$2 \cdot 68$ If $a_{2} \neq 0$, and $r, s$ are the roots of $a_{0}+a_{1} x+a_{2} x^{2}=0$, then the equation $a_{0}+a_{1} \dot{x}+a_{2} x^{2}=a_{0}\left(1-\frac{x}{r}\right)\left(1-\frac{x}{s}\right)$ holds in which of the following cases:
(A) For all $x$ values, $a_{0} \neq 0$.
(B) For all $x$ values.
(C) Only when $x=0$.
(D) Only when $x=r$ or $x=s$.
(E) Only when $x=r$ or $x=s, a_{0} \neq 0$.
(16th American High School Mathematics Examination, 1965)
[Solution] From $a_{2} x^{2}+a_{1} x+a_{0}$
$$
\begin{array}{l}
=a_{2}\left(x^{2}+\frac{a_{1}}{a_{2}} x+\frac{a_{0}}{a_{2}}\right) \\
=a_{2}\left[x^{2}-(r+s) x+r s\right] \\
=a_{2}(r-x)(s-x),
\end{array}
$$

If $r s \neq 0$, and $a_{2} \neq 0$, then $a_{0} \neq 0$.
$$
\begin{aligned}
\therefore \quad & a_{2} x^{2}+a_{1} x+a_{0} \\
= & a_{2} r s\left(1-\frac{x}{r}\right)\left(1-\frac{x}{s}\right) \\
= & a_{2} \cdot \frac{a_{0}}{a_{2}}\left(1-\frac{x}{r}\right)\left(1-\frac{x}{s}\right) \\
= & a_{0}\left(1-\frac{x}{r}\right)\left(1-\frac{x}{s}\right) .
\end{aligned}
$$

Therefore, the answer is $(A)$.