20. If $b$ is a real number, then the number of negative real roots of $f(x)=x^{4}+(6-b)x^{3}+(11-6b)x^{2}+(6-11b)x-6b$ could be ( ).
(A) dependent on $b$
(B) 1 or 3
(C) 1
(D) 0
(E) *
20. (B).
$$
\begin{array}{l}
x^{4}+(6-b) x^{3}+(11-6 b) x^{2}+(6-11 b) x-6 b \\
=\left(x^{3}+6 x^{2}+11 x+6\right)(x-b) \\
=(x+1)(x+2)(x+3)(x-b) . \\
\because b>0, \\
\therefore f(x)=0 \text { has 3 negative real roots: }-1,-2,-3 .
\end{array}
$$