What does it mean when polynomials have closed, exact complex solutions, but not exact real solutions? I was watching this introduction to peturbation theory. His first example is solving
$$x^5 + x = 1$$
for which he claims there is no exact real solution. I asked WolframAlpha what it thought.
It gives an inexact decimal solution $x \approx -0.75488...$ and some exact complex solutions
$$x = -\sqrt[3]{-1}$$
$$x = (-1)^\frac{2}{3}$$


Is there some deep reason as to why the complex roots would have exact forms but not the real root?
Could we have an $n$-degree polynomial with $a$ exact solutions and $b$ inexact solutions, for arbitrary $a+b=n$?
Can the exact and inexact solutions be distributed arbitrarily between the real line and the rest of the complex plane?
Can we say anything in general, or is this just a fluke for this particular polynomial?
 A: The unique real zero of $x^5+x-1$ is in exact form
$$
\frac{(100 + 12\sqrt{69}\;)^{1/3}}{6} + \frac{2}{3(100 + 12\sqrt{69}\;)^{1/3}} - \frac{1}{3}
$$
So this is not an instance of the "Casus irreducibilis", where a real zero cannot be expressed in radicals without using complex numbers.
A: It is possible to create a polynomial with roots that are irrational in the reals but, for any complex root $z = a + bi$, $a$ and $b$ are rational and therefore exact.  One can also create examples that are opposite, like this:
$$(x + \pi)(x + i) = 0$$
and
$$(x + \pi i)(x + 1)$$
as examples.
So to the original question, can we have a polynomial with $a$ exact and $b$ inexact solutions, sure you can, just build it up.  Create it from first order polynomials multiplied together, in which $a$ are exact and $b$ have irrational roots.
In general, it is impossible to know the location of the roots for an arbitrary polynomial of degree greater than 5, so one couldn't do this construction using only the polynomial's coefficients.  You would have to start by placing the roots and then expanding it in order to know what the polynomial is in a standard form.
A: $$x^5+x-1=\left(x^2-x+1\right) \left(x^3+x^2-1\right)$$
For the quadratic term, the roots are
$$x_1=\frac{1+i \sqrt{3}}{2} \qquad \text{and} \qquad x_2=\frac{1-i \sqrt{3}}{2}$$
For the cubic, there is only one real root which, using the hyperbolic method, is
$$x_3=\frac{1}{3} \left(2 \cosh \left(\frac{1}{3} \cosh
   ^{-1}\left(\frac{25}{2}\right)\right)-1\right)$$
