jmc

$x^4+x^3+2x^2+x+1=(x^2+1)(x^2+x+1)$. We apply the polynomial generalization of the Chinese Remainder Theorem. Indeed, $p(x)=(x^{2008}+x^{2007}+x^{2006})+\cdots +(x^4+x^3+x^2)+x+1\equiv x+1(\mathrm{mod}{x}^2+x+1)$ since $x^{n+2}+x_{n+1}+x^n=x^{n-2}(x^2+x+1)\equiv 0(\mathrm{mod}{x}^2+x+1)$. Also, $p(x)=(x^{2008}+x^{2006})+(x^{2007}+x^{2005})+\cdots +(x^4+x^2)+(x^3+x)+1\equiv 1(\mathrm{mod}{x}^2+1)$ using similar reasoning. Hence $p(x)\equiv x+1(\mathrm{mod}{x}^2+x+1),p(x)\equiv 1(\mathrm{mod}{x}^2+1)$, and by CRT we have $p(x)\equiv -x^2(\mathrm{mod}{x}^4+x^3+2x^2+x+1)$. Then $|r(2008)|\equiv 2008^2\equiv \boxed{64}(\mathrm{mod}1000)$.