16. Let $P(x)$ be a cubic polynomial, and
$$
P(0)=k, P(1)=2 k, P(-1)=3 k \text {. }
$$

Then $P(2)+P(-2)=(\quad)$.
(A) 0
(B) $k$
(C) $6 k$
(D) $7 k$
(E) $14 k$
$$
\begin{array}{l}
P(x)=a_{3} x^{3}+a_{2} x^{2}+a_{1} x+a_{0}\left(a_{3} \neq 0\right) . \\
\text { Then }\left\{\begin{array}{l}
P(0)=a_{0}=k, \\
P(1)=a_{3}+a_{2}+a_{1}+a_{0}, \\
P(-1)=-a_{3}+a_{2}-a_{1}+a_{0}
\end{array}\right. \\
\Rightarrow\left\{\begin{array}{l}
a_{0}=k, \\
a_{2}=\frac{3 k}{2}
\end{array}\right. \\
\Rightarrow P(2)+P(-2) \\
\quad=8 a_{3}+4 a_{2}+2 a_{1}+a_{0}+(-8) a_{3}+4 a_{2}+ \\
\quad(-2) a_{1}+a_{0} \\
=8 a_{2}+2 a_{0}=14 k .
\end{array}
$$