If $ x^4+x^3+x^2+x+1=0$ then what is the value of $x^5$ If $$x^4+x^3+x^2+x+1=0$$ then what's the value of $x^5$ ?? 
I thought it would be $-1$ but it does not satisfy the equation

The left side of the equation is $P= (x^4 +x^2) + (x^3 + x) +1 = 2$ 
In the equation $x^2(x^2+1) +1 > 0$, there is no such real number for $x$. Logically, a false statement implies anything, so $x^5$ = anything you wish . 
M Andrews' correct argument above can be modified to see that $x^5-1 =(x-1)(x^4+x^3+x^2+x+1)$, so the roots of $P=0$ are the 4 other complex 5th roots of unity (which are not x=1) namely 
$x=e^{2m\pi i}/5$, so $m=1,2,3,4$, and so on.
For instance, for $m=1$ $x= \cos(2\pi/5) + i \sin(2\pi/5)$ . So $x^5 =1 $, provided you allow complex numbers as roots of P.
