Cubic equation

Graph of a 3rd degree function; the zeros ( y = 0) are where the graph intersects the x -axis . This graph has three real zeros.

Cubic equations are third degree polynomial equations , that is, algebraic equations of form

${\displaystyle A\cdot x^{3}+B\cdot x^{2}+C\cdot x+D=0\quad {\text{mit}}\quad A,B,C,D\in \mathbb {C} \quad {\text{und}}\quad A\neq 0}$.

According to the fundamental theorem of algebra, a cubic equation always has three complex solutions , which can also coincide. With their help, the equation can be represented in factorized form : ${\displaystyle x_{1},x_{2},x_{3}}$

${\displaystyle A\cdot (x-x_{1})\cdot (x-x_{2})\cdot (x-x_{3})=0\,.}$

In the case of real coefficients, the left side of the cubic equation geometrically describes a cubic parabola in the - plane, i.e. the graph of a cubic function . Its zeros , i.e. its points of intersection with the -axis, are the real solutions of the cubic equation. According to the intermediate value theorem, the function graph always has at least one real zero, but at most three. ${\displaystyle x}$ ${\displaystyle y}$ ${\displaystyle x}$

Possible solutions

Guessing a solution

Procedure

A solution is known exactly, it can be 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 thus the remaining solutions of the cubic equation are obtained. However, this procedure is only practicable for a rational solution . Even with the irreducible equation , the method with the still relatively simple solution is no longer practicable, since the coefficients of the remaining quadratic equation become very complicated. In these cases, the solutions are easier to determine using the Cardan formula below . ${\displaystyle x_{1}}$ ${\displaystyle (x-x_{1})}$ ${\displaystyle x_{1}}$ ${\displaystyle x^{3}-6x-6=0}$ ${\displaystyle x={\sqrt[{3}]{2}}+{\sqrt[{3}]{4}}}$

If all the coefficients of the cubic equation are integers , one can try to guess a rational solution, that is, to find it by trial and error. If the leading coefficient is equal to 1, you can try the whole number divisors of the last coefficient (also negative values!). If it differs from one, then all fractions whose numerator is a divisor of and whose denominator is a divisor of must be tried out. The theorem about rational zeros guarantees that with this finite effort one can find a rational zero, if one exists. If the coefficients are rational, then integral coefficients can be obtained by multiplying the equation by the main denominator of all the coefficients. ${\displaystyle A}$ ${\displaystyle D}$ ${\displaystyle A}$ ${\displaystyle D}$ ${\displaystyle A}$

example

As rational solutions to the cubic equation

${\displaystyle 3x^{3}-8x^{2}-11x+10=0}$

only the integer divisors of the last coefficient as well as are possible . Indeed, a solution is what one becomes convinced of through engagement. Polynomial division returns ${\displaystyle \pm 1,\pm 2,\pm 5,\pm 10}$ ${\displaystyle \pm {\tfrac {1}{3}},\pm {\tfrac {2}{3}},\pm {\tfrac {5}{3}},\pm {\tfrac {10}{3}}}$ ${\displaystyle x_{1}={\tfrac {2}{3}}}$

${\displaystyle (3x^{3}-8x^{2}-11x+10):\left(x-{\tfrac {2}{3}}\right)=3x^{2}-6x-15}$

and with the quadratic formula there are further solutions . ${\displaystyle x_{2,3}=1\pm {\sqrt {6}}}$

Algebraic determination

→ Main article : Cardan formulas

Reduction of the equation to a normal form

There are a number of equivalent transformations of the cubic equation by linear transformation of the argument, which make it possible to simplify this for the subsequent solution procedure ( Tschirnhaus transformation ). The polynomial can first be normalized by dividing by . ${\displaystyle A\neq 0}$

${\displaystyle x^{3}+a\cdot x^{2}+b\cdot x+c=0\quad {\text{mit}}\quad a={\frac {B}{A}},\;b={\frac {C}{A}}\;{\text{und}}\;c={\frac {D}{A}}}$

Linear transformation of the argument with the help of substitution results in the following term: ${\displaystyle x=\alpha \cdot z+\beta }$

${\displaystyle z^{3}+{\frac {3\beta +a}{\alpha }}\cdot z^{2}+{\frac {3\beta ^{2}+2a\beta +b}{\alpha ^{2}}}\cdot z+{\frac {\beta ^{3}+a\beta ^{2}+b\beta +c}{\alpha ^{3}}}=0}$

By choosing , the square term can be eliminated and the reduced form of the cubic equation is obtained: ${\displaystyle \beta =-{\tfrac {a}{3}}}$

${\displaystyle z^{3}+{\frac {p}{\alpha ^{2}}}\cdot z+{\frac {q}{\alpha ^{3}}}=0\quad {\text{mit}}\quad p=b-{\frac {a^{2}}{3}}\quad {\text{und}}\quad q={\frac {2a^{3}}{27}}-{\frac {ab}{3}}+c}$

The reduced form with can now be solved with the help of Cardan's formulas and the solutions of the original equation can be determined by subsequent substitution. This makes the totality of real and complex solutions accessible. ${\displaystyle \alpha =1}$

Analytical determination of the real solutions of the real equation

In the event that the original polynomial only has real coefficients, the discriminant can be used to check whether only real solutions exist: ${\displaystyle \Delta }$

${\displaystyle \Delta :=\left({\frac {q}{2}}\right)^{2}+\left({\frac {p}{3}}\right)^{3}}$

Is , then all solutions are real. Otherwise there is exactly one real solution, the other two are complex non-real and conjugate to each other. ${\displaystyle \Delta \leq 0}$

The case p = 0

Case 1:   ${\displaystyle p=0}$

Here you choose and receive . A single real solution results after back substitution . ${\displaystyle \alpha ={\tfrac {1}{3}}{\sqrt[{3}]{a^{3}-27c}}}$ ${\displaystyle z^{3}=1}$ ${\displaystyle x={\tfrac {1}{3}}\left({\sqrt[{3}]{a^{3}-27c}}-a\right)}$

Sub-case 1a:   and ${\displaystyle p=0}$ ${\displaystyle q=0}$

The only real solution and has the multiplicity of 3. ${\displaystyle z=0}$ ${\displaystyle x=-{\tfrac {a}{3}}}$

The cases with p ≠ 0

A solution strategy for the remaining solutions, which manages without the use of complex numbers, is the following:
The reduced form is transformed by substitution with the help of a suitable trigonometric or hyperbolic function in such a way that it can be reduced to known addition theorems .

Suitable functions are:

function ${\displaystyle f}$	Range of values	Addition theorem	${\displaystyle \sigma }$	cubic equation	case
${\displaystyle \cos }$	${\displaystyle |f(\eta )|\leq 1}$	${\displaystyle \cos(3\eta )=4\cos ^{3}(\eta )-3\cos(\eta )}$	${\displaystyle -1}$	${\displaystyle f(\eta )^{3}+{\tfrac {3}{4}}\sigma f(\eta )={\tfrac {1}{4}}f(3\eta )}$	2
${\displaystyle \cosh }$	${\displaystyle f(\eta )\geq 1}$	${\displaystyle \cosh(3\eta )=4\cosh ^{3}(\eta )-3\cosh(\eta )}$	${\displaystyle -1}$	${\displaystyle f(\eta )^{3}+{\tfrac {3}{4}}\sigma f(\eta )={\tfrac {1}{4}}f(3\eta )}$	3
${\displaystyle -\cosh }$	${\displaystyle f(\eta )\leq -1}$	${\displaystyle (-\cosh )(3\eta )\;=\;4(-\cosh )^{3}(\eta )-3(-\cosh )(\eta )}$	${\displaystyle -1}$	${\displaystyle f(\eta )^{3}+{\tfrac {3}{4}}\sigma f(\eta )={\tfrac {1}{4}}f(3\eta )}$	3
${\displaystyle \sinh }$	arbitrarily real	${\displaystyle \sinh(3\eta )=4\sinh ^{3}(\eta )\,+3\sinh(\eta )}$	${\displaystyle +1}$	${\displaystyle f(\eta )^{3}+{\tfrac {3}{4}}\sigma f(\eta )={\tfrac {1}{4}}f(3\eta )}$	4th

The listed addition theorems are parameterized in such a way that they can be converted into the same cubic equation that can be obtained with the reduced form of the given equation

${\displaystyle f(\eta )^{3}+{\frac {p}{\alpha ^{2}}}\cdot f(\eta )+{\frac {q}{\alpha ^{3}}}=0}$

can be brought to cover. With the help of the settlement one gets immediately by comparing coefficients ${\displaystyle \sigma :=\operatorname {sgn}(p)}$

${\displaystyle {\frac {3}{4}}\sigma ={\frac {p}{\alpha ^{2}}}\;\Longleftrightarrow \;\alpha =2{\sqrt {\frac {|p|}{3}}}}$     and     . ${\displaystyle -{\frac {1}{4}}f(3\eta )={\frac {q}{\alpha ^{3}}}=q{\frac {3\sigma }{4p}}{\frac {1}{2}}{\sqrt {\frac {3}{|p|}}}={\frac {q}{8}}{\sqrt {\frac {27}{|p^{3}|}}}}$

Thus, the original coefficients and express: ${\displaystyle \eta }$ ${\displaystyle p}$ ${\displaystyle q}$

${\displaystyle f(3\eta )=\Gamma \;\Longleftrightarrow \;\eta ={\frac {1}{3}}f^{\langle -1\rangle }\left(\Gamma \right)}$,

where is set and denotes an associated arc or area function . The final solution of the cubic equation can then be determined by back substitution. Off , and thus one obtains ${\displaystyle \Gamma :=-{\tfrac {q}{2}}{\sqrt {\tfrac {27}{|p^{3}|}}}=-\operatorname {sgn}(q){\sqrt {\left|{\tfrac {27\Delta }{p^{3}}}-1\right|}}}$ ${\displaystyle f^{\langle -1\rangle }}$ ${\displaystyle \alpha =2{\sqrt {\tfrac {|p|}{3}}}={\tfrac {2}{3}}{\sqrt {|a^{2}-3b|}}}$ ${\displaystyle \beta =-{\tfrac {a}{3}}}$ ${\displaystyle z=f(\eta )}$

${\displaystyle x=\alpha z+\beta ={\tfrac {1}{3}}\left(2{\sqrt {|a^{2}-3b|}}\cdot f(\eta )-a\right)}$.

First of all, the sign of determines the choice of the substitution function , secondly , which must be in the real value range of . ${\displaystyle p}$ ${\displaystyle f}$ ${\displaystyle \Gamma }$ ${\displaystyle f}$

Case 2:     (from which     and     follows): ${\displaystyle \Delta \leq 0}$ ${\displaystyle p<0}$ ${\displaystyle \left|\Gamma \right|\leq 1}$

Substitution with , corresponds to ${\displaystyle z:=\cos {\eta }}$ ${\displaystyle \cos {3\eta }=\Gamma .}$

There are three possible solutions

${\displaystyle x_{k}={\tfrac {1}{3}}\left(2{\sqrt {a^{2}-3b}}\cdot \cos {\eta _{k}}-a\right)}$with and ${\displaystyle \eta _{k}={\tfrac {1}{3}}\left(\arccos {\left(\Gamma \right)}+2k\pi \right)}$ ${\displaystyle k\in \{0;1;2\}}$

Sub-case 2a:     (from which     follows): ${\displaystyle \Delta =0}$ ${\displaystyle \left|\Gamma \right|=1}$

There are only two solutions. The reduced form is simplified to . The two solutions and can now be read directly from the linear factors . Leads to the same result , i.e. or . Corresponding is and . The latter solution has the multiplicity of 2. ${\displaystyle 0=z^{3}-{\tfrac {3}{4}}z\mp {\tfrac {1}{4}}=(z\mp 1)\left(z\pm {\tfrac {1}{2}}\right)^{2}}$ ${\displaystyle z_{1}=\pm 1}$ ${\displaystyle z_{2}=\mp {\tfrac {1}{2}}}$ ${\displaystyle 3\eta =\pm \operatorname {arccos} (\pm 1)\in \{0,\pi \}}$ ${\displaystyle \eta \in \left\{0,\pm {\tfrac {2\pi }{3}}\right\}}$ ${\displaystyle \eta \in \left\{\pi ,\pm {\tfrac {\pi }{3}}\right\}}$ ${\displaystyle x_{1}={\tfrac {1}{3}}\left(2{\sqrt {a^{2}-3b}}-a\right)}$ ${\displaystyle x_{2}=-{\tfrac {1}{3}}\left({\sqrt {a^{2}-3b}}+a\right)}$

Case 3:     and     (from which     and     follows): ${\displaystyle \Delta >0}$ ${\displaystyle p<0}$ ${\displaystyle |\Gamma |>1}$ ${\displaystyle q\neq 0}$

Substitution with , corresponds to , so ${\displaystyle z:=\left(-\operatorname {sgn}(q)\cosh \right)(\eta )}$ ${\displaystyle \left(-\operatorname {sgn}(q)\cosh \right)(3\eta )=\Gamma =-{\tfrac {q}{2}}{\sqrt {\tfrac {27}{|p^{3}|}}}}$ ${\displaystyle \cosh(3\eta )=|\Gamma |.}$

First, you have two solutions , which are thrown into one again because of . So: with . ${\displaystyle 3\eta =\pm \operatorname {arcosh} \left(|\Gamma |\right)}$ ${\displaystyle \cosh(\pm \eta )=\cosh \eta }$ ${\displaystyle x=-{\tfrac {1}{3}}\left(2{\sqrt {a^{2}-3b}}\cdot \operatorname {sgn}(q)\cosh {\eta }+a\right)}$ ${\displaystyle \eta ={\tfrac {1}{3}}\operatorname {arcosh} \left(\left|\Gamma \right|\right)}$

Borderline case 3a:     and     (from which it     follows): ${\displaystyle \Delta =0}$ ${\displaystyle p<0}$ ${\displaystyle \Gamma =\pm 1}$

${\displaystyle 3\eta =\pm \operatorname {arcosh} (1)=0}$, so and . Note: The other two (purely imaginary) solutions of are represented by the use of discarded Real: . The result is as in sub-case 2a: and . ${\displaystyle \eta =0}$ ${\displaystyle x_{1}={\tfrac {1}{3}}\left(2{\sqrt {a^{2}-3b}}-a\right)}$

${\displaystyle 3\eta =\pm 2\pi \mathrm {i} }$ ${\displaystyle \cosh(3\eta )=1}$ ${\displaystyle \cosh }$ ${\displaystyle \cosh(\eta )=\cosh \left(\pm {\tfrac {2\pi \mathrm {i} }{3}}\right)=-{\tfrac {1}{2}}}$ ${\displaystyle z_{1}=-\operatorname {sgn}(q)\cosh(0)=-\operatorname {sgn}(q)}$ ${\displaystyle z_{2}=-\operatorname {sgn}(q)\cosh \left(\pm {\tfrac {2\pi \mathrm {i} }{3}}\right)={\tfrac {\operatorname {sgn}(q)}{2}}}$

Case 4:     and   : ${\displaystyle \Delta >0}$ ${\displaystyle p>0}$

Substitution with , corresponds to ${\displaystyle z:=\sinh {\eta }}$ ${\displaystyle \sinh {3\eta }=\Gamma .}$

The result is:

${\displaystyle x={\frac {1}{3}}\left(2{\sqrt {3b-a^{2}}}\cdot \sinh {\eta }-a\right)}$ With ${\displaystyle \eta ={\frac {1}{3}}\operatorname {arsinh} {(\Gamma )}}$

The result is a real solution.

Fast numerical calculation

The method of Deiters and Macías-Salinas first puts the cubic function into the form and then uses the Laguerre-Nair-Samuelson inequality to find bounds for the solutions, ${\displaystyle f(x)=x^{3}+a_{2}x^{2}+a_{1}x+a_{0}}$

${\displaystyle x_{\mathrm {u,o} }:=x_{\mathrm {infl} }\pm {\frac {2}{3}}{\sqrt {d}}}$.

Where , and is the abscissa value of the inflection point. Then the following cases have to be distinguished: ${\displaystyle d:=a_{2}^{2}-3a_{1}}$ ${\displaystyle x_{\mathrm {infl} }:=-a_{2}/3}$

1. ${\displaystyle f(x_{\mathrm {infl} })=0}$: Then the turning point is the first solution . ${\displaystyle x_{1}=x_{\mathrm {infl} }}$
2. ${\displaystyle d=0}$: Then there is a solution. ${\displaystyle x_{1}=x_{\mathrm {infl} }-{\sqrt[{3}]{f(x_{\mathrm {infl} })}}}$
3. Otherwise an approximate solution is determined iteratively . This is done based on the start value ${\displaystyle x_{1}}$

${\displaystyle x_{1,\mathrm {init} }={\begin{cases}x_{\mathrm {u} }&{\text{wenn }}d>0\land f(x_{\mathrm {infl} })>0\\x_{\mathrm {infl} }&{\text{wenn }}d<0\\x_{\mathrm {o} }&{\text{wenn }}d>0\land f(x_{\mathrm {infl} })<0\end{cases}}}$

with the Halley method :

${\displaystyle x_{1}\;\;\;\leftarrow \;x_{1}-{\frac {f(x_{1})f^{\prime }(x_{1})}{f^{\prime }(x_{1})^{\,2}-{\frac {1}{2}}f(x_{1})f^{\prime \prime }(x_{1})}}}$.

Then the quadratic function (with a small one , the amount of which depends on the achieved accuracy) is formed by polynomial division , the zeros of which (in the case ) can be calculated directly: ${\displaystyle g(x)={\bigl (}f(x)-e{\bigr )}/(x-x_{1})}$ ${\displaystyle e:=f(x_{1})}$ ${\displaystyle e=0}$

${\displaystyle g(x)=x^{2}+b_{1}x+b_{0}}$with and . ${\displaystyle b_{1}=x_{1}+a_{2}}$ ${\displaystyle b_{0}=b_{1}x_{1}+a_{1}}$

If implemented carefully (see the revised additional information on the original publication), this method is 1.2 to 10 times faster on modern processors (2014, architecture x86-64) than the Cardanic formulas evaluated for comparable accuracy .

See also

• Linear equation
• Quartic equation
• Omar Chajjam
• Cardan formulas

Web links

Commons : Cubic functions  - collection of images, videos and audio files

• Online tool for computing nth order polynomials
• Cubic Equation - JavaScript
• Calculations with examples from Joachim Mohr

Sources and literature

• Peter Gabriel : Matrices, Geometry, Linear Algebra . Birkhäuser, Basel 1996, ISBN 3-7643-5376-7 .

Individual evidence

1. ↑ UK Deiters, R. Macías-Salinas: Calculation of densities from cubic equations of state: revisited . In: Ind. Eng. Chem. Res. Volume 53 , 2014, p. 2529-2536 , doi : 10.1021 / ie4038664 .
2. ^ Paul Samuelson: How Deviant Can You Be? . In: Journal of the American Statistical Association . 63, No. 324, 1968, pp. 1522-1525. doi : 10.2307 / 2285901 .
   S. a. Samuelson's inequality on Wikipedia, accessed 2016-06-10
3. ↑ Cubic rootfinder using Halley's method: C / C ++ program code (PDF) University of Cologne. Retrieved July 5, 2019.