How to show $x^{10}-2x^9+3x^8-...-10x+11=0$ has no real root? I tried to solve below equation $$x^{10}-2x^9+3x^8-...-10x+11=0$$ I plot the graph and see there is no real root , But I get stuck how to show analytical .Can some one help me or give an idea ? 
Thanks in advance.
This is graph of the function $$f(x)=x^{10}-2x^9+3x^8-...-10x+11$$
https://www.desmos.com/calculator/6k1lj498ra
 A: If you multiply the given polynomial by $(1+x)^2$ you get $q(x)=x^{12}+12x+11$.
Of course we have a double root at $x=-1$ and $q(x)$ is a convex function since $q''(x)\geq 0$.
Since $x=-1$ is the only real root of $q'$, $q(-1)=0$ is an absolute minimum for $q(x)$ and $q(x)$ does not vanish over $\mathbb{R}\setminus\{-1\}$. It follows that the original polynomial has no real zeroes.
A: Let's consider the polynomial from the link, $p(x)=x^{10}-2x^9+3x^8-4x^7+5x^6-6x^5+7x^4-8x^3+9x^2-10x+11$. Then, $q(x)=(x+1)^2\,p(x)=x^{12}+12x+11$. This is certainly positive for large enough $x$. The only zero of the derivative is $x=-1$, where $q(x)=0$. This must be the global minimum of $q$, so it takes only non-negative values, and is $0$ only for $x=-1$. This means $p(x)>0$ for all $x$.
A: There is some confusion over the sign of the coefficients. But it might prove useful to consider $$x^8(x-1)^2+2x^6(x-1)^2+3x^4(x-1)^2+4x^2(x-1)^2 =$$$$=x^{10}-2x^9+x^8+2x^8-4x^7+2x^6+3x^6-6x^5+3x^4+4x^4-8x^3+4x^2 =$$$$=x^{10}-2x^9+3x^8-4x^7+5x^6-6x^5+7x^4-8x^3+4x^2$$which is clearly non-negative, or some variant of this. You should be able to conclude from there.
