How to solve a quintic polynomial equation?

Solution 1:

If you haven't reached that part of Galois theory yet, there is still a "simple" elementary way to test if a quintic is solvable. First, reduce it to depressed form (without the $x^4$ term).

Method 1:

Theorem (by Watson, 1930s): "Given an irreducible quintic with rational coefficients,

$$x^5 + 10c x^3 + 10d x^2 + 5 e x + f = 0\tag1$$

If the sextic,

$$3125p^6 - 625(3c^2 + e)p^4 + 25(15c^4 + 8c d^2 - 2c^2e + 3e^2 - 2d f)p^2 + p\sqrt{D} + (-25c^6 - 40c^3d^2 - 16d^4 + 35c^4e + 28c d^2e - 11c^2e^2 + e^3 - 2c^2d f - 2d e f + c f^2)=0$$

and discriminant $D$,

$$D=-3200c^3d^2e^2 - 2160d^4e^2 + 6400c^4e^3 + 5760c d^2e^3 - 2560c^2e^4 + 256e^5 + 5120c^3d^3f + 3456d^5f - 11520c^4d e f - 10080c d^3e f + 4480c^2d e^2f - 640d e ^3f + 3456c^5f^2 + 2640c^2d^2f^2 - 1440c^3e f^2 + 360d^2e f^2 + 160c e^2f^2 - 120c d f^3 + f^4$$

has a root $p$ such that $p^2$ is rational, then $(1)$ is a solvable quintic."

See "Commentary on an unpublished lecture by G. N. Watson on solving the quintic" by Bruce Berndt. This is easily implemented in Mathematica and is useful when dealing with parametric quintics.

Method 2:

If you are in a rush and just want to determine if a particular equation is solvable, you can find the order of its Galois group using this online Magma calculator. For example, to test the solvable but irreducible $x^5-5x+12=0$, copy and paste the command,

Z := Integers(); P < x > := PolynomialRing(Z); f := x^5-5*x+12; G, R := GaloisGroup(f); G;

One then finds the order is $10$, hence that quintic is solvable. All groups with order $<60$ are, though there are solvable groups with order $>60$.

Note: Don't forget the asterisk (*) between the numerical coefficient and the variable, like this: 5*x.

Addendum on Method 1:

We can reduce Method 1 to a simple rational root test by "root squaring".

Render the sextic equation above as

$\alpha p^6 + \beta p^4 + \gamma p^2 + p\sqrt{D} + \delta = 0$

Separate the odd degree term from the even degree ones:

$\alpha p^6 + \beta p^4 + \gamma p^2 + \delta = - p\sqrt{D}$

And square. Only even powers of $p$ now appear so we have a polynomial equation in $p^2$::

$(\alpha p^6 + \beta p^4 + \gamma p^2 + \delta)^2=p^2D$

$\alpha^2(p^2)^6 + 2\alpha \beta(p^2)^5 + (2\alpha \gamma + \beta^2)(p^2)^4 + 2(\alpha \delta + \beta \gamma)(p^2)^3 + (2\beta \delta + \gamma^2)(p^2)^2 + (2 \gamma \delta - D)(p^2) + \delta^2 = 0$

You may then apply the standard rational root test to this equation. If it passes, the conditions for solvability of the quintic are satisfied.