# polynomial

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• Jul 27th 2010, 12:21 PM
rtrt1
polynomial
(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 PM
eumyang
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 PM
chiph588@
Rational Root Theorem. This tells you there are no linear factors.
• Jul 27th 2010, 12:54 PM
Also sprach Zarathustra
Maybe by using calculus...
• Jul 27th 2010, 03:54 PM
melese
Quote:

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...
• Jul 27th 2010, 03:59 PM
chiph588@
Quote:

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.
• Jul 27th 2010, 04:55 PM
melese
Quote:

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 $\displaystyle P(x)=0$.