5.44 Let \( P(x) = x^{4} + a x^{3} + b x^{2} + c x + d \), where \( a, b, c, d \) are constants, and \( P(1) = 1993, P(2) = 3986, P(3) = 5979 \).

Try to 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
$$\begin{aligned}
& \frac{1}{4}[P(11) + P(-7)] \\
= & \frac{1}{4}[Q(11) + 1993 \times 11 + Q(-7) + 1993 \times (-7)] \\
= & \frac{1}{4}[Q(11) + Q(-7)] + 1993,
\end{aligned}$$

but $Q(11) = 10 \times 9 \times 8 \times (11-r)$
$$Q(-7) = (-8) \times (-9) \times (-10) \times (-7-r),$$

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