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?
How to factor $x^5 - x + 1$
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abstract-algebra
polynomials
factoring
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5It 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. – 2012-07-09
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0$$(x^5 - x + 1) = (x^2 + \alpha x + \beta)(x^2 + \gamma x + \delta) (x + \xi)$$ where $\alpha, \beta, \gamma, \delta, \xi \in \mathbb{R}$ and $\xi < 0$. – 2012-07-09
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0@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. – 2012-07-09
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0Indeed, $\xi \approx 1.16$. – 2012-07-09
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0http://www.wolframalpha.com/input/?i=factor%5Bx%5E5-x%2B1%5D – 2012-07-09
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0@MarkDominus Oh yeah. I wanted to write $\xi > 0$. – 2012-07-09
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0@Marvis You had me worried for a minute. – 2012-07-09