<s>[INST] What is the Galois group of $x^6+x^5-8x^4-5x^3+19x^2+4x-11=0$? A while ago I happened on this equation: $$f(x)=x^6+x^5-8x^4-5x^3+19x^2+4x-11=0.$$
$f$ does not have linear factors, as can be seen by drawing its graph. I am not certain about whether it has quadratic or cubic factors over $\mathbb{Q}$, but over $\mathbb{Q}(\sqrt{5}$) it factors into $$(x^3+\frac{1}{2}(1-\sqrt{5})x^2+\frac{1}{2}(-\sqrt{5}-7)x+\frac{1}{2}(1+3\sqrt{5}))*$$ $$(x^3+\frac{1}{2}(1+\sqrt{5})x^2+\frac{1}{2}(\sqrt{5}-7)x+\frac{1}{2}(1-3\sqrt{5})). $$
This seems to say that one can first extend $\mathbb{Q}$ to $\mathbb{Q}(\sqrt{5})$, then extend somehow to include the roots of the two cubics. I do note that the polynomial $f(x^2-3)$ is divisible by $f(x)$ with no remainder. This means that if $r$ is a root of $f(x)$; i.e. a root of either of these two cubics, then $r^2-3$ is also a root. In fact I constructed the polynomial by evaluating $g(g(g(x)))-x$, where $g(x)=x^2-3$, and throwing out the quadratic factor. 
I evaluated numerically the roots of the equation and came up with
(1) -2.22918
(2) 1.96925
(3) 0.877962
(4) -2.10553
(5) 1.43326
(6) -0.94576
$g$ permutes these roots; in fact, $g$ as a permutation is (123)(456), using the labels above. So $g$ is an element of the Galois group. There is also an order-2 mapping of the roots, namely the mapping $h$ that takes 1 to 1 and $\sqrt{5}$ to $-\sqrt{5}$; the non-zero element of the Galois group of $\mathbb{Q}(\sqrt{5})$ over $\mathbb{Q}$. I have tried generating subgroups of $S_6$ by using $g$ and some triple transposition (for $h$) for several possibilities of the triple transposition, and in each case I came up with $\mathbb{Z}_6$ as the Galois group. But if that's the case, then a cubic extension of $\mathbb{Q}(\sqrt{5})$ to an extension field $E$ would split the equation; $E$ would be the splitting field. But that then says the roots of one of the cubics above can be written as combinations of the roots of the others using only numbers from $\mathbb{Q}(\sqrt{5})$ as coefficients, but I don't see how this is done. It still seems to me the Galois group could be a group of order 18. How does one show that this is not the case?
 [/INST] Irreducible over the rationals. Galois group as Robert said; I added two degree six polynomials where the roots are also real of modest absolute value, but come out very pretty, method originally due to Gauss.
================================================================
? f = x^6 + x^5 - 8 * x^4 - 5 * x^3 + 19 * x^2 + 4 * x - 11
%13 = x^6 + x^5 - 8*x^4 - 5*x^3 + 19*x^2 + 4*x - 11
? factor(f)
%14 = 
[x^6 + x^5 - 8*x^4 - 5*x^3 + 19*x^2 + 4*x - 11 1]

? polgalois(f)
%2 = [18, -1, 1, "F_18(6) = [3^2]2 = 3 wr 2"]
? poldisc(f)
%3 = 2415125
? factor(poldisc(f))
%4 = 
[5 3]

[139 2]
===================================================================
? f = x^6 + x^5 - 5 * x^4 - 4 * x^3 + 6 * x^2 + 3 * x - 1
%5 = x^6 + x^5 - 5*x^4 - 4*x^3 + 6*x^2 + 3*x - 1
? 
? polgalois(f)
%6 = [6, -1, 1, "C(6) = 6 = 3[x]2"]
? poldisc(f)
%7 = 371293
? factor(poldisc(f))
%8 = 
[13 5]

===========================================================
? 
? 
? f = x^6 + x^5 - 15 * x^4 - 28 * x^3 + 15 * x^2 + 38 * x - 1
%9 = x^6 + x^5 - 15*x^4 - 28*x^3 + 15*x^2 + 38*x - 1
? polgalois(f)
%10 = [6, -1, 1, "C(6) = 6 = 3[x]2"]
? poldisc(f)
%11 = 8390618797
? factor(poldisc(f))
%12 = 
[11 2]

[37 5]
============================================================

One of them has roots $2  \cos \left( \frac{2\pi}{13} \right),2  \cos \left( \frac{4\pi}{13} \right), $ and so on. The last one has a root $2  \cos \left( \frac{2\pi}{37} \right) +2  \cos \left( \frac{20\pi}{37} \right) +2  \cos \left( \frac{22\pi}{37} \right), $ the other five are also sums of three cosine terms.
</s>