2. Let $P(x)=x^{4}+a x^{3}+b x^{2}+c x+d, a, b, c, d$ be constants, $P(1)=1993, P(2)=3986$, $P(3)=5979$, try to calculate $\frac{1}{4}[P(11)+P(-7)]$.
2. Let $1993=n$, and let $Q(x)=P(x)-n x$, then we have $Q(1)=Q(2)=Q(3)=0$.

Thus $Q(x)=(x-1)(x-2)(x-3)(x-r)$.
$$
\begin{array}{l}
\frac{1}{4}[P(11)+P(-7)]=\frac{1}{4}[Q(11)+Q(-7)]+n \\
=\frac{1}{4}[10 \times 9 \times 8 \times(8-r)+(-8)(-9)(-10)(-7-r)]+n \\
=\frac{1}{4} \times 10 \times 9 \times 8 \times 15+1993=4693 .
\end{array}
$$