Polynomial long division
Cubic equations are 3rd degree polynomial equations and have the form:
In order to solve cubic equations and equations of higher degree one needs the Polynomial long division.
First, a zero is guessed by trial and error and then, using polynomial long division, one can transform the equation into a quadratic equation.
Method
 Guess a zero $x_N$
 Polynomial long division: Divide equation by $(xx_N)$
 Solve the quadratic equation
Hint
Example
Solve cubic equation: $x^319x30=0$

Guess a zero
The first zero must be found by trial and error.
Use different values for $x$ until 0 results.
$x^319x30=0$
$x=1$:
$1^319\cdot130=48$ $\neq0$ =>no zero
$x=1$:
$(1)^319\cdot(1)30=12$ $\neq0$ =>no zero
$x=2$:
$(2)^319\cdot(2)30=0$ =>zero at $x_{1}=2$ 
Polynomial long division
The function is divided by $(xx_1)$. Use polynomial long division for this.
$(x^319x30):(x+2)$First calculate $x^3:x$ and write down the result.
Now $x^2$ is multiplied by $(x+2)$. The result is written in the second line and receives a minus.
Both lines are now added together with the remainder written underneath.
Similarly as before, one now calculates $2x^2:x$. Write the result to the right and multiply it by $(x+2)$ to put it in the line below.
The two lines are subtracted again.
Finally, $15x:x$ is calculated, multiplied back and deducted. The rest is 0 and the polynomial division is done.

Solve quadratic equation
The new quadratic equation can now be solved e. g. with the pqformula.
$x^22x15=0$
$x_{2,3} = \frac{p}{2} \pm\sqrt{(\frac{p}{2})^2q}$
$x_{2,3} = 1 \pm\sqrt{1^2+15}$
$x_{2,3} = 1 \pm\sqrt{16}$
$x_{2,3} = 1 \pm4$
$x_2=5$ and $x_3=3$
The solutions of the initial equation are $x_1=2$, $x_2=5$ and $x_3=3$