14. Let $c \in \mathbf{C}$, the polynomial
$$
p(x)=\left(x^{2}-2 x+2\right)\left(x^{2}-c x+4\right)\left(x^{2}-4 x+8\right)
$$

has exactly four distinct complex roots. Then $|c|=(\quad)$.
(A) 2
(B) $\sqrt{6}$
(C) $2 \sqrt{2}$
(D) 3
(E) $\sqrt{10}$
14. E.

Notice,
$$
\begin{array}{l}
x^{2}-2 x+2=(x-1)^{2}+1 \Rightarrow x=1 \pm \mathrm{i}, \\
x^{2}-4 x+8=(x-2)^{2}+4 \Rightarrow x=2 \pm 2 \mathrm{i} .
\end{array}
$$

Since $p(x)$ has exactly four distinct complex roots, the roots of $x^{2}-c x+4$ could be $1 \pm \mathrm{i}, 2 \pm 2 \mathrm{i}$.

Since the product of the roots is 4, the roots of $x^{2}-c x+4$ could be $1+\mathrm{i}, 2-2 \mathrm{i}$ or $1-\mathrm{i}, 2+2 \mathrm{i}$.
Therefore, $c=3 \pm \mathrm{i} \Rightarrow|c|=\sqrt{10}$.