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As I understand it $x^5 - x + 1$ is not solvable by radicals. But it splits over $\mathbb{C}$, so how does it factor into linear factors?

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It will factor like $$(x-c_1)(x-c_2)(x-c_3)(x-c_4)(x-c_5)$$ where each $c_i$ is a complex number with no nice closed form. – Alex Becker Jul 9 '12 at 20:25
$$(x^5 - x + 1) = (x^2 + \alpha x + \beta)(x^2 + \gamma x + \delta) (x + \xi)$$ where $\alpha, \beta, \gamma, \delta, \xi \in \mathbb{R}$ and $\xi < 0$. – user17762 Jul 9 '12 at 20:26
@marvis: Shouldn't $\xi$ be positive? We know that the polynomial has a root in $(-\infty, 0)$, since $P(-\infty) < 0 < P(0) = 1$, and if this root is $r$ then the polynomial is divisible by $(x-r)$ and so has the form $P'(x)(x+(-r))$ where $-r$ is positive. – MJD Jul 9 '12 at 20:32
Indeed, $\xi \approx 1.16$. – gt6989b Jul 9 '12 at 20:41

By Descartes' Rule of Signs, there is one real root. Find it. The result will give you a product of a linear factor times a quartic, which can be solved by radicals. You may also use Jacobi theta functions to find the roots, since it is in Bring's (reduced) form.

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See Marvis' comment. – user02138 Jul 9 '12 at 20:26

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