$h(x)=x^2+x+1=0$
$\Rightarrow x^2+\frac{1}{2}x+\frac{1}{2}x+\frac{1}{4}+\frac{3}{4}=0$
$\Rightarrow x(x+\frac{1}{2})+\frac{1}{2}(x+\frac{1}{2})+\frac{3}{4}=0$
$\Rightarrow (x+\frac{1}{2})(x+\frac{1}{2})+\frac{3}{4}=0$
$\Rightarrow (x+\frac{1}{2})^2+\frac{3}{4}=0$
$\Rightarrow (x+\frac{1}{2})^2=\frac{-3}{4}$ (vô lí)
-Vậy: đa thức h(x) ko có nghiệm.