Algebra – Solving a Polynomial Equation in Closed Form

abstract-algebraalgebra-precalculusgalois-theorypolynomialsradicals

How can we solve the following quintic in closed form ?
$$16x^5-200x^3-200x^2+25x+30=0$$

What is special about this equation? What can we say about solvability? I'm not particularly into math. I saw this equation in a discussion group. Here are some ideas about the equation from non-mathematicians .

  1. There is no closed-form formula for $5$-degree equations. Therefore there is no solution. In my opinion this is wrong. For example, $x^5=1$ can be easily solved. The true version if that idea would be there is no general solution formula . (by radicals)

  2. The equation is not factored. Therefore, there is no closed-form solution. This idea is also wrong in my opinion. Because the factors do not have to be rational numbers.

  3. Wolfram Alpha does not give closed form for the equation. Therefore, it cannot be solved. I cannot comment on this idea.

I found that the quintic is irreducible over $\Bbb Q$.

I've heard a little bit about Galois theory. But I don't have the mathematical knowledge. If possible, could you help to find out if this equation is solvable? The problem author argues that the equation has a solution.

I tried WA software several times. But, it does not succeed.

Best Answer

This polynomial $f$ has Galois group $F_5$, the Frobenius group of order 20. Since this group is solvable, then $f$ is solvable by radicals. Here are Magma commands showing this.

R<x> := PolynomialRing(Rationals());
f := 16*x^5-200*x^3-200*x^2+25*x+30;
G, data := GaloisGroup(f);
TransitiveGroupDescription(G);
IsSolvable(G);

which give as output

F(5) = 5:4
true

The docstring for the command SolveByRadicals describes how it gives solutions to a solvable polynomial:

Compute a splitting field as a radical tower and the roots of $f$ as elements in this tower. Return also the non-trivial roots of unity used.

(See here for more details.) We enter

K, roots := SolveByRadicals(f : Name := "K.");
K:Maximal;

which yields output

   K<K.1>
     |
     |
  $1<K.2>
     |
     |
  $2<K.3>
     |
     |
  $3<K.4>
     |
     |
     Q

K  : K.1^5 + 1/512*(-1875*K.4 - 3125)*K.3*K.2 - 2560000000
$1 : K.2^2 + 15118284881920000*K.4 - 34359738368000000
$2 : K.3^2 + 32*K.4 + 160
$3 : K.4^2 - 5

Denote by $\alpha_i$ the elements K.i for $i = 1, \ldots, 4$. Then this output is saying that we have a chain of simple radical extensions $$ K = K_1 \supseteq K_2 \supseteq K_3 \supseteq K_4 \supseteq \mathbb{Q} $$ where $K_i = K_{i+1}(\alpha_i)$ and the $\alpha_i$ satisfy the equations \begin{align*} 0&={\alpha_1}^5+1/512(-1875\alpha_4-3125)\alpha_3\alpha_2-2560000000\\ 0&={\alpha_2}^2+15118284881920000\alpha_4-34359738368000000\\ 0&={\alpha_3}^2+32\alpha_4+160\\ 0&={\alpha_4}^2-5\,.\end{align*} Or in other words, \begin{align*} \alpha_4 &= \sqrt{5}\\ \alpha_3 &= \sqrt{-32 \alpha_4 - 160}\\ \alpha_2 &= \sqrt{-15118284881920000 \alpha_4 + 34359738368000000}\\ \alpha_1 &= \sqrt[5]{1/512 (1875 \alpha_4 + 3125) \alpha_3 \alpha_2 + 2560000000} \end{align*} up to choosing the correct roots of unity when taking these roots.

To express the roots of $f$ in terms of these generators, we enter [K!r : r in roots];, which yields

[
(1/268435456000000000*(-3*K.4 - 5)*K.3*K.2 + 1/128000000)*K.1^4 + 1/640000*K.1^3 + (1/33554432000000*(-3*K.4 - 5)*K.3*K.2 + 1/16000)*K.1^2 + 1/80*K.1,
((1/268435456000000000*(2*K.4 + 5)*K.3 + 1/13421772800000000*(K.4 + 3))*K.2 + (1/4096000000*(K.4 - 1)*K.3 + 1/512000000*(-K.4 - 1)))*K.1^4 + (-1/10240000*K.3 + 1/2560000*(K.4 - 1))*K.1^3 + ((1/67108864000000*(-K.4 - 5)*K.3 + 1/838860800000*(K.4 + 2))*K.2 + (1/256000*K.3 + 1/64000*(K.4 - 1)))*K.1^2 + (1/2560*(-K.4 + 1)*K.3 + 1/320*(-K.4 - 1))*K.1,
((1/536870912000000000*(-K.4 - 5)*K.3 + 1/6710886400000000*(K.4 + 2))*K.2 + (1/2048000000*K.3 + 1/512000000*(K.4 - 1)))*K.1^4 + (1/20480000*(K.4 - 1)*K.3 + 1/2560000*(-K.4 - 1))*K.1^3 + ((1/33554432000000*(2*K.4 + 5)*K.3 + 1/1677721600000*(-K.4 - 3))*K.2 + (1/512000*(-K.4 + 1)*K.3 + 1/64000*(-K.4 - 1)))*K.1^2 + (-1/1280*K.3 + 1/320*(K.4 - 1))*K.1,
((1/268435456000000000*(2*K.4 + 5)*K.3 + 1/13421772800000000*(-K.4 - 3))*K.2 + (1/4096000000*(-K.4 + 1)*K.3 + 1/512000000*(-K.4 - 1)))*K.1^4 + (1/10240000*K.3 + 1/2560000*(K.4 - 1))*K.1^3 + ((1/67108864000000*(-K.4 - 5)*K.3 + 1/838860800000*(-K.4 - 2))*K.2 + (-1/256000*K.3 + 1/64000*(K.4 - 1)))*K.1^2 + (1/2560*(K.4 - 1)*K.3 + 1/320*(-K.4 - 1))*K.1,
((1/536870912000000000*(-K.4 - 5)*K.3 + 1/6710886400000000*(-K.4 - 2))*K.2 + (-1/2048000000*K.3 + 1/512000000*(K.4 - 1)))*K.1^4 + (1/20480000*(-K.4 + 1)*K.3 + 1/2560000*(-K.4 - 1))*K.1^3 + ((1/33554432000000*(2*K.4 + 5)*K.3 + 1/1677721600000*(K.4 + 3))*K.2 + (1/512000*(K.4 - 1)*K.3 + 1/64000*(-K.4 - 1)))*K.1^2 + (1/1280*K.3 + 1/320*(K.4 - 1))*K.1
]

You can try this code for yourself using the Magma online calculator.

Related Question