Cubic function

This article discusses cubic equations in one variable. For a discussion of cubic equations in two variables, see elliptic curve.

Graph of a cubic function with its 3 roots, i.e. where the curve crosses the x-axis (y = 0). It has 2 critical points.

In mathematics, a cubic function is a function of the form

$f(x)=ax^{3}+bx^{2}+cx+d,\,$

where a is nonzero; or in other words, a polynomial of degree three. The derivative of a cubic function is a quadratic function. The integral of a cubic function is a quartic function.

Setting ƒ(x) = 0 and assuming a ≠ 0 produces a cubic equation of the form:

$ax^{3}+bx^{2}+cx+d=0.\,$

Usually, the coefficients a, b,c, d are real numbers. However, most of the theory is also valid if they belong to a field of characteristic other than 2 or 3.

Solving a cubic equation amounts to finding the roots of a cubic function.

History

Cubic equations were known to the ancient Indians and ancient Greeks since the 5th century BC, and even earlier to the ancient Babylonians who were able to solve certain cubic equations, and ancient Egyptians, who dealt with the problem of doubling the cube, and attempted to solve it using compass and straightedge constructions. Hippocrates, Menaechmus and Archimedes are believed to have come close to solving this problem using intersecting conic sections, though historians such as Reviel Netz dispute whether the Greeks were thinking about cubic equations or just problems that can lead to cubic equations.

Two-dimensional graph of a cubic, the polynomial ƒ(x) = 2x³ − 3x² − 3x + 2.

In the 11th century, the Persian poet-mathematician, Omar Khayyám (1048–1131), made significant progress in the theory of cubic equations. In an early paper he wrote regarding cubic equations, he discovered that a cubic equation can have more than one solution, that it cannot be solved using earlier compass and straightedge constructions, and found a geometric solution which could be used to get a numerical answer by consulting trigonometric tables. In his later work, the Treatise on Demonstration of Problems of Algebra, he wrote a complete classification of cubic equations with general geometric solutions found by means of intersecting conic sections.

In the 12th century, another Persian mathematician, Sharaf al-Dīn al-Tūsī (1135–1213), wrote the Al-Mu'adalat (Treatise on Equations), which dealt with eight types of cubic equations with positive solutions and five types of cubic equations which may not have positive solutions. He used what would later be known as the "Ruffini-Horner method" to numerically approximate the root of a cubic equation. He also developed the concepts of a derivative function and the maxima and minima of curves in order to solve cubic equations which may not have positive solutions. He understood the importance of the discriminant of the cubic equation and used an early version of Cardano's formula to find algebraic solutions to certain types of cubic equations.

In the early 16th century, the Italian mathematician Scipione del Ferro (1465–1526) found a method for solving a class of cubic equations, namely those of the form x³ + mx = n. In fact, all cubic equations can be reduced to this form if we allow m and n to be negative, but negative numbers were not known to him at that time. Del Ferro kept his achievement secret until just before his death, when he told his student Antonio Fiore about it.

In 1530, Niccolò Tartaglia (1500–1557) received two problems in cubic equations from Zuanne da Coi and announced that he could solve them. He was soon challenged by Fiore, which led to a famous contest between the two. Each contestant had to put up a certain amount of money and to propose a number of problems for his rival to solve. Whoever solved more problems within 30 days would get all the money. Tartaglia received questions in the form x³ + mx = n, for which he had worked out a general method. Fiore received questions in the form x³ + mx² = n, which proved to be too difficult for him to solve, and Tartaglia won the contest.

Later, Tartaglia was persuaded by Gerolamo Cardano (1501–1576) to reveal his secret for solving cubic equations. In 1539, Tartaglia did so only on the condition that Cardano would never reveal it and that if he did reveal a book about cubics, that he would give Tartaglia time to publish. Some years later, Cardano learned about Ferro's prior work and published Ferro's method in his book Ars Magna in 1545, meaning Cardano gave Tartaglia 6 years to publish his results (with credit given to Tartaglia for an independent solution). Cardano's promise with Tartaglia stated that he not publish Tartaglia's work, and Cardano felt he was publishing del Ferro's, so as to get around the promise. Nevertheless, this led to a challenge to Cardano by Tartaglia, which Cardano denied. The challenge was eventually accepted by Cardano's student Lodovico Ferrari (1522–1565). Ferrari did better than Tartaglia in the competition, and Tartaglia lost both his prestige and income .

Cardano noticed that Tartaglia's method sometimes required him to extract the square root of a negative number. He even included a calculation with these complex numbers in Ars Magna, but he did not really understand it. Rafael Bombelli studied this issue in detail and is therefore often considered as the discoverer of complex numbers.

Roots of a cubic function

The nature of the roots

Every cubic equation with real coefficients has at least one solution x among the real numbers; this is a consequence of the intermediate value theorem. We can distinguish several possible cases using the discriminant,

$\Delta =-4b^{3}d+b^{2}c^{2}-4ac^{3}+18abcd-27a^{2}d^{2}.\,$

The following cases need to be considered:

• If Δ < 0, then the equation has one real root and a pair of imaginary complex conjugate roots.
• If Δ > 0, then all roots are real and unequal.
• If Δ = 0, then the equation has three real roots and at least two are equal.

  It may be that the equation has a double real root and another distinct single real root; alternatively, all three roots coincide yielding a triple real root. A possible way to decide between these subcases is to compute the resultant of the cubic and its second derivative: a triple root exists if and only if this resultant vanishes.

See also: multiplicity of a root of a polynomial

General formula of roots

For the general cubic equation $ax^{3}+bx^{2}+cx+d=0$, the general formulas for the roots, in terms of the coefficients, are as follows:

${\begin{aligned}x_{1}=&-{\frac {b}{3a}}\\&-{\frac {1}{3a}}{\sqrt[{3}]{\frac {2b^{3}-9abc+27a^{2}d+{\sqrt {\left(2b^{3}-9abc+27a^{2}d\right)^{2}-4\left(b^{2}-3ac\right)^{3}}}}{2}}}\\&-{\frac {1}{3a}}{\sqrt[{3}]{\frac {2b^{3}-9abc+27a^{2}d-{\sqrt {\left(2b^{3}-9abc+27a^{2}d\right)^{2}-4\left(b^{2}-3ac\right)^{3}}}}{2}}}\\x_{2}=&-{\frac {b}{3a}}\\&+{\frac {1+i{\sqrt {3}}}{6a}}{\sqrt[{3}]{\frac {2b^{3}-9abc+27a^{2}d+{\sqrt {\left(2b^{3}-9abc+27a^{2}d\right)^{2}-4\left(b^{2}-3ac\right)^{3}}}}{2}}}\\&+{\frac {1-i{\sqrt {3}}}{6a}}{\sqrt[{3}]{\frac {2b^{3}-9abc+27a^{2}d-{\sqrt {\left(2b^{3}-9abc+27a^{2}d\right)^{2}-4\left(b^{2}-3ac\right)^{3}}}}{2}}}\\x_{3}=&-{\frac {b}{3a}}\\&+{\frac {1-i{\sqrt {3}}}{6a}}{\sqrt[{3}]{\frac {2b^{3}-9abc+27a^{2}d+{\sqrt {\left(2b^{3}-9abc+27a^{2}d\right)^{2}-4\left(b^{2}-3ac\right)^{3}}}}{2}}}\\&+{\frac {1+i{\sqrt {3}}}{6a}}{\sqrt[{3}]{\frac {2b^{3}-9abc+27a^{2}d-{\sqrt {\left(2b^{3}-9abc+27a^{2}d\right)^{2}-4\left(b^{2}-3ac\right)^{3}}}}{2}}}\end{aligned}}$

Monic formula of roots

For a monic polynomial (dividing by the leading coefficient), renaming the coefficients as $x^{3}+ax^{2}+bx+c=0$, this reduces to

${\begin{aligned}x_{1}=&-{\frac {a}{3}}\\&-{\frac {1}{3}}{\sqrt[{3}]{\frac {2a^{3}-9ab+27c+{\sqrt {\left(2a^{3}-9ab+27c\right)^{2}-4\left(a^{2}-3b\right)^{3}}}}{2}}}\\&-{\frac {1}{3}}{\sqrt[{3}]{\frac {2a^{3}-9ab+27c-{\sqrt {\left(2a^{3}-9ab+27c\right)^{2}-4\left(a^{2}-3b\right)^{3}}}}{2}}}\\x_{2}=&-{\frac {a}{3}}\\&+{\frac {1+i{\sqrt {3}}}{6}}{\sqrt[{3}]{\frac {2a^{3}-9ab+27c+{\sqrt {\left(2a^{3}-9ab+27c\right)^{2}-4\left(a^{2}-3b\right)^{3}}}}{2}}}\\&+{\frac {1-i{\sqrt {3}}}{6}}{\sqrt[{3}]{\frac {2a^{3}-9ab+27c-{\sqrt {\left(2a^{3}-9ab+27c\right)^{2}-4\left(a^{2}-3b\right)^{3}}}}{2}}}\\x_{3}=&-{\frac {a}{3}}\\&+{\frac {1-i{\sqrt {3}}}{6}}{\sqrt[{3}]{\frac {2a^{3}-9ab+27c+{\sqrt {\left(2a^{3}-9ab+27c\right)^{2}-4\left(a^{2}-3b\right)^{3}}}}{2}}}\\&+{\frac {1+i{\sqrt {3}}}{6}}{\sqrt[{3}]{\frac {2a^{3}-9ab+27c-{\sqrt {\left(2a^{3}-9ab+27c\right)^{2}-4\left(a^{2}-3b\right)^{3}}}}{2}}}\\\end{aligned}}$

Introducing the following notation

${\begin{aligned}m&=2a^{3}-9ab+27c\\k&=a^{2}-3b\\n&=m^{2}-4k^{3}=(2a^{3}-9ab+27c)^{2}-4\left(a^{2}-3b\right)^{3}\\\omega _{1}&={\frac {-1+i{\sqrt {3}}}{2}}\\\omega _{2}&={\frac {-1-i{\sqrt {3}}}{2}}\end{aligned}}$

simplifies the above equations to

${\begin{aligned}x_{1}&=-{\frac {1}{3}}\left(a+{\sqrt[{3}]{\frac {m+{\sqrt {n}}}{2}}}+{\sqrt[{3}]{\frac {m-{\sqrt {n}}}{2}}}\,\right)\\x_{2}&=-{\frac {1}{3}}\left(a+\omega _{2}{\sqrt[{3}]{\frac {m+{\sqrt {n}}}{2}}}+\omega _{1}{\sqrt[{3}]{\frac {m-{\sqrt {n}}}{2}}}\,\right)\\x_{3}&=-{\frac {1}{3}}\left(a+\omega _{1}{\sqrt[{3}]{\frac {m+{\sqrt {n}}}{2}}}+\omega _{2}{\sqrt[{3}]{\frac {m-{\sqrt {n}}}{2}}}\,\right)\\\end{aligned}}$

In this form, they can be identified as the inverse discrete Fourier transform of the triple:

$\left(-a,-{\sqrt[{3}]{\frac {m+{\sqrt {n}}}{2}}},-{\sqrt[{3}]{\frac {m-{\sqrt {n}}}{2}}}\right)$

which is the perspective taken in the method of Lagrange resolvents.

Cardano's method

The solutions can be found with the following method due to Scipione del Ferro and Tartaglia, published by Gerolamo Cardano in 1545.

We first divide the standard equation by the leading coefficient to arrive at an equation of the form

$x^{3}+ax^{2}+bx+c=0\qquad (1).$

The substitution $x=t-a/3\,$ eliminates the quadratic term, giving the so-called depressed cubic

$t^{3}+pt+q=0\,\qquad (2)$

where

$p=b-{\frac {a^{2}}{3}}\qquad {\mbox{and}}\qquad q=c+{\frac {2a^{3}-9ab}{27}}.$

We introduce two variables u and v linked by the condition

$u+v=t\,$

and substitute this in the depressed cubic (2), giving

$u^{3}+v^{3}+(3uv+p)(u+v)+q=0\qquad (3)\,$.

At this point Cardano imposed a second condition for the variables u and v

$3uv+p=0\,$

which, combined with (3) (by substituting for v) gives

$u^{6}+qu^{3}-{p^{3} \over 27}=0\,.$

This can be seen as a quadratic equation in u³. When we solve this equation, we find that

$u^{3}=-{q \over 2}\pm {\sqrt {{q^{2} \over 4}+{p^{3} \over 27}}}$

and thus

$u={\sqrt[{3}]{-{q \over 2}\pm {\sqrt {{q^{2} \over 4}+{p^{3} \over 27}}}}}\qquad (4)$

Since t = v + u, t = x + a/3, and v = −p/3u, we find

$x=-{\frac {p}{3u}}+u-{a \over 3}.$

Note that there are six possibilities in computing u with (4), since there are two possibilities for the square root ($\pm$), and three for the cubic root (the principal root and the principal root multiplied by ${\tfrac {-1}{2}}\pm i{\tfrac {\sqrt {3}}{2}}$). The sign of the square root however does not affect the resulting t (a simple calculation shows that −p/3u = v), although care must be taken in three special cases to avoid divisions by zero:

First, if p = q = 0, then we have the triple real root

$t=0.\,$

Second, if p = 0 and q ≠ 0, then

$u=0{\text{ and }}v=-{\sqrt[{3}]{q}}.$

Third, if p ≠ 0 and q = 0 then

$u={\sqrt {p \over 3}}\qquad {\mbox{and}}\qquad v=-{\sqrt {p \over 3}},$

in which case the three roots are

$t=u+v=0,\qquad t=\omega _{1}u-{p \over 3\omega _{1}u}={\sqrt {-p}},\qquad t={u \over \omega _{1}}-{\omega _{1}p \over 3u}=-{\sqrt {-p}},$

where

$\omega _{1}=e^{i{\frac {2\pi }{3}}}=-{\tfrac {1}{2}}+i{\tfrac {\sqrt {3}}{2}}.$

Summary

In summary, for the cubic equation

$x^{3}+ax^{2}+bx+c=0\$

the solutions for x are given by

$x=u-{p \over 3u}-{a \over 3}$

where

$p=b-{a^{2} \over 3}$

$q=c+{2a^{3}-9ab \over 27}$

$u={\sqrt[{3}]{-{q \over 2}\pm {\sqrt {{q^{2} \over 4}+{p^{3} \over 27}}}}}$

The expression above for u can generate up to three values (there are three cubic roots related by a factor which is one of the two complex cubic roots of one, and two square roots of any sign ; but these 6 expressions can generate only 3 pairs). This also applies to the final solutions for x.

Alternate method

An alternate method to obtain the same results is as follows.

We know that

$u^{3}=-{\frac {q}{2}}+{\sqrt {{\frac {q^{2}}{4}}+{\frac {p^{3}}{27}}}}\qquad {\mbox{or}}\qquad -{\frac {q}{2}}-{\sqrt {{\frac {q^{2}}{4}}+{\frac {p^{3}}{27}}}}.$

Since u and v must satisfy

$-u^{3}-v^{3}=q\qquad {\mbox{and}}\qquad -uv={\frac {p}{3}}$,

it can be shown that

$u^{3}=-{\frac {q}{2}}+{\sqrt {{\frac {q^{2}}{4}}+{\frac {p^{3}}{27}}}}\quad \iff \quad v^{3}=-{\frac {q}{2}}-{\sqrt {{\frac {q^{2}}{4}}+{\frac {p^{3}}{27}}}}$

Writing out the three cube roots we get

$u=\left\{{\begin{aligned}&{\sqrt[{3}]{-{\frac {q}{2}}+{\sqrt {{\frac {q^{2}}{4}}+{\frac {p^{3}}{27}}}}}}\\&\left(-{\frac {1}{2}}+i{\frac {\sqrt {3}}{2}}\right){\sqrt[{3}]{-{\frac {q}{2}}+{\sqrt {{\frac {q^{2}}{4}}+{\frac {p^{3}}{27}}}}}}\\&\left(-{\frac {1}{2}}-i{\frac {\sqrt {3}}{2}}\right){\sqrt[{3}]{-{\frac {q}{2}}+{\sqrt {{\frac {q^{2}}{4}}+{\frac {p^{3}}{27}}}}}}\\\end{aligned}}\right.{\text{ and }}v=\left\{{\begin{aligned}&{\sqrt[{3}]{-{\frac {q}{2}}-{\sqrt {{\frac {q^{2}}{4}}+{\frac {p^{3}}{27}}}}}}\\&\left(-{\frac {1}{2}}+i{\frac {\sqrt {3}}{2}}\right){\sqrt[{3}]{-{\frac {q}{2}}-{\sqrt {{\frac {q^{2}}{4}}+{\frac {p^{3}}{27}}}}}}\\&\left(-{\frac {1}{2}}-i{\frac {\sqrt {3}}{2}}\right){\sqrt[{3}]{-{\frac {q}{2}}-{\sqrt {{\frac {q^{2}}{4}}+{\frac {p^{3}}{27}}}}}}\\\end{aligned}}\right.$

Remembering t = u + v we get only three possible values for t because only three combinations of u and v are possible if $-uv={\frac {p}{3}}$ is to remain valid as it must — so

$t=\left\{{\begin{aligned}&{\sqrt[{3}]{-{\frac {q}{2}}+{\sqrt {{\frac {q^{2}}{4}}+{\frac {p^{3}}{27}}}}}}+{\sqrt[{3}]{-{\frac {q}{2}}-{\sqrt {{\frac {q^{2}}{4}}+{\frac {p^{3}}{27}}}}}}\\&\left(-{\frac {1}{2}}+i{\frac {\sqrt {3}}{2}}\right){\sqrt[{3}]{-{\frac {q}{2}}+{\sqrt {{\frac {q^{2}}{4}}+{\frac {p^{3}}{27}}}}}}+\left(-{\frac {1}{2}}-i{\frac {\sqrt {3}}{2}}\right){\sqrt[{3}]{-{\frac {q}{2}}-{\sqrt {{\frac {q^{2}}{4}}+{\frac {p^{3}}{27}}}}}}\\&\left(-{\frac {1}{2}}-i{\frac {\sqrt {3}}{2}}\right){\sqrt[{3}]{-{\frac {q}{2}}+{\sqrt {{\frac {q^{2}}{4}}+{\frac {p^{3}}{27}}}}}}+\left(-{\frac {1}{2}}+i{\frac {\sqrt {3}}{2}}\right){\sqrt[{3}]{-{\frac {q}{2}}-{\sqrt {{\frac {q^{2}}{4}}+{\frac {p^{3}}{27}}}}}}\\\end{aligned}}\right.$

and x is obtained from

$x=t-{\frac {a}{3}}$

Note that the above methods do apply if p and q are complex. This solution avoids the addition of an inverted cubic radical in the solution, and also resolves the ambiguity of signs for the square roots in the first solution given above.

Summary

To simplify the expressions above, it is customary to define this resolution in several steps by defining intermediate variables. Let

$A={a \over 3}$ and $B={b \over 3}$

$P={p \over 3}=B-A^{2}$

$Q={q \over 2}={c \over 2}+A^{3}-3AB$

Then the discriminant of the quadratic equation of $u^{3}\$ or $v^{3}\$ is

$D={q^{2} \over 4}+{p^{3} \over 27}=Q^{2}+P^{3}$,

Let's also define a constant that represents a generator for the three cubic roots of unity:

$\zeta =e^{i{2\pi \over 3}}=\cos({2\pi \over 3})+i.\sin({2\pi \over 3})=-{1 \over 2}+i.{{\sqrt {3}} \over 2}$

Then the solutions for x = t - A can be simply defined for k in 0, 1, 2 in :

${\begin{aligned}x_{k}&=t_{k}-A\\&=\zeta ^{k}.{\sqrt[{3}]{-Q+{\sqrt {D}}}}+\zeta ^{2k}.{\sqrt[{3}]{-Q-{\sqrt {D}}}}-A\\&=-\left(\zeta ^{k}.{\sqrt[{3}]{Q-{\sqrt {D}}}}+\zeta ^{2k}.{\sqrt[{3}]{Q+{\sqrt {D}}}}+A\right)\\\end{aligned}}$

where $-Q\pm {\sqrt {D}}$ are the two possible values for $u^{3}\$ or $v^{3}\$.

In the case p and q are both real, the following cases can be distinguished, according to the sign of the discriminant.

1. If D is strictly positive then there is one real and two complex roots.
2. If D is null then there is one real root (a triple root) or two real roots (a single and a double root.)
3. If D is strictly negative then there are three real roots (Casus irreducibilis).

Lagrange resolvents

Description

Lagrange resolvents, introduced by Joseph Louis Lagrange in his paper Réflexions sur la résolution algébrique des équations, reduced the solution of a cubic equation to the solution of an auxiliary quadratic polynomial, the "resolving equation" of the original equation, by a change of variables from the roots $x_{0},x_{1},x_{2}$ to the Lagrange resolvents $r_{0},r_{1},r_{2}.$ In modern terms, the resolvents are the discrete Fourier transform of the original roots.

In the modern language of Galois theory, Lagrange exploited the fact that the symmetric group S₃ has the cyclic group of order three $C_{3}$ as a normal subgroup, thus allowing one to decompose the permutation group of the roots as $C_{3}\to S_{3}\to C_{2}=S_{2}=S_{3}/C_{3}$ – the $C_{3}$ corresponds to the discrete Fourier transform (of order 3), which the $C_{2}=S_{2}$ corresponds to the quadratic resolving polynomial.

In the same way, the symmetric group of order four S₄ has a Klein four-group as normal subgroup, with quotient a symmetric group of order three S₃, which allows one to solve a quartic in terms of a cubic resolving polynomial.

However, this method does not work for polynomials of degree five or greater, as the resolving polynomial has higher degree than the original polynomial (for a quintic the resolving polynomial has degree 24). This is explained by the Abel–Ruffini theorem, which proves that such polynomials cannot be solved by radicals.

Procedure

Suppose that r₀, r₁ and r₂ are the roots of equation (1), and define $\zeta =(-1+i{\sqrt {3}})/2$, so that ζ is a primitive third root of unity. We now set

$s_{0}=r_{0}+r_{1}+r_{2},\,$

$s_{1}=r_{0}+\zeta r_{1}+\zeta ^{2}r_{2},\,$

$s_{2}=r_{0}+\zeta ^{2}r_{1}+\zeta r_{2}.\,$

This is the discrete Fourier transform of the roots: observe that while the coefficients of the polynomial are symmetric in the roots, in this formula an order has been chosen on the roots, so these are not symmetric in the roots. The roots may then be recovered from the three s_i by inverting the above linear transformation via the inverse discrete Fourier transform, giving

$r_{0}=(s_{0}+s_{1}+s_{2})/3,\,$

$r_{1}=(s_{0}+\zeta ^{2}s_{1}+\zeta s_{2})/3,\,$

$r_{2}=(s_{0}+\zeta s_{1}+\zeta ^{2}s_{2})/3.\,$

We already know the value s₀ = −a, so we only need to seek values for the other two.

The $s_{i}$ are not symmetric in the roots – $s_{0}$ is invariant, while the two non-trivial cyclic permutation of the roots sends $s_{1}$ to $\zeta s_{1}$ and $s_{2}$ to $\zeta ^{2}s_{2}$, or $s_{1}$ to $\zeta ^{2}s_{1}$ and $s_{2}$ to $\zeta s_{2}$ (depending on which permutation), while transposing $r_{1}$ and $r_{2}$ switches $s_{1}$ and $s_{2}$ – other transpositions switch these roots and multiplying them by a power of $\zeta .$

Thus, if one takes the cubes, the factors of $\zeta$ become factors of $\zeta ^{3}=1,$ so every cyclic permutation leaves the cubes invariant, and a transposition of two roots exchanges s₁³ and s₂³ – in other words, $S_{3}$ (permutation of the roots) acts as a permutation group on $\{s_{0}^{3},s_{1}^{3},s_{2}^{3}\},$ specifically by the sign permutation exchanging the last two, meaning that the action factors through the sign map $S_{3}\to C_{2}=S_{3}/C_{3}$ corresponding to the cyclic group of order 3 being a normal subgroup.

Hence the polynomial

$(z-s_{1}^{3})(z-s_{2}^{3})\qquad (5)$

is invariant under permutations of the roots, and so has coefficients expressible in terms of (1). Using calculations involving symmetric functions or alternatively field extensions, we can calculate (5) to be

${z}^{2}+\left(-9\,ba+2\,{a}^{3}+27\,c\right)z+\left({a}^{2}-3\,b\right)^{3}.$

The roots of this quadratic equation are

${\frac {9}{2}}\,ab-{a}^{3}-{\frac {27}{2}}\,c\pm {\frac {1}{2}}\,{\sqrt {D}},$

where D is the discriminant. Taking cube roots give us s₁ and s₂, from which we can recover the roots r_i of (1).

Factorization

If r is any root of (1), then we may factor using r to obtain

$\left(x-r\right)\left(x^{2}+(a+r)x+b+ar+r^{2}\right)=x^{3}+ax^{2}+bx+c.$

Hence if we know one root we can find the other two by solving a quadratic equation, giving

${\frac {1}{2}}\left(-a-r\pm {\sqrt {-3r^{2}-2ar+a^{2}-4b}}\right)$

for the other two roots.

Root-finding formula

The formula for finding the roots of a cubic function, based on Cardano's method, is fairly complicated. Therefore, it is common to use the rational root test or a numerical solution instead.

If we have

$f(x)=ax^{3}+bx^{2}+cx+d=a(x-x_{1})(x-x_{2})(x-x_{3})\,$

with $a,b,c,d\in {\mathbb {R}}$ and $a\neq 0$, let

$q={\frac {3ac-b^{2}}{9a^{2}}}$

and

$r={\frac {9abc-27a^{2}d-2b^{3}}{54a^{3}}},$

we define the discriminant:

$\Delta =q^{3}+r^{2}.\,$

There are two distinct cases:

• $\Delta >0\,$

in which case there is one real root and 2 imaginary. We define:

$s={\sqrt[{3}]{r+{\sqrt {q^{3}+r^{2}}}}}$

and

$t={\sqrt[{3}]{r-{\sqrt {q^{3}+r^{2}}}}}.$

• $\Delta \leq 0\,$

in which case we have 3 real roots. We express the complex quantity $r+i{\sqrt {-\Delta }}$ in polar form:

$r+i{\sqrt {-\Delta }}=\rho e^{i\theta }=(\rho ,\theta )\,$

$\rho ={\sqrt {-q^{3}}}\,$

$\theta =\arccos(r/{\rho })\,$

and we define:

$s=({\sqrt[{3}]{\rho }},\theta /3)\,$

and

$t=({\sqrt[{3}]{\rho }},-\theta /3).\,$

In both cases, the solutions are

$x_{1}=s+t-{\frac {b}{3a}},$

$x_{2}=-{\frac {1}{2}}(s+t)-{\frac {b}{3a}}+{\frac {\sqrt {3}}{2}}(s-t)i,$

$x_{3}=-{\frac {1}{2}}(s+t)-{\frac {b}{3a}}-{\frac {\sqrt {3}}{2}}(s-t)i.$

Solution in terms of Chebyshev radicals

If we have a cubic equation which is already in depressed form, we may write it as $\,x^{3}-3px-q=0$. Substituting $x={\sqrt {p}}z$ we obtain $z^{3}-3z-p^{-{\frac {3}{2}}}q=0$ or equivalently

$z^{3}-3z=p^{-{\frac {3}{2}}}q\ .$

From this we obtain solutions to our original equation in terms of the Chebyshev cube root $C_{1 \over 3}$ as

$r_{0}={\sqrt {p}}\,C_{1 \over 3}(p^{-{\frac {3}{2}}}q),\,$

$r_{1}=-{\sqrt {p}}\,C_{1 \over 3}(-p^{-{\frac {3}{2}}}q),\,$

$r_{2}=-r_{0}-r_{1}\ .$

If now we start from a general equation

$x^{3}+ax^{2}+bx+c=0\qquad (1)$

and reduce it to the depressed form under the substitution x = t − a/3, we have $\,p=(a^{2}-3b)/9$ and $\,q=-(2a^{3}-9ab+27c)/27$, leading to

$t_{a;b;c}=p^{-{\frac {3}{2}}}q=-{\frac {2a^{3}-9ab+27c}{(a^{2}-3b)^{3/2}}}.$

This gives us the solutions to (1) as

$r_{0}={\sqrt {p}}\,C_{1 \over 3}(t_{a;b;c})-{a \over 3},\,$

$r_{1}=-{\sqrt {p}}\,C_{1 \over 3}(-t_{a;b;c})-{a \over 3},\,$

$r_{2}=-r_{0}-r_{1}-a\ .$

The case of a cubic equation with real coefficients

Suppose the coefficients of (1) are real. If s is the quantity q/r from the section on real roots, then s = t²; hence 0 < s < 4 is equivalent to −2 < t < 2, and in this case we have a polynomial with three distinct real roots, expressed in terms of a real function of a real variable, quite unlike the situation when using cube roots. If s > 4 then either t > 2 and $C_{1 \over 3}(t)$ is the sole real root, or t < −2 and $-C_{1 \over 3}(-t)$ is the sole real root. If s < 0 then the reduction to Chebyshev polynomial form has given a t which is a pure imaginary number; in this case $iC_{1 \over 3}(-it)-iC_{1 \over 3}(it)$ is the sole real root. We are now evaluating a real root by means of a function of a purely imaginary argument; however we can avoid this by using the function

$S_{1 \over 3}(t)=iC_{1 \over 3}(-it)-iC_{1 \over 3}(it)=2\operatorname {sinh} \left(\operatorname {arcsinh} \left({t \over 2}\right)/3\right),\,$

which is a real function of a real variable with no singularities along the real axis. If a polynomial can be reduced to the form $x^{3}+3x-t$ with real t, this is a convenient way to solve for its roots.

Derivative

The derivative, $f'(x)=3ax^{2}+2bx+c$, will yield $x={\frac {-b\pm {\sqrt {b^{2}-3ac\ }}}{3a}}$ when $f'(x)=0\,$. Bearing its resemblance to the quadratic formula, this formula can be used to find the critical points of a cubic function. It turns out that, if $b^{2}-3ac>0\,$, then the cubic function will have two critical points — a local maximum and a local minimum; if $b^{2}-3ac=0\,$, then there is one critical point, and it will yield the inflection point; if $b^{2}-3ac<0\,$, then there are no critical points; the cubic is then strictly monotonic.

Bipartite cubics

The graph of

$y^{2}=x(x-a)(x-b)\,$

where $0<a<b$ is called a bipartite cubic. This is from the theory of elliptic curves.

One can graph a bipartite cubic on a graphing device by graphing the function

$f(x)={\sqrt {x(x-a)(x-b)}}\,$

corresponding to the upper half of the bipartite cubic. It is defined on

$(0,a)\cup (b,+\infty ).\,$

See also

• Linear equation
• Quadratic equation
• Quartic equation
• Quintic equation
• Polynomial
• Newton's method
• Spline (mathematics)
• Viete

References

• W. S. Anglin; & J. Lambek (1995). "Mathematics in the Renaissance", in The heritage of Thales, Ch. 24. Springers.
• Lucye Guilbeau (1930). "The History of the Solution of the Cubic Equation", Mathematics News Letter 5 (4), p. 8-12.
• R.W.D. Nickalls (1993). A new approach to solving the cubic: Cardan's solution revealed, The Mathematical Gazette, 77:354–359.

External links

• Calculator for solving Cubics (also solves Quartics and Quadratics)
• Tartaglia's work (and poetry) on the solution of the Cubic Equation at Convergence
• Cubic Equation Solver.
• Quadratic, cubic and quartic equations on MacTutor archive.
• "Cubic Formula". PlanetMath.
• Cardano solution calculator as java applet at some local site. Only takes natural coefficients.
• Graphic explorer for cubic functions With interactive animation, slider controls for coefficients
• On Solution of Cubic Equations at Holistic Numerical Methods Institute
• Dave Auckly, Solving the quartic with a pencil American Math Monthly 114:1 (2007) 29--39
• "Cubic Equation" by Eric W. Weisstein, The Wolfram Demonstrations Project, 2007.
• The affine equivalence of cubic polynomials at Dynamic Geometry Sketches