How can one feel comfortable with non-solvable algebraic numbers?
The nice thing about solvable numbers is this idea that they have a formula. You can manipulate the formula as if it were actually a number using some algebra formalism that you probably have felt comfortable with for a while. For instance $\sqrt{3+\sqrt{6}}+2$ is an algebraic number. What do you get if you add it to $7$? Well $\left(\sqrt{3+\sqrt{6}}+2\right)+7=\sqrt{3+\sqrt{6}}+9$ seems like a decent answer. As a side note: there actually some reasonably hard algorithmic questions along these lines, but I'll assume they don't worry you. :-)
We'd like to be able to manipulate other algebraic numbers with similar comfort. The first method I was taught is pretty reasonable:
Kronecker's construction: If $x$ really is an algebraic number, then it is the root of some irreducible polynomial $x^n - a_{n-1} x^{n-1} - \ldots - a_1 x - a_0$. But how do we manipulate $x$? It's almost silly: we treat it just like $x$, and add and multiply as usual, except that $x \cdot x^{n-1}$ needs to be replaced by $a_{n-1} x^{n-1} + \ldots + a_1 x + a_0$, and division is handled by replacing $1/x$ with $( x^{n-1} - a_{n-1} x^{n-2} - \ldots - a_2 x - a_1)/a_0$. This is very similar to "integers mod n" where you replace big numbers by their remainder mod n,. In fact this is just replacing a polynomial in $x$ with its remainder mod $x^n - a_{n-1} x^{n-1} - \ldots - a_1 x - a_0$.
I found it somewhat satisfying, but in many ways it is very mysterious. We use the same symbol for many different algebraic numbers; each time we have to keep track of the $f(x)$ floating in the background. Also it raises deep questions about how to tell two algebraic numbers apart. Luckily more or less all of these questions have clean algorithmic answers, and they are described in Cohen's textbooks CCANT (A Course in Computational Algebraic Number Theory, Henri Cohen, 1993).
Companion matrices: But years later, it still bugged me. Then I studied splitting fields of group representations. The crazy thing about these fields is that they are subrings of matrix rings. So “numbers” were actually matrices. You've probably seen some tricks like this $$\mathbb{C} = \left\{ \begin{bmatrix} a & b \\ -b & a \end{bmatrix} : a,b \in \mathbb{R} \right\}$$ where we can make a bigger field out of matrices over a smaller field. It turns out that is always true: If $K \leq F$ are fields, then $F$ is a $K$-vector space, and the function $f:F \to M_n(K) : x \mapsto ( y \mapsto xy )$ is an injective homomorphism of fields, so that $f(F)$ is a field isomorphic to $F$ but whose “numbers” are just $n \times n$ matrices over $K$, where $n$ is the dimension of $F$ as a $K$-vector space (and yes $n$ could be infinite if you want, but it's not).
That might seem a little complicated, but $f$ just says "what does multiplying look like?" For instance if $\mathbb{C} = \mathbb{R} \oplus \mathbb{R} i$ then multiplying $a+bi$ sends $1$ to $a+bi$ and $i$ to $-b+ai$. The first row is $[a,b]$ and the second row $[-b,a]$. Too easy.
Ok, fine, but that assumes you already know how to multiply, and perhaps you are not yet comfortable enough to multiply non-solvable algebraic numbers! Again we use the polynomial $x^n - a_{n-1} x^{n-1} - \ldots - a_1 x - a_0$, but this time as a matrix. We use the same rule, viewing $F=K \oplus Kx \oplus Kx^2 \oplus \ldots \oplus Kx^{n-1}$ and ask what $x$ does to each basis element: well $x^i$ is typically sent to $x^{i+1}$. It's only the last one that things get funny:
$$f(x) = \begin{bmatrix} 0 & 1 & 0 & 0 & \ldots & 0 & 0 \\ 0 & 0 & 1 & 0 & \ldots & 0 & 0 \\
0 & 0 & 0 & 1 & \ldots & 0 & 0 \\
& & & & \ddots & & \\
0 & 0 & 0 & 0 & \ldots & 1 & 0 \\
0 & 0 & 0 & 0 & \ldots & 0 & 1 \\
a_0 & a_1 & a_2 & a_3 & \ldots & a_{n-2} & a_{n-1}
\end{bmatrix}$$
So this fancy “number” $x$ just becomes a matrix, most of whose entries are $0$. For instance $x^2 - (-1)$ gives the matrix $i = \left[\begin{smallmatrix} 0 & 1 \\ -1 & 0 \end{smallmatrix}\right]$.
This nice part here is that different algebraic numbers can actually have different matrix representations. The dark part is making sure that if you have two unrelated algebraic numbers that they actually multiply up like a field. You see $M_n(K)$ has many subfields, but is not itself a field, so you have to choose matrices that both lie within a subfield. Now for splitting fields and centralizer fields and all sorts of handy dandy fancy fields, you absolutely can make sure everything you care about comes from the field. Starting from just a bunch of polynomials though, you need to be careful and find a single polynomial that works for both. This is called the primitive element theorem.
This also lets you see the difference between eigenvalues in the field $K$ and eigenvalues (“numbers”) in the field $F$: the former are actually numbers, or multiples of the identity matrix, while the latter are full-fledged matrices that happen to lie in a subfield. If you ever studied the “real form” of the eigenvalue decomposition with $2\times 2$ blocks, those $2 \times 2$ blocks are exactly the $\begin{bmatrix}a&b\\-b&a\end{bmatrix}$ complex numbers.
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.
Best Answer
Unless you have reason to believe otherwise, it is likely that this polynomial cannot be factored algebraically. However, you can gain some insight into its roots, including the following (you're welcome to prove some or all of these claims yourself):
Since the coefficients of the polynomial are all positive, its real roots can only be negative.
Since its derivative is strictly positive over the real numbers, it is strictly increasing and therefore only has one real root.
As a result, its other 4 roots must come as two complex conjugate pairs, i.e. the roots can be written as $a_1, a_2 \pm i b_2, a_3 \pm i b_3$.
Using Vieta's formulas, we can produce some relationships between the roots, starting with $a_1 + a_2 + a_3 = -\frac{1}{3}$.