Example 5 Factorize:
$$
(x+1)(x+2)(x+3)-6 \times 7 \times 8 \text {. }
$$
(1987, Sichuan Province Junior High School Mathematics Competition)
Solution: Let $f(x)=(x+1)(x+2)(x+3)-$ $6 \times 7 \times 8$. Obviously, $f(5)=0$. By the Factor Theorem, $f(x)$ has a factor $(x-5)$. Therefore, we can assume
$$
\begin{array}{l}
(x+1)(x+2)(x+3)-6 \times 7 \times 8 \\
=(x-5)\left(x^{2}+a x+b\right) .
\end{array}
$$

Taking $x=-1$, we get $-6 \times 7 \times 8=-6(1-a+b)$;
Taking $x=-2$, we get $-6 \times 7 \times 8=-7(4-2 a+$
$b)$.
Solving these, we get $a=11, b=66$.