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Graph of a polynomial of degree 5, with 4 critical points

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

$g(x)=ax^5+bx^4+cx^3+dx^2+ex+f,\,$

where a, b, c, d, e and f are members of a field, typically the rational numbers, the real numbers or the complex numbers, and a is nonzero. In other words, a quintic function is defined by a polynomial of degree five.

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

$ax^5+bx^4+cx^3+dx^2+ex+f=0.\,$

If a is zero but one of the other coefficients is non-zero, the equation is classified as either a quartic equation, cubic equation, quadratic equation or linear equation.

Because they have an odd degree, normal quintic functions appear similar to normal cubic functions when graphed, except they may possess an additional local maximum and local minimum each. The derivative of a quintic function is a quartic function.

## Finding roots of a quintic equation

Finding the roots of a given polynomial has been a prominent mathematical problem.

Solving linear, quadratic, cubic and quartic equations by factorization into radicals is fairly straightforward, no matter whether the roots are rational or irrational, real or complex; there are also formulae that yield the required solutions. However, there is no formula for general quintic equations over the rationals in terms of radicals; this is known as the Abel–Ruffini theorem, first published in 1824, which was one of the first applications of group theory in algebra. This result also holds for equations of higher degrees. An example quintic whose roots cannot be expressed by radicals is $x^5 - x + 1 = 0.$ This quintic is in Bring–Jerrard normal form.

As a practical matter, exact analytic solutions for polynomial equations are often unnecessary, and so numerical methods such as Laguerre's method or the Jenkins-Traub method are probably the best way of obtaining solutions to general quintics and higher degree polynomial equations that arise in practice. However, analytic solutions are sometimes useful for certain applications, and many mathematicians have tried to develop them.

### Solvable quintics

Some fifth-degree equations can be solved by factorizing into radicals; for example, $x^5 - x^4 - x + 1 = 0\,$, which can be written as $(x^2 + 1) (x + 1) (x - 1)^2 = 0\,$, or $x^5 -2 = 0\,$, which has $\sqrt[5]{2}\,$ as solution. Other quintics like $x^5 - x + 1 = 0\,$ cannot be solved by radicals. Évariste Galois developed techniques for determining whether a given equation could be solved by radicals which gave rise to group theory and Galois theory. Applying these techniques, Arthur Cayley found a general criterion for determining whether any given quintic is solvable.[1] This criterion is the following.[2]

Given the equation

$ax^5+bx^4+cx^3+dx^2+ex+f=0,$

the Tschirnhaus transformation $x=y-\frac{b}{5a}$, which depresses the quintic, gives the equation

$y^5+p y^3+q y^2+r y+s=0$,

where

\begin{align} p &= \frac{5ac-2b^2}{5a^2}\\ q &= \frac{25a^2d-15abc+4b^3}{25a^3}\\ r &= \frac{125a^3e-50a^2bd+15ab^2c-3b^4}{125a^4}\\ s &= \frac{3125 a^4f-625a^3 be+125a^2b^2 d-25ab^3 c+4 b^5}{3125a^5}. \end{align}

Both equations are solvable by radicals if and only if either they are factorisable in equations of lower degrees with rational coefficients or the polynomial $P^2-1024z\Delta$, named Cayley resolvent, has a rational root in z, where

$P =z^3-z^2(20r+3p^2)- z(8p^2r - 16pq^2- 240r^2 + 400sq - 3p^4)$
${} - p^6 + 28p^4r- 16p^3q^2- 176p^2r^2- 80p^2sq + 224prq^2- 64q^4$
${} + 4000ps^2 + 320r^3- 1600rsq$

and

$\Delta=-128p^2r^4+3125s^4-72p^4qrs+560p^2qr^2s+16p^4r^3+256r^5+108p^5s^2$
${} -1600qr^3s+144pq^2r^3-900p^3rs^2+2000pr^2s^2-3750pqs^3+825p^2q^2s^2$
${} +2250q^2rs^2+108q^5s-27q^4r^2-630pq^3rs+16p^3q^3s-4p^3q^2r^2.$

In 1888, George Paxton Young[3] described how to solve a solvable quintic equation, without providing an explicit formula; Daniel Lazard writes out a three-page formula (Lazard (2004)).

During the second half of 19th century, John Stuart Glashan, George Paxton Young, and Carl Runge found that any irreducible quintic with rational coefficients in Bring-Jerrard form,

$x^5 + ax + b = 0\,$

is solvable by radicals if and only if either a = 0 or it is of the following form:

$x^5 + \frac{5\mu^4(4\nu + 3)}{\nu^2 + 1}x + \frac{4\mu^5(2\nu + 1)(4\nu + 3)}{\nu^2 + 1} = 0$

where $\mu$ and $\nu$ are rational. In 1994, Blair Spearman and Kenneth S. Williams gave an alternative,

$x^5 + \frac{5e^4( 4c + 3)}{c^2 + 1}x + \frac{-4e^5(2c-11)}{c^2 + 1} = 0.$

The relationship between the 1885 and 1994 parameterizations can be seen by defining the expression

$b = \frac{4}{5} \left(a+20 \pm 2\sqrt{(20-a)(5+a)}\right)$

where

$a = \frac{5(4\nu+3)}{\nu^2+1}$

and using the negative case of the square root yields, after scaling variables, the first parametrization while the positive case gives the second. It is then a necessary (but not sufficient) condition that the irreducible solvable quintic

$z^5 + a\mu^4z + b\mu^5 = 0\,$

with rational coefficients must satisfy the simple quadratic curve

$y^2 = (20-a)(5+a)\,$

for some rational $a, y$.

The substitution $c=-m/l^5$, $e=1/l$ in Spearman-Williams parameterization allows to not exclude the special case a = 0, giving the following result:

If a and b are rational numbers, the equation $x^5+ax+b=0$ is solvable by radicals if either its left hand side is a product of polynomials of degree less than 5 with rational coefficients or there exist two rational numbers l and m such that

$a=\frac{5 l (3 l^5-4 m)}{m^2+l^{10}}\qquad b=\frac{4(11 l^5+2 m)}{m^2+l^{10}}$.

### Examples of solvable quintics

A quintic is solvable using radicals if the Galois group of the quintic (which is a subgroup of the symmetric group S(5) of permutations of a five element set) is a solvable group. In this case the form of the solutions depends on the structure of this Galois group.

A simple example is given by the equation $x^5-5x^4+30x^3-50x^2+55x-21=0\,,$ whose Galois group is the group F(5) generated by the cyclic permutations (1 2 3 4 5) and (1 2 4 3); the only real solution is

$x=1+\sqrt[5]{2}-\sqrt[5]{4}+\sqrt[5]{8}-\sqrt[5]{16}\,.$

However, for other solvable Galois groups, the form of the roots can be much more complex. For example, the equation $x^5-5x+12=0\,$ has Galois group D(5) generated by "(1 2 3 4 5)" and "(1 4)(2 3)" and the solution requires more symbols to write. Define

$a = \sqrt{2/\phi}$
$b = \sqrt{2\phi}$
$c = 5^{1/4}\,,$

where $\phi = (1+\sqrt{5})/2$ is the golden ratio, then the only real solution $x = -1.84208\dots$ is given by

$-(5^{1/4})x = \sqrt[5]{(a+c)^2(b-c)} + \sqrt[5]{(-a+c)(b-c)^2} + \sqrt[5]{(a+c)(b+c)^2} - \sqrt[5]{(-a+c)^2(b+c)} \,,$

or, equivalently, by

$x = y_1^{1/5}+y_2^{1/5}+y_3^{1/5}+y_4^{1/5}\,,$

where the yi are the four roots of the quartic equation

$y^4+4y^3+\frac{4}{5}y^2-\frac{8}{5^3}y-\frac{1}{5^5}=0\,.$

In general, if an equation P(x) = 0 of prime degree p with rational coefficients is solvable in radicals, then each root is the sum of p-th roots of the roots of an auxiliary equation Q(y) = 0 of degree (p-1), also with rational coefficients, that can be used to solve the former. However these p-th roots may not be computed independently (this would provide pp roots instead of p). Thus a correct solution needs to express all these p-roots in term of one of them. Galois theory shows that this is always theoretically possible, even if the resulting formula may be to large to be of any use.

It is possible that some of the roots of Q(y) = 0 are rational (like in the above example with F(5) Galois group) or some are zero. When it is the case the formula for the roots is much simpler, like for the solvable de Moivre quintic

$x^5+5ax^3+5a^2x+b = 0\,,$

where the auxiliary equation has two zero roots and reduces, by factoring them out, to the quadratic equation

$y^2+by-a^5 = 0\,,$

such that the five roots of the de Moivre quintic are given by

$x_k = \omega^k\sqrt[5]{y_i} -\frac{a}{\omega^k\sqrt[5]{y_i}},$

where yi is any root of the auxiliary quadratic equation and ω is any of the four primitive 5th roots of unity. This can be easily generalized to construct a solvable septic and other odd degrees, not necessarily prime.

Here is a list of known solvable quintics:

There are infinitely many solvable quintics in Bring-Jerrard form which have been parameterized in preceding section.

Up to the scaling of the variable, there are exactly five solvable quintics of the shape $x^5+ax^2+b$, which are[4] (where s is a scaling factor):

$x^5-2s^3x^2-\frac{s^5}{5}$
$x^5-100s^3x^2-1000s^5$
$x^5-5s^3x^2-3s^5$
$x^5-5s^3x^2+15s^5$
$x^5-25s^3x^2-300s^5$

Paxton Young (1888) gave a number of examples, some of them being reducible, having a rational root:

 $x^5-10x^3-20x^2-1505x-7412$ $x^5+\frac{625}{4}x+3750$ $x^5-\frac{22}{5}x^3-\frac{11}{25}x^2+\frac{462}{125}x+\frac{979}{3125}$ $x^5+20x^3+20x^2+30x+10$ Solution: $\sqrt[5]{2}-\sqrt[5]{2}^2+\sqrt[5]{2}^3-\sqrt[5]{2}^4$ $x^5+320x^2-1000x+4288$ Reducible: −8 is a root $x^5+40x^2-69x+108$ Reducible: −4 is a root $x^5-20x^3+250x-400$ $x^5-5x^3+\frac{85}{8}x-13/2$ $x^5+\frac{20}{17}x+\frac{21}{17}$ $x^5-\frac{4}{13}x+\frac{29}{65}$ $x^5+\frac{10}{13}x+\frac{3}{13}$ $x^5+110(5x^3+60x^2+800x+8320)$ $x^5-20 x^3 -80 x^2 -150 x -656$ $x^5 -40 x^3 +160 x^2 +1000 x -5888$ $x^5-50x^3-600x^2-2000x-11200$ $x^5+110(5 x^3 + 20x^2 -360 x +800)$ $x^5- 20 x^3 +320 x^2 +540 x + 6368$ Reducible : -8 is a root $x^5-20 x^3 -160 x^2 -420 x -8928$ Reducible : 8 is a root $x^5-20 x^3 +170 x + 208$

An infinite sequence of solvable quintics may be constructed, whose roots are sums of n-th roots of unity, with n = 10 k + 1 being a prime number:

 $x^5+x^4-4x^3-3x^2+3x+1$ Roots: $2\cos(\frac{2k\pi}{11})$ $x^5+x^4-12x^3-21x^2+x+5$ Root: $\sum_{k=0}^5 e^\frac{2i\pi 6^k }{31}$ $y^5+y^4-16y^3+5y^2+21y-9$ Root: $\sum_{k=0}^7 e^\frac{2i\pi 3^k }{41}$ $y^5+y^4-24y^3-17y^2+41y-13$ Root: $\sum_{k=0}^{11} e^\frac{2i\pi (21)^k }{61}$ $y^5+y^4-28y^3+37y^2+25y+1$ Root: $\sum_{k=0}^{13} e^\frac{2i\pi (23)^k }{71}$

There are also two parameterized families of solvable quintics: The Kondo–Brumer quintic,

$x^5+(a-3)x^4+(-a+b+3)x^3+(a^2-a-1-2b)x^2+bx+a = 0\,$

and the family depending on the parameters $a, l, m$

$x^5-5p(2x^3 + ax^2 + bx)-pc = 0\,$

where

$p=\frac{l^2(4m^2+a^2)-m^2}{4}, \qquad b=l(4m^2+a^2)-5p-2m^2, \qquad c=\frac{b(a+4m)-p(a-4m)-a^2m}{2}$

If the Galois group of a quintic is not solvable, then the Abel-Ruffini theorem tells us that to obtain the roots it is necessary to go beyond the basic arithmetic operations and the extraction of radicals. About 1835, Jerrard demonstrated that quintics can be solved by using ultraradicals (also known as Bring radicals), the real roots of $t^5 + t - a=0\,$ for real numbers $a\,$. In 1858 Charles Hermite showed that the Bring radical could be characterized in terms of the Jacobi theta functions and their associated elliptic modular functions, using an approach similar to the more familiar approach of solving cubic equations by means of trigonometric functions. At around the same time, Leopold Kronecker, using group theory developed a simpler way of deriving Hermite's result, as had Francesco Brioschi. Later, Felix Klein came up with a method that relates the symmetries of the icosahedron, Galois theory, and the elliptic modular functions that feature in Hermite's solution, giving an explanation for why they should appear at all, and developed his own solution in terms of generalized hypergeometric functions.[5] Similar phenomena occur in degree 7 (septic equations) and 11, as studied by Klein and discussed in icosahedral symmetry: related geometries.

## References

1. ^ A. Cayley. On a new auxiliary equation in the theory of equation of the fifth order, Philosophical Transactions of the Royal Society of London (1861).
2. ^ This formulation of Cayley's result is extracted from Lazard (2004) paper.
3. ^ George Paxton Young. Solvable Quintics Equations with Commensurable Coefficients American Journal of Mathematics 10 (1888), 99–130 at JSTOR
4. ^ http://www.math.harvard.edu/~elkies/trinomial.html
5. ^ (Klein 1888); a modern exposition is given in (Tóth 2002, Section 1.6, Additional Topic: Klein's Theory of the Icosahedron, p. 66)
• Charles Hermite, "Sur la résolution de l'équation du cinquème degré",Œuvres de Charles Hermite, t.2, pp. 5–21, Gauthier-Villars, 1908.
• Felix Klein, Lectures on the Icosahedron and the Solution of Equations of the Fifth Degree, trans. George Gavin Morrice, Trübner & Co., 1888. ISBN 0-486-49528-0.
• Leopold Kronecker, "Sur la résolution de l'equation du cinquième degré, extrait d'une lettre adressée à M. Hermite", Comptes Rendus de l'Académie des Sciences," t. XLVI, 1858 (1), pp. 1150–1152.
• Blair Spearman and Kenneth S. Williams, "Characterization of solvable quintics $x^5 + ax + b$", American Mathematical Monthly, Vol. 101 (1994), pp. 986–992.
• Ian Stewart, Galois Theory 2nd Edition, Chapman and Hall, 1989. ISBN 0-412-34550-1. Discusses Galois Theory in general including a proof of insolvability of the general quintic.
• Jörg Bewersdorff, Galois theory for beginners: A historical perspective, American Mathematical Society, 2006. ISBN 0-8218-3817-2. Chapter 8 (The solution of equations of the fifth degree) gives a description of the solution of solvable quintics $x^5 + cx + d$.
• Victor S. Adamchik and David J. Jeffrey, "Polynomial transformations of Tschirnhaus, Bring and Jerrard," ACM SIGSAM Bulletin, Vol. 37, No. 3, September 2003, pp. 90–94.
• Ehrenfried Walter von Tschirnhaus, "A method for removing all intermediate terms from a given equation," ACM SIGSAM Bulletin, Vol. 37, No. 1, March 2003, pp. 1–3.
• Daniel Lazard, "Solving quintics in radicals", in Olav Arnfinn Laudal, Ragni Piene, The Legacy of Niels Henrik Abel, pp. 207–225, Berlin, 2004,. ISBN 3-540-43826-2. available at http://www.loria.fr/publications/2002/A02-R-449/A02-R-449.ps (broken link)
• Tóth, Gábor (2002), Finite Möbius groups, minimal immersions of spheres, and moduli
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