Example 4 Let $p(x)=x^{4}+a x^{3}+b x^{2}+c x+d$, where $a, b, c, d$ are constants, $p(1)=1993$, $p(2)=3986$, $p(3)=5979$. Try to calculate $\frac{1}{4}[p(11)+p(-7)]$.
Analysis Notice that $1993, 3986, 5979$ are all multiples of 1993, so $p(x)-1993 x$ has roots $x=1$, 2, 3. Since $p(x)$ is a 4th-degree polynomial, let the 4th root be $r$, then we have $p(x)-1993 x=(x-1)(x-2)(x-3)(x-r)$. Substituting $x=11$ and $x=-7$, we can solve the problem.

Solution Let $Q(x)=p(x)-1993 x$, then by the problem statement, $Q(1)=Q(2)=Q(3)=0$.
Since $Q(x)$ is a 4th-degree polynomial, we can set $Q(x)=(x-1)(x-2)(x-3)(x-r)$,
$$
\text{Thus } \begin{aligned}
& \frac{1}{4}[p(11)+p(-7)] \\
= & \frac{1}{4}[Q(11)+Q(-7)]+1993 \\
= & \frac{1}{4}[10 \times 9 \times 8 \times(11-r)+(-8) \times(-9) \times(-10) \times(-7-r)]+1993 \\
= & \frac{1}{4} \times 10 \times 9 \times 8 \times 18+1993=5233 .
\end{aligned}
$$