Since $P$ needs have no real roots and $\lim_{x \to \infty} P(x) = +\infty$, it means we need have $P(x) > 0$ for all real $x$. Since we then need $P(1) > 0$, it means the number of negative coefficients can be at most $1005$ (out of the total $2011$ coefficients). Let us now build an example with $1005$ negative coefficients. Take $P_0(x) = \sum_{k=0}^{501} (x^{2010-2k} - x^{2010-2k-1} - x^{2k+1} + x^{2k}) + x^{1006} - x^{1005} + x^{1004}$. But $x^{1006} - x^{1005} + x^{1004} = x^{1004}(x^2-x+1) \geq 0$, while $x^{2010-2k} - x^{2010-2k-1} - x^{2k+1} + x^{2k} = x^{2k}(x-1)^2(x^{2010-4k-2} + \cdots + 1)$. But a polynomial $x^{2m+1} - 1$ has only one real root (root $1$), so polynomial $x^{2m} + \cdots + 1$ has no real roots, hence is positively defined. Therefore all terms in the sum above are non-negative, and may be zero only at values $0$ and/or $1$. However, $P_0(0) = P_0(1) = 1$, so $P_0$ has no real roots. Note. I copied here my proof from the other, subsequent post on same topic. EDIT. True (according with following post), the polynomial $x^{2011} + 1$ has only one real root (root $-1$), so polynomial $x^{2010} - x^{2009} + \cdots + x^2 - x + 1$ has no real roots.

goodar2006 wrote: suppose that polynomial $p(x)=x^{2010}\pm x^{2009}\pm...\pm x\pm 1$ does not have a real root. what is the maximum number of coefficients to be $-1$?(14 points) At first, let $x=1$, we see that there doesn't may more than $1005$ $-1^s.$ Now, look at $p(x)=x^{2010}-x^{2009}+x^{2008}-x^{2007}+...- x+1$ If $x\le0$, each term is positive. If$0<x<1:$ $x^{i-1}>x^i$ and $x^{2010}>0$ So $\sum_{i=0}^{i=1004}x^{2i}>\sum_{i=0}^{i=1004}x^{2i+1}$ and $x^2010>0$ and add it, we find $p(x)>0$ If $x\ge1$: $x^{i-1}\lex^i$ and $1>0$ and we find that $\sum_{i=1}^{i=1005}x^{2i} \le \sum_{i=1}^{i=1005}x^{2i-1}$ and $1>0$, add them and we find $p(x)>0$ So the maximum is $1005.$ This is my solution.

An interesting problem