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#1 2021-12-15 21:53:37

Registered: 2005-06-28
Posts: 35,550

Cubic equations

In algebra, a cubic equation in one variable is an equation of the form

in which a is nonzero.

The solutions of this equation are called roots of the cubic function defined by the left-hand side of the equation. If all of the coefficients a, b, c, and d of the cubic equation are real numbers, then it has at least one real root (this is true for all odd-degree polynomial functions). All of the roots of the cubic equation can be found by the following means:

* algebraically, that is, they can be expressed by a cubic formula involving the four coefficients, the four basic arithmetic operations and nth roots (radicals). (This is also true of quadratic (second-degree) and quartic (fourth-degree) equations, but not of higher-degree equations, by the Abel–Ruffini theorem.)
* trigonometrically
* numerical approximations of the roots can be found using root-finding algorithms such as Newton's method.

The coefficients do not need to be real numbers. Much of what is covered below is valid for coefficients in any field with characteristic other than 2 and 3. The solutions of the cubic equation do not necessarily belong to the same field as the coefficients. For example, some cubic equations with rational coefficients have roots that are irrational (and even non-real) complex numbers.

If the coefficients of a cubic equation are rational numbers, one can obtain an equivalent equation with integer coefficients, by multiplying all coefficients by a common multiple of their denominators. Such an equation

with integer coefficients, is said to be reducible if the polynomial of the left-hand side is the product of polynomials of lower degrees. By Gauss's lemma, if the equation is reducible, one can suppose that the factors have integer coefficients.

Finding the roots of a reducible cubic equation is easier than solving the general case. In fact, if the equation is reducible, one of the factors must have degree one, and thus have the form

with q and p being coprime integers. The rational root test allows finding q and p by examining a finite number of cases (because q must be a divisor of a, and p must be a divisor of d).

Thus, one root is

and the other roots are the roots of the other factor, which can be found by polynomial long division. This other factor is

(The coefficients seem not to be integers, but must be integers if p / q is a root.)

Then, the other roots are the roots of this quadratic polynomial and can be found by using the quadratic formula.

It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.


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