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 $x^5 - x + 1$
1
$\begingroup$
abstract-algebra
polynomials
factoring
-
0@Marvis You had me worried for a minute. – 2012-07-09
1 Answers
3
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.
-
0See Marvis' comment. – 2012-07-09