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The polynomial is $f(t)=t^5-4t+2$.

I can show it's irreducible with Eisenstein's Criterion, and I know I need to show that it has exactly two complex (non-real) roots to prove that it's not solvable, using the follow result:

Let $p>1$ be prime and $f\in\mathbb{Q}[t]$ irreducible with degree $p$. If $f$ has exactly two non-real complex roots, then the Galois group of $f$ is isomorphic to $S_p$. Hence, with $p\geq5$, we have that $f$ is not solvable by radicals over $\mathbb{Q}$.

But I don't know a way to show it has exactly two roots, I know that, using Descartes' rule of signs, we can show that it has at least two complex roots (because it has 2 or 0 positive and one negative real root). Is there something I can use on top of Descartes' rule to make this work? If not, what's the other way?

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    $\begingroup$ A plot of $f(t),t\in [-2,2]$ will give immediately the number of real roots. $\endgroup$
    – reuns
    Apr 27 at 18:30
  • $\begingroup$ found mathcounterexamples.net/… $\endgroup$
    – Will Jagy
    Apr 27 at 20:42
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\begin{align} \lim_{x\to+\infty}f(x) &= +\infty\\ f(1) &= -1\\ f(0) &= 2\\ \lim_{x\to-\infty}f(x) &= -\infty\\ \end{align} Therefore $f(x)$ has three real roots. You have shown that $f$ has two complex non-real roots, so you found all roots of $f$. Then, by the lemma you have proveded it follows that the Galois group of $f$ is $S_5$ and therefore $f$ is not solvable by radicals over $\mathbb{Q}$.

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Since $f'(t)=5t^4-4$, $f$ is increasing on $\left(-\infty,-\sqrt[4]{\frac45}\,\right]$ and on $\left[\sqrt[4]{\frac45},\infty\right)$ and decreasing on $\left[-\sqrt[4]{\frac45},\sqrt[4]{\frac45}\,\right]$. But$$f\left(-\sqrt[4]{\frac45}\right)>0\quad\text{and}\quad f\left(\sqrt[4]{\frac45}\right)<0.$$So, $f$ has exactly one real root on each of the three intervals mentioned abov. Since none of them is a double root, it has exactly two complex non-real roots.

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There is no ambiguity about the number of real roots, as the second derivative $20 x^3$ is negative for negative $x$ and positive for positive $x.$ In the graph, I was taught to call thse conditions concave down and concave up. The first derivative also has evident roots, $\pm \sqrt[4] \frac{4}{5}.$ It is enough to plot points when $x$ is an integer, although I put in a few more

enter image description here

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