Our objective is to find a real root of the cubic equation

The other two roots (real or complex) can then be found by polynomial division and the quadratic formula. The solution proceeds in two steps. First, the cubic equation is "depressed"; then one solves the depressed cubic.

to the cubic equation, to obtain:

Multiplying out and simplifying, we obtain the "depressed" cubic

Let's try this for the example

2*x*^{3}-30*x*^{2}+162*x*-350=0.

Our substitution will be

How to do this had been discovered earlier by Scipione dal Ferro (1465-1526).

We will find *s* and *t* so that

3st |
= | A |
(1) |

s^{3}-t^{3} |
= | B. |
(2) |

It turns out that

(*s*-*t*)^{3}+3*st* (*s*-*t*)=*s*^{3}-*t*^{3}.

This is true since we can simplify the left side by using the binomial formula to:

(*s*^{3}-3*s*^{2}*t*+3*st*^{2}-*t*^{3})+(3*s*^{2}*t*-3*st*^{2})=*s*^{3}-*t*^{3}.

How can we find *s* and *t* satisfying (1) and (2)? Solving the first equation for *s* and substituting into (2) yields:

Simplifying, this turns into the "tri-quadratic" equation

which using the substitution

From this, we can find a value for

Let's do the computation for our example

We need

3st |
= | 6 | (3) |

s^{3}-t^{3} |
= | 20. | (4) |

Solving for

which multiplied by -

Using the quadratic formula, we obtain that

We will discard the negative root, then take the cube root to obtain

By Equation (4),

Our solution

The solution to our original cubic equation

2*x*^{3}-30*x*^{2}+162*x*-350=0

is given by

Shortly after the discovery of a method to solve the cubic equation, Lodovico Ferraria (1522-1565), a student of Cardano, found a similar method to solve the quartic equation.

This section is loosely based on a chapter in the book *Journey Through Genius*
by William Dunham.

Then find the other two roots. Which of the roots equals our solution

into a depressed cubic.

(This is for practice purposes only; to make the computations a little less messy, the root will turn out to be an integer, so one could use the Rational Zero test instead.)

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