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.

Example

Solve cubic equation: $x^3-19x-30=0$

1. Guess a zero

   The first zero must be found by trial and error.
   Use different values for $x$ until 0 results.
   $x^3-19x-30=0$
   
   $x=1$:
   $1^3-19\cdot1-30=-48$ $\neq0$ =>no zero
   
   $x=-1$:
   $(-1)^3-19\cdot(-1)-30=-12$ $\neq0$ =>no zero
   
   $x=-2$:
   $(-2)^3-19\cdot(-2)-30=-0$ =>zero at $x_{1}=-2$

2. Polynomial long division

   The function is divided by $(x-x_1)$. Use polynomial long division for this.
   
   $(x^3-19x-30):(x+2)$
   
   

   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.

   
   

3. Solve quadratic equation

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