14. If the polynomial
$$
z^{6}-10 z^{5}+A z^{4}+B z^{3}+C z^{2}+D z+16
$$

has all roots as positive integers, then the value of $B$ is ( ).
(A) -88
(B) -80
(C) -64
(D) -41
(E) -40
14. A.

Let the six positive integer roots of the polynomial be $x_{i}(i=1,2$, $\cdots, 6), x_{1} \leqslant x_{2} \leqslant \cdots \leqslant x_{6}$.
Then by Vieta's formulas, we have
$$
\begin{array}{l}
\left\{\begin{array}{l}
x_{1}+x_{2}+\cdots+x_{6}=10, \\
x_{1} x_{2} \cdots x_{6}=16=2^{4}
\end{array}\right. \\
\Rightarrow x_{1}=x_{2}=1, x_{3}=x_{4}=x_{5}=x_{6}=2 \text {. } \\
\text { Therefore, } z^{6}-10 z^{5}+A z^{4}+B z^{3}+C z^{2}+D z+16 \\
=(z-1)^{2}(z-2)^{4} \\
=\left(z^{2}-2 z+1\right)\left(z^{4}-8 z^{3}+24 z^{2}-32 z+16\right) \text {. } \\
\end{array}
$$

Thus, $B=-32-48-8=-88$.