you can use this method.
Except there's a problem. Real cubic equations partition naturally into two main cases (plus a borderline case that lies in between). Either the equation has one (non-repeated) real root, or it has three real roots. (The other possibilities belong to the borderline case.) If the cubic equation has one real root, then the Cardano-Tartaglia method "works" and gives an unambiguous answer. I put the word "works" in quotation marks, since the arithmetic is normally horrendous, and even rational roots tend to be expressed in absurdly complicated ways involving nested roots.
But in the other cases, the case of three distinct real roots, the Cardano-Tartaglia method doesn't really work. The quadratic equation that $u^3$ and $v^3$ are supposed to be roots of has no real roots at all. Now one of the ways to deal with this is to try to force $u^3$ and $v^3$ have meaning anyway - and that is precisely the route by which complex numbers entered mathematics. (It took cubic equations to require complex numbers, not quadratic equations.)
But there's another way, and for this, we can thank another 16th century mathematician, François Viète (name often Latinized to Vieta). Yes, the same guy with the formulas for the coefficients in terms of the roots. His method applies only to those cases in which the cubic has three distinct real roots - that is, precisely the cases that trouble the Cardano-Tartaglia method - and it's intended as a more-or-less numerical method for which one needs a good table of trigonometric functions.
The key to Viète's method lies in the trigonometric identity $\cos 3\theta=4\cos^3\theta-3\cos\theta.$
Step 1: In the original equation, replace $x$ by $y+c$ for the appropriate constant $c$ so that we can get rid of the $x^2$ term and write the equation as $y^3-py=q.$ (This is the same as the first step for Cardano-Tartaglia as well.)

If we assume that $p(x)=x^3+Ax^2+Bx+C$ then we have $c = -A/3$.

Step 2: Substitute $y=z\sqrt{4p/3\,}$ and then simplify so that the equation reads $4z^3-3z=w.$
Step 3: Let $z=\cos\theta.$ This gives us $\cos 3\theta=w,$ so $3\theta=\arccos(w).$
Step 4: Divide by 3, but acknowledge that there are three possibilities: $\theta=\frac13\arccos(w)+\frac{2\pi j}3,$ for $j=0,1,2.$
Step 5: Backtrack: compute $z=\cos\theta$ from these three possibilities, then $y$ from $z$ and $x$ from $y.$

How does this work for the equation at hand? Start with $x^3-15x^2+125=0.$
Let $x=y+5.$ Substitute, collect terms, and simplify. We get $y^3-75y=125.$
Let $y=10z.$ This gives us:
  $$\begin{align*}1000z^3-750z & =125 \\
4z^3-3z & =\frac12 \end{align*}$$
Let $z=\cos\theta.$ We get:
  $$ \begin{align*}
\cos3\theta & =\frac12 \\
    3\theta & = \frac{\pi}3 
\end{align*} $$
So $\displaystyle \theta=\frac{\pi}9$ or $\displaystyle \theta=\frac{7\pi}9$ or $\theta=\cfrac{13\pi}9.$
$z=\cos\cfrac{\pi}9$ or $z=\cos\cfrac{7\pi}9$ or $z=\cos\cfrac{13\pi}9=\cos\frac{5\pi}9.$
Finally, $x=10\cos\cfrac{\pi}9+5$ or $x=10\cos\cfrac{7\pi}9+5$ or $x=10\cos\cfrac{5\pi}9+5$ 
I haven't checked systematically to see if this matches the other forms of the answer given in the posts above, but it should. The world is full of trig identities.

A note: in general, one cannot construct the solutions to cubic equations using compass and straightedge (Plato/Euclid rules). But Viète's solution suggest that if we could only trisect angles, we could construct the solutions of those cubics with three real roots - and if we allow ourselves the use of the Archimedes marked straightedge, then we can trisect angles.
The Archimedes marked straightedge can also be used to compute cube roots and therefore construct the solutions to cubics with one real root by following Cardano-Tartaglia - but that's a whole different subject.

Similar techniques exists for both 4th and 5th order polynomials. See Is there a general formula for solving 4th degree equations (quartic)? for an example of the former. 
