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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

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    $\begingroup$ Looks like coefficients are alternating, so why $...+10x+11$ at the end? $\endgroup$ – Ennar Dec 2 '17 at 18:14
  • $\begingroup$ Your link has $-10x+11$ $\endgroup$ – Mark Bennet Dec 2 '17 at 18:21
  • $\begingroup$ And what exactly did you try? $\endgroup$ – Professor Vector Dec 2 '17 at 18:21
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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.

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  • $\begingroup$ Nice solution. Perhaps one can add a term $5(x-1)^{2} + 6$ to both members ? $\endgroup$ – Panurge Dec 2 '17 at 18:35
  • $\begingroup$ Well, I guess the intended, final form of this is $(x^8 + 2x^6 + 3x^4 + 4x^2 + 5)(x-1)^2+6$, right? $\endgroup$ – Professor Vector Dec 2 '17 at 19:36
  • $\begingroup$ @Panurge Indeed that can be done, but I left some work for OP, and the signs at the end had been left ambiguous $\endgroup$ – Mark Bennet Dec 2 '17 at 19:49
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    $\begingroup$ @ProfessorVector You could do that, or just leave it as a rather more explicit sum of squares. When I put this the signs of the terms had not been clarified and I thought to leave something for OP to do. $\endgroup$ – Mark Bennet Dec 2 '17 at 19:51
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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.

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  • $\begingroup$ This works for general n. $\endgroup$ – marty cohen Dec 2 '17 at 19:54
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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 $x$.

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