5. 44 Let $P(x)=x^{4}+a x^{3}+b x^{2}+c x+d$, where $a, b, c, d$ are constants, and $P(1)=1993, P(2)=3986, P(3)=5979$.
Calculate $-\frac{1}{4}[P(11)+P(-7)]$.
[Solution] Let $Q(x) = P(x) - 1993x$,
then $Q(1) = Q(2) = Q(3) = 0$,
so $Q(x) = (x-1)(x-2)(x-3)(x-r)$.
Thus, from $P(x) = Q(x) + 1993x$ we get
$$
\begin{array}{l} 
\frac{1}{4}[P(11) + P(-7)] \\
= \frac{1}{4}[Q(11) + 1993 \times 11 + Q(-7) + 1993 \times (-7)] \\
= \frac{1}{4}[Q(11) + Q(-7)] + 1993, \\
\text{but} \quad Q(11) = 10 \times 9 \times 8 \times (11-r) \\
Q(-7) = (-8) \times (-9) \times (-10) \times (-7-r),
\end{array}
$$

Therefore, $Q(11) + Q(-7) = 10 \times 9 \times 8 \times 18 = 12960$,
$$
\frac{1}{4}[P(11) + P(-7)] = 3240 + 1993 = 5233.
$$