Cubic Equation

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All cubic equations are represented in the following format: $f{x}^3+g{x}^2+hx+j=0$

Dividing all of the coefficients by f renders the following form: $x^3+a{x}^2+bx+c=0$

The Integral Root Theorem, which is a special case of the Rational Root Theorem, tells us that if $a$, $b$ and $c$ are all integers and if the equation has any integral roots all of those roots will be factors of the constant term c in this manner: $p\cdot q=c$

The Integral Root Theorem actually states that both p and q have to be integers.

Let's try to solve the following cubic equation: $x^3-6x^2-\mathrm{24}x+\mathrm{64}=0$

Using the Integral Root Theorem the roots should be one or more of the following:
$\pm 1,\pm 2,\pm 4,\pm 8,\pm \mathrm{16},\pm \mathrm{32},\pm \mathrm{64}$

Let's check if we can find a root using these guesses. First we try the candidate $\pm 1$
+1: ${\mathrm{(1)}}^3-6{\mathrm{(1)}}^2-\mathrm{24}\mathrm{(1)}+\mathrm{64}=\mathrm{35}$
-1: ${\mathrm{(-1)}}^3-6{\mathrm{(-1)}}^2-\mathrm{24}\mathrm{(-1)}+\mathrm{64}=\mathrm{81}$

Then we try the candidate $\pm 2$
+2: ${\mathrm{(2)}}^3-6{\mathrm{(2)}}^2-\mathrm{24}\mathrm{(2)}+\mathrm{64}=0$

We've found our first root $x_1=2$ . We need not go any further. Now we continue utilizing Polynomial Long Division.

This is how to divide the polynomial $(x^3-6x^2-\mathrm{24}x+\mathrm{64})$ by the polynomial $(x-2)$

                  x² - 4x - 32
x - 2 x³ - 6x² - 24x - 64
      -(x³ - 2x²)
               -4x² - 24x +64
            -(-4x² + 8x)
                       -32x + 64
                     -(-32x + 64)
                               0

In case you don't understand: Check this out.

This means that $x^3-6x^2-\mathrm{24}x+\mathrm{64}=(x-2)(x^2-4x-\mathrm{32})$

Now we can use my guide for Quadratic Equations provided in this page in order to get the other two roots. The roots of:

$$x^2-4x-\mathrm{32}=0$$

Using the following formula $$x=-\frac{p}{2}\pm \sqrt{\frac{p^2}{4}-q}$$

We get $$x=-\frac{\mathrm{(-4)}}{2}\pm \sqrt{\frac{{\mathrm{(-4)}}^2}{4}+\mathrm{32}}$$

$$x=2\pm 6$$

So $$x_2=8$$ and $$x_3=\mathrm{-4}$$ The first root we obtained earlier. It was $$x_1=2$$