# Cubic Equation

Coefficients may be real or complex numbers.

Third-degree equations, also known as cubic equations, are polynomial equations where the highest degree is three. They have a general form `ax^3 + bx^2 + cx + d = 0`, where `a`, `b`, `c`, and `d` are constants, with `a` being non-zero. These equations play a crucial role in various scientific and engineering fields.

## How to use this calculator ?

This calculator is a cubic equation (ax^{3}+bx

^{2}+cx+d=0) solver.

It calculates the exact solutions when they exist and also gives a numerical approximation of them.

### Coefficients input

Here are some hints to help you enter the coefficients of the equation.- Accepted inputs are,
- integers, example: 5, -7
- fractions, example: 1/3 or -2/9
- decimal values, example: 3.9 or -9.65
- constants, example: pi or e
- common functions, for example: sin(pi/5)
- square root operator, example : input sqrt(3) or 3^0.5 for `sqrt(3)`
- complex numbers, example : 1+i ou -i

- To enter a product of two factors, use the * operator. For example: enter 2*pi and not 2pi.

## Theoretical Foundations

A third-degree equation can be presented in different forms, but its most common representation is `ax^3 + bx^2 + cx + d = 0`. Historically, the solutions to these equations have been explored since antiquity, leading to significant developments in algebra.

## General Resolution of Third-Degree Equations

The Cardano method is a classical approach for solving third-degree equations. Consider a general equation `ax^3 + bx^2 + cx + d = 0`. The first step involves eliminating the quadratic term through a variable substitution. We set `x = y - b/(3a)`, which transforms the original equation into a new form `y^3 + py + q = 0`, where `p` and `q` are expressions based on `a`, `b`, `c`, and `d`. Next, we introduce two new variables, `u` and `v`, such that `y = u + v`. Substituting `y` in the transformed equation, we get an equation in `u` and `v` that leads to a system of equations. This system can be solved to find `u` and `v`, often using the method of substitution or methods for solving systems of non-linear equations. Finally, the roots of the original cubic equation are found by expressing `x` in terms of `u` and `v` and reversing the initial substitution. This method, although complex in its calculations, is very effective for finding the exact roots of third-degree equations.

## Example of Resolution with Cardano's Method

Consider the equation `x^3 - 3x^2 - 4x + 12 = 0`. This equation was specifically chosen because it has integer solutions that can be found by simple methods like trial and error. The objective here is to unfold Cardano's method to solve this equation and verify that the obtained solutions correspond to the integer roots that can be guessed. This demonstrates the validity of Cardano's method and helps understand its application, even for equations that seem to have simple solutions at first glance.

To apply Cardano's method, we start by eliminating the quadratic term. We set `x = y + 1` (since `-b/(3a) = 1`). Substituting in the original equation, we obtain `y^3 - 7y + 6 = 0`.

Next, we introduce `u` and `v` such that `y = u + v`, transforming the equation into `u^3 + v^3 + (3uv - 7)(u + v) + 6 = 0`. By choosing `3uv - 7 = 0`, we find:

`u^3 + v^3 + 6 = 0` and `uv = 7/3`. These two equations form a system that allows us to find `u` and `v`.

By replacing `v` with `7/(3u)` in the first equation, we obtain `u^3 + (7/(3u))^3 + 6 = 0`. Multiplying the entire equation by `u^3`, we get a polynomial equation in `u^3`: `u^6 + 6u^3 + (7/3)^3 = 0`. This equation can be solved as a quadratic equation by setting `z = u^3`, giving `z^2 + 6z + (7/3)^3 = 0`.

The solutions of this equation are:

`z_1=-10/9*i*sqrt(3) - 3`

`z_2=10/9*i*sqrt(3) - 3`

Use this equation calculator to verify the calculations.

We find the values of `u` using `z=u^3`. This involves extracting the cubic roots of `z_1` and `z_2`, giving us the possible values for `u` with the nth root extractor of z1 and the nth root extractor of z2

Then, `v` can be found using the relation `uv = 7/3`. Finally, the values of `x` are obtained by replacing `y` with `u + v` in `x = y + 1`.

Using the following calculator expression calculator, with a bit of patience, we arrive at the solutions 2, 3, and -2, which can be verified here.

## Resolving a Third-Degree Equation Knowing a Root

When a root `r` of a cubic equation `ax^3 + bx^2 + cx + d = 0` is known, several methods can be used to simplify the equation.

### Method of Euclidean Division

The Euclidean division allows dividing the cubic equation by `(x - r)`. This operation reduces the equation to a quadratic form `ex^2 + fx + g = 0`. This new equation can then be solved using standard methods for second-degree equations.

This calculator can be used to perform the division of two polynomials: Euclidean division of polynomials.

### Method of Factorization

Another approach is to factorize the cubic equation in the form `(x - r)(ax^2 + bx + c)`. By expanding this expression, we find the original equation, which allows us to deduce the values of `a`, `b`, and `c`. This factorized form makes it easier to solve the remaining second-degree equation.

This calculator can be useful in this case: factorize a polynomial.

**Example 1:** Take the equation `x^3 - 3x^2 - 4x + 12 = 0` and assume that we know `x = 3` is a root. We can factorize the equation as `(x - 3)(ax^2 + bx + c) = 0`. By expanding this expression, we get `ax^3 + (b - 3a)x^2 + (c - 3b)x - 3c = 0`. By comparing with the original equation, we can establish a system of equations to find `a`, `b`, and `c`. In this case, we find that `a = 1`, `b = 0`, and `c = 4`, so the factorized equation becomes `(x - 3)(x^2 - 4) = 0`.

### Forced Factorization

**Example:** Consider again the equation `x^3 - 3x^2 - 4x + 12 = 0` and assume we know a root, `x = 2`. To apply forced factorization, we start by rearranging the equation to make `(x - 2)` appear. We can rewrite the equation as:

`x^3 - 3x^2 - 4x + 12 = 0`

we add and then subtract `2*x^2` to "force" the factor `x-2`

`x^2*(x-2) + 2*x^2 - 3x^2 - 4x + 12 = 0`

`x^2*(x-2) - x^2 - 4x + 12 = 0`

in the same way we add and then subtract `2*x` to "force" the factor `x-2`

`x^2*(x-2) - x*(x-2) -2*x - 4x + 12 = 0`

`x^2*(x-2) - x*(x-2) -6*x + 12 = 0`

`x^2*(x-2) - x*(x-2) -6*(x - 2) = 0`

`(x-2)*(x^2 - x -6) = 0`

Resolving the equation is thus facilitated as it is reduced to a second-degree equation.

## Graphical Analysis

Graphical representation of a cubic equation can help visualize the roots and the nature of the equation. The graph of `ax^3 + bx^2 + cx + d = 0` shows the points where the curve intersects the x-axis, which are the roots of the equation.

To visualize the graph of a function (here the third-degree polynomial in question), one can also study its properties such as the sense of variation, singular points, local maxima and minima, and inflection points. This analysis not only helps find where the function intersects the x-axis (the roots) but also to understand the overall behavior of the function. For example, studying the first and second derivatives can reveal points where the slope of the curve changes direction or where the curvature of the graph changes, thus providing a deeper understanding of the structure of the cubic equation. Drawing the graph of the equation helps visualize these characteristics and better understand the nature of the function.

We can refer to polynomial calculators to verify the calculations.

To visualize the graph of a function (here the third-degree polynomial in question), we can use: graph of f(x)=x^3 - 3x^2 - 4x + 12.

## Conclusion

Understanding third-degree equations and their resolution methods is essential for many scientific and technical fields. This guide offers a comprehensive overview of the different methods and applications of these equations.

## See also

Quadratic Equation Solver

Equation and Inequation Calculators

Mathematics Calculators