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factorize x^5 - x + 1 = 0 i'm not sure how to factor beyond the 3rd/4th degree. how do i do this one?

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fairly sure this will not factor over the rationals, since a quick graph shows it has only one real zero
but how does a "quick graph [that] shows it has one zero" show anything? y=x^5 also has one zero as well. not sure what the point is
\(y=x^5\) is factored

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if you can find the zeros then you can factor. i am assuming you mean factor over the rational numbers, although maybe i am mistaken about that
i dont spose theres a simple easy way to "complete a quintic"?
not according to galois
you can try newtons method of trial and error to get closer and closer to a real root
hmmm - do you really want to factorise this or are you looking for the solutions? if its the solutions you want then use the method suggested by @amistre64
the only factorisation I can think of here is as follows:\[\begin{align} x^5-x+1&=0\\x(x^4-1)+1&=0\\ x(x^2-1)(x^2+1)+1&=0\\ x(x-1)(x+1)(x^2+1)+1&=0\\ \end{align}\]but I don't think that helps towards getting to the solution. :(
since f(-1)>0 and f(-2)<0 one root is between -1 and -2 by the Bolzano theorem..
since it is odd degree poly. we can say that it has at least one real root.. but I do not know how to say it has only one real root without looking its graph and I also wanna learn it..
wow man look at the roots!!^5-x%2B1%3D0^5-x%2B1

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