The equation $x^3+6x^2+11x+6=0$ has: 
$\textbf{(A)}\ \text{no negative real roots}\qquad\textbf{(B)}\ \text{no positive real roots}\qquad\textbf{(C)}\ \text{no real roots}\\ \textbf{(D)}\ \text{1 positive and 2 negative roots}\qquad\textbf{(E)}\ \text{2 positive and 1 negative root}$
By the rational root theorem, $1, -1, 2, -2, 3, -3, 6, -6$ are possible rational roots. Because $x^3+6x^2+11x+6>0$ for $x>0$, so there are no positive roots. We try $-1, -2, -3, -6$, so $x=-1, x=-3, x=-2$, so there are no positive real roots; $\fbox{B}$