TITLE: Is there a systematic way of solving cubic equations?
QUESTION [7 upvotes]: In my text book, to solve cubic equations, I need to 

find by trial & error what $f(a)$ will make the equation 0. The factor will be $(x-a)$
then the other factor will be $Ax^2+Bx+C$ then I can solve using compare coefficient

But for te first part, is there any other way apart from trail & error? Or is there some technique to guess the factor of an equation? 
The question I am currently doing happens to be $2x^3+3x+4=9$ 
UPDATE:  $2x^3+3x+4=9$ should be $2x^3+3x+4=9x^2$ 
UPDATE 2
So I am going to try substituting 
$x=\pm{1}, \pm\frac{1}{2}, \pm{2}, \pm{4}$
When $x=-\frac{1}{2}$, one of the provided answers, 
$2(−0.5)^3−9(−0.5)^2+3(−0.5)+4=-18.75\neq 0$ 
did I make a mistake?

REPLY [3 votes]: Yes there is, but it won't be much use in an exam:
Given the cubic equation:


For the general cubic equation (1) with real coefficients, the general
  formula for the roots, in terms of the coefficients, is as follows if
  $(2 b^3-9 a b c+27 a^2 d)^2-4 (b^2-3 a c)^3=-27 a^2 \Delta>0$, i.e. if there are
  two non real roots:

However, this formula is wrong if the operand of the square root is
  negative or if the coefficients belong to a field which is not
  contained in the field of the real numbers: When this operand is real
  and positive, the cubic roots are real and well defined. In the other
  case, the square root is not real and one has to choose, once for all
  a determination for it, for example the one with positive imaginary
  part. For extracting the cubic roots we have also to choose a
  determination for the cubic roots, and this gives nine possible values
  for the first root of an equation which has only three roots.
A correct solution may be obtained by remarking that the proof of
  above formula shows that the product of the two cubic roots is
  rational. This gives the following formula in which $\sqrt{ }$ or
  $\sqrt[3]{ }$ stand for any determination of the square or cubic root,
  if $b^2-3ac \mbox{ } \neq \mbox{ } 0$.

If $Q \mbox{ } \neq \mbox{ } 0$ and $b^2 − 3ac = 0$, the sign of Q has to
  be chosen for having $C \mbox{ } \neq \mbox{ } 0$.
If $Q = 0$ and $b^2 − 3ac = 0$, the three roots are equal:
$$x_1=x_2=x_3=-\frac{b}{3a}$$.
If $Q = 0$ and $b^2-3ac \mbox{ } \neq \mbox{ } 0$, above expression for
  the roots is correct but misleading, hiding the fact that no radical
  is needed to represent the roots. In fact, in this case, there is a
  double root
$$x_1=x_2=\frac{bc-9ad}{2(3ac-b^2)}$$.
and a simple root
$$x_3=\frac{9a^2d-4abc+b^3}{a(3ac-b^2)}$$.
(continued)
source: Wikipedia