[Solution 1] Given $f(x)=[(x+1)(x+4)][(x+2)(x+3)]+35$
$$
=\left(x^{2}+5 x+4\right)\left(x^{2}+5 x+6\right)+35,
$$

If we let $x^{2}+5 x+4=t$, then the above expression can be transformed into
$$
\varphi(t)=t(t+2)+35=t^{2}+2 t+35.
$$

Therefore, $\varphi(t)_{\min }=\frac{4 a c-b^{2}}{4 a}=\frac{4 \cdot 35 \cdot 1-2^{2}}{4}=34$.
When $t=-\frac{b}{2 a}=-1$, we know that $x^{2}+5 x+4=-1$,
which means $x^{2}+5 x+5=0$ has real roots.
Thus, $f(x)_{\min }=34$.
Therefore, the answer is $(B)$.
[Solution 2]
$$
\begin{aligned}
f(x) & =\left(x^{2}+5 x+4\right)\left(x^{2}+5 x+6\right)+35 \\
& =\left(x^{2}+5 x+4\right)\left[\left(x^{2}+5 x+4\right)+2\right]+35 \\
& =\left(x^{2}+5 x+4\right)^{2}+2\left(x^{2}+5 x+4\right)+1+34 \\
& =\left(x^{2}+5 x+5\right)^{2}+34,
\end{aligned}
$$

Since $\left(x^{2}+5 x+5\right)^{2} \geqslant 0$, then $f(x)_{\text {min }}=34$.
Therefore, the answer is $(B)$.
$13 \cdot 30$ The minimum value of the function $f(x)=(x+1)(x+2)(x+3)(x+4)+35$ is
(A) $33 \frac{7}{8}$.
(B) 34 .
(C) $34 \frac{1}{16}$.
(D) 35 .
(E) 59 .
(China Jiangsu Province Xuzhou Junior High School Mathematics Competition, 1985)