Example 5 Given $P(x)=x^{5}+a_{1} x^{4}+a_{2} x^{3}+$ $a_{3} x^{2}+a_{4} x+a_{5}$, and when $k=1,2,3,4$, $P(k)$ $=k \times 1$ 997. Then $P(10)-P(-5)=$ $\qquad$
Solution: Let $Q(x)=P(x)-1997 x$, then when $k=$ $1,2,3,4$, $Q(k)=P(k)-1997 k=0$.
Therefore, $1,2,3,4$ are roots of $Q(x)=0$.
Thus, we can set $Q(x)=(x-1)(x-2)(x-$ $3)(x-4)(x-r)$, then
$$
\begin{aligned}
& P(10)=Q(10)+1997 \times 10 \\
= & 9 \times 8 \times 7 \times 6 \times(10-r)+1997 \times 10, \\
& P(-5)=Q(-5)+1997 \times(-5) \\
= & (-6) \times(-7) \times(-8) \times(-9) \\
\times & (-5-r)+1997 \times(-5) . \\
\therefore & P(10)-P(-5) \\
= & 9 \times 8 \times 7 \times 6 \times 5+1997 \times 15 \\
= & 75315 .
\end{aligned}
$$