# Show that the following polynomial in $\mathbb{Q}[t]$ is irreducible and not solvable over $\mathbb{Q}$

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?

• A plot of $f(t),t\in [-2,2]$ will give immediately the number of real roots. Apr 27 at 18:30
• Apr 27 at 20:42

\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 $$\mathbb{Q}$$.
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.
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