# Is there something like Cardano's method for a SOLVABLE quintic.

So there is no quadratic formula equivalent for a GENERAL fifth degree equation, but is there an equivalent formula for a SOLVABLE fifth degree equation.

• Galois theory shows how in principle one can use a chain of subgroups establishing that the Galois group is solvable to construct a chain of intermediate fields, each obtained by extension by radicals, ending in a splitting field for your polynomial. The explicit details depend on that chain, and can be quite messy. In fact even the general formula for the quartic is extremely messy if you write it out in full. – Marc van Leeuwen Mar 8 '12 at 8:44

Marc is correct, the Galois group for a solvable irreducible quintic can only have order $5,10$ or $20,$ since these are the only orders of solvable subgroups of $S_5$ which have order divisible by $5,$ so in some ways the possible structures for the Galois group of a solvable quintic are less messy than the structure of the symmetric group $S_4.$ Following the arguments in Galois Theory texts for the relevant Galois groups should be feasible( and is probably already done explicitly somewhere). It might help to note that such an irreducible quintic has either $1$ or $5$ real roots.

• If OP's earlier questions are anything to go by, OP is not just (perhaps not at all) interested in irreducible quintics, but in reducible ones, even (perhaps especially) quintics that split completely over the rationals. The possible structures for the Galois group of a solvable and not-necessarily-irreducible quintic include all those messy subgroups of $S_4$, as well as $S_3$, $S_2$, and $S_1$. I think there are over a dozen different kinds of solvable quintic if you include the reducible ones, which means OP needs a lot of formulas, not just one. – Gerry Myerson Mar 8 '12 at 10:58
• @Gerry Myerson : Yes, if you allow reducible quintics, then it's different matter, but you can of course (in principle) solve the irreducible factors by radicals using the formula for quadratics,cubic and quartics, you don't need any "new" Galois theory. So the irreducible quintic is the only case not covered by the "standard" theory. – Geoff Robinson Mar 8 '12 at 11:10

Yes, there is a general expression for solutions of solvable quintics. A published account, including the relevant formulas, appears in

David S. Dummit. Solving solvable quintics, Math. Comp., 57 (195), (1991), 387–401. MR1079014 (91j:12005). Corrigenda, Math. Comp., 59 (199), (1992), 309. MR1166516.

The corrigenda is actually a microfiche supplement containing an appendix to the paper, that was inadvertently not published with the paper itself. It contains some of the formulas needed for the explicit computation of the roots. To give you an idea of the complexity of the task, the appendix runs for $43$ pages.

From the paper:

It is well known that an irreducible quintic with coefficients in the rational numbers $\mathbb Q$ is solvable by radicals if and only if its Galois group is contained in the Frobenius group $F_{20}$ of order $20$, i.e., if and only if the Galois group is isomorphic to $F_{20}$ , to the dihedral group $D_{10}$ of order $10$, or to the cyclic group $\mathbb Z/5\mathbb Z$.

Explicitly associated to an irreducible quintic $f$ there is a resolvent sextic that Dummit calls $f_{20}$, and $f$ is solvable by radicals iff $f_{20}$ has a rational root, in which case $f_{20}$ factors as a linear term times an irreducible quintic. When this happens, explicit criteria are given for which of the three possibilities above actually occurs, together with the relevant formulas. The paper includes examples showing all three cases are possible.

For any solvable polynomial equation $a_px^p+a_{p-1}x^{p-1}+\cdots+a_1x+a_0=0$ of odd prime degree $p$, the roots have the form:

$x=-\frac{a_{p-1}}{pa_p}+\omega\sqrt[p]{r_1}+\omega^2\sqrt[p]{r_2}+\cdots+\omega^{p-1}\sqrt[p]{r_{p-1}}$

Where $r_1,\cdots,r_{p-1}$ are solutions to a polynomial of degree $p-1$ with coefficients in the same base field as the original equation.

A to determine which polynomial, one method is to set each of these expressions equal to a different root and deriving elementary symmetric polynomials in the $r_i$, and solving a linear system.

Example: $x^5-5x+12$ with one real root and Galois Group $D_5$.

We know that the roots are

$x_1=a+b+c+d$

$x_2=a\text{ cis }\frac{2\pi}{5}+b\text{ cis }\frac{4\pi}{5}+c\text{ cis }\frac{6\pi}{5}+d\text{ cis }\frac{8\pi}{5}$

$x_3=a\text{ cis }\frac{4\pi}{5}+b\text{ cis }\frac{8\pi}{5}+c\text{ cis }\frac{2\pi}{5}+d\text{ cis }\frac{6\pi}{5}$

$x_4=a\text{ cis }\frac{6\pi}{5}+b\text{ cis }\frac{2\pi}{5}+c\text{ cis }\frac{8\pi}{5}+d\text{ cis }\frac{4\pi}{5}$

$x_5=a\text{ cis }\frac{8\pi}{5}+b\text{ cis }\frac{6\pi}{5}+c\text{ cis }\frac{4\pi}{5}+d\text{ cis }\frac{2\pi}{5}$

Here, $a,b,c,d$ are fifth-roots of solutions to a quartic.

If you ask Mathematica to solve the quintic in closed form, it will give you root objects. The first one is real, the next two are one complex conjugate pair, and the last two are the other complex conjugate pair.

The expressions of the pair $(x_1,x_4)$ are complex conjugates, and so are expressions for $(x_2,x_3)$. This gives you a limited number of possible candidates for which $x$ corresponds to which root object that Mathematica gave.

For each candidate assignment, ask mathematica to simplify fully with "FullSimplify", the elementary symmetric polynomials in $a^5,b^5,c^5,d^5$ and look for the case where only rational numbers are spat out. These are $\pm$ coefficients of the auxiliary quartic with $a^5,b^5,c^5,d^5$ as the roots. Mathematica may take a while in this step, so be patient.

In our case, the correct assignments are $x_1=r_1$, $x_2=r_3$, $x_3=r_4$, $x_4=r_5$, $x_5=r_2$.

Here, $r_i$ are the root objects Mathematica gave.

The elementary symmetric polynomials in $a^5,b^5,c^5,d^5$ come out:

$e_1=-4, e_2=\frac{4}{5}$, $e_3=\frac{8}{125}$, $e_4=-\frac{1}{3125}$.

The quartic: $u^4+4u^3+\frac{4}{5}u^2-\frac{8}{125}u-\frac{1}{3125}=0$

Solving it and matching up the expressions correctly to $a^5,b^5,c^5,d^5$, we are done.

Note that mathematica evaluates $n$th roots of negative real numbers to complex numbers with argument $\frac{2\pi}{n}$, instead of evaluating them to be equal to negative real when $n$ is odd. You must account for that to check your work by flipping sign on the entire expression in the fifth root and the sign in front of the fifth root if the inside expression is negative real.

Final answer (accounting Mathematica's $n$th root preferences):

$x= -\omega\sqrt{1+\frac{2}{5}\sqrt{5}-\frac{3}{25}\sqrt{125+55\sqrt{5}}}+\omega^2\sqrt{-1+\frac{2}{5}\sqrt{5}+\frac{3}{25}\sqrt{125-55\sqrt{5}}}-\omega^3\sqrt{1-\frac{2}{5}\sqrt{5}+\frac{3}{25}\sqrt{125-55\sqrt{5}}}-\omega^4\sqrt{1+\frac{2}{5}\sqrt{5}+\frac{3}{25}\sqrt{125+55\sqrt{5}}}$

Theoretically, these steps can be repeated for any quintic, and replicated on any polynomial of odd prime degree. The only obstacle is in solving the auxiliary polynomial of degree $p-1$, beginning with $p=7$ onwards, which would require finding an intermediate field in which it factors. This can be done by assuming factorization and solving a system in Mathematica.