Cubic function

Cubic equations are polynomial equations of the third degree, ie algebraic equations of the form

A cubic equation has according to the fundamental theorem of algebra always three complex solutions, which may also coincide. With their help, the equation can be represented in factored form:

In the case of real coefficients, the left side of the cubic equation geometrically describes a cubic parabola in - plane, which is the graph of a cubic function. The zeros, so its intersection with the axis, the real solutions of the cubic equation. The function graph has according to the intermediate value theorem is always at least one real zero, but no more than three.

Solutions

Rates of a solution

If you know a solution exactly, so you can the cubic polynomial using the polynomial or the Horner scheme by dividing and thus obtains a quadratic equation. This can be solved with the help of a solution formula, and so receives the remaining solutions of the cubic equation. This method is practical only for a rational solution. As early as the irreducible equation, the method is no longer feasible with the relatively simple solution, since the coefficients of the remaining quadratic equation are very complicated. In these cases, the solutions with the below mentioned Cardano formula to determine easier.

If all the coefficients of the cubic equation integer, so you can try to guess a rational solution, that is, to find by trial and error. Is the leading coefficient A from the amount equal to 1, so you can see the whole number factor of the last coefficient D by taste (negative values ​​). If A is different from one, then all fractions whose numerator, be tried a divisor of D and whose denominator is a divisor of A. The theorem on rational zeros guarantees that you will find with this effort finite rational zero point, if one exists. The coefficients rationally, one can achieve integral coefficients by multiplying the equation by the denominator of all the coefficients.

Example

A rational solution of the cubic equation

Get only the integer divisor of the last coefficient in question as well. In fact, a solution of which is satisfied by inserting. polynomial provides

And with the quadratic solution formula result as other solutions.

Reducing the equation to a normal shape

There are a number of equivalent transformations of the cubic equation by linear transformation of the argument, that allow them to simplify the following dissolution method ( Tschirnhaus transformation). By dividing by the polynomial can first be normalized.

Through linear transformation of the argument using the substitution results in the following expression:

By choosing can eliminate the quadratic term and we obtain the reduced form of the cubic equation:

With the reduced form can now be resolved by means of the formulas Cardano and subsequent back-substitution, the solutions of the initial equation can be determined. In this way the whole of the real and complex solutions is available.

Analytical determination of the real solutions

By calculation of the discriminant of the reduced form can be first checked whether only real solutions are present:

Is so all solutions are real. This automatically includes that the original polynomial even had real coefficients.

Solution of the special case p = 0

Case 1: and

Case 2: and

Solution by cases

A suitable solution strategy for the remaining solutions, which does not require the use of complex numbers, is the following: The reduced form is formed by substituting with an appropriate trigonometric or hyperbolic function so that it can be attributed to the well-known addition formulas. Here, the respective domain is observed. This approach will initially be shown with an example:

It is the well-known addition theorem

By substitution can be the reduced form of the cubic equation in a similar term transfer:

By comparing coefficients, one obtains directly

From this it is now the term for the original coefficients and represent:

The final solution of the cubic equation can then be determined by back-substitution. Off and obtained thus

The choice of the substitution function, however, still the value range must be considered. To find all solutions, the range of values ​​must therefore be covered by three different functions, so that a total of three additional cases arise that must be considered:

Case 3: substitution and

Case 4: substitution and

Case 5: substitution and