1. polynomial

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))

2. Hmm, are you the OP in this thread from the Math Is Fun forum?
http://www.mathisfunforum.com/viewtopic.php?pid=146808

3. Rational Root Theorem. This tells you there are no linear factors.

4. Maybe by using calculus...

5. Originally Posted by chiph588@
Rational Root Theorem. This tells you there are no linear factors.
Thanks for the link... The answer follows almost immediately, leaving only one case to check...

6. Originally Posted by melese
Thanks for the link... The answer follows almost immediately, leaving only one case to check...
One case? Aren't there three? A quadratic, cubic, or quartic factor? I attempted to solve these with long division, but it got messy real quick.

7. Originally Posted by chiph588@
One case? Aren't there three? A quadratic, cubic, or quartic factor? I attempted to solve these with long division, but it got messy real quick.
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 $\displaystyle P(x)=0$.