(Angry)How can I prove that you can not solve the polynomial : x^8-x^7+x^5-x^4+x^3-x+1

Over Q

(No use by Eisenstein theorem , without using polynomial Cyaklotomi))

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- Jul 27th 2010, 12:21 PMrtrt1polynomial
(Angry)How can I prove that you can not solve the polynomial : x^8-x^7+x^5-x^4+x^3-x+1

Over Q

(No use by Eisenstein theorem , without using polynomial Cyaklotomi)) - Jul 27th 2010, 12:40 PMeumyang
Hmm, are you the OP in this thread from the Math Is Fun forum?

http://www.mathisfunforum.com/viewtopic.php?pid=146808

(Wondering) - Jul 27th 2010, 12:52 PMchiph588@
Rational Root Theorem. This tells you there are no linear factors.

- Jul 27th 2010, 12:54 PMAlso sprach Zarathustra
Maybe by using calculus...

- Jul 27th 2010, 03:54 PMmelese
- Jul 27th 2010, 03:59 PMchiph588@
- Jul 27th 2010, 04:55 PMmelese
Here's what I meant.

According to the Rational Root Theorem:

Let $\displaystyle P(x)=a_nx^n+a_{n-1}x^{n-1}+...+a_1x^1+a_0=0 $ be a polynomial equation with integer coefficients. If $\displaystyle P(x) $ has a rational root $\displaystyle p/q $, with $\displaystyle p, q$ relatively prime integers, then $\displaystyle p|a_0 $ and $\displaystyle q|a_n $. (Statement found in PlanetMath: rational root theorem)

In our case, $\displaystyle P(x)=x^8-x^7+x^5-x^4+x^3-x+1=0 $ and $\displaystyle a_n=a_0=1 $. Assume there is a rational root $\displaystyle p/q $, then $\displaystyle p|a_0=1 $ and $\displaystyle q|a_n=1$. Now, $\displaystyle p=q=\pm 1$, and so $\displaystyle p/q=\pm 1$. (Two case after all!)

Checking the case we have $\displaystyle P(1)=1 $ and $\displaystyle P(-1)=1 $, contradicting $\displaystyle P(x)=0 $.