$29 \cdot 49$ For some real number $r$, the polynomial $8 x^{3}-4 x^{2}-42 x+45$ is divisible by $(x-r)^{2}$. The number closest to $r$ is
(A) 1.22.
(B) 1.32.
(C) 1.42.
(D) 1.52.
(E) 1.62.
(31st American High School Mathematics Examination, 1980)
[Solution] Since the given polynomial can be divided by $(x-r)^{2}$, the remainders when it is divided by $(x-r)$ and $(x-r)^{2}$ must be zero. Using synthetic division, we have
$$
\begin{aligned}
& 8 x^{3}-4 x^{2}-42 x+45 \\
= & (x-r)\left[8 x^{2}+(8 r-4) x+\left(8 r^{2}-4 r-42\right)\right]+8 r^{3}-4 r^{2} \\
& -42 r+45,
\end{aligned}
$$

The remainder $8 r^{3}-4 r^{2}-42 r+45=0$, and since $(x-r)^{2}$ can divide the original polynomial, then $x$
$-r$ can divide the quotient above, i.e.,
$$
\begin{aligned}
& 8 x^{2}+(8 r-4) x+\left(8 r^{2}-4 r-42\right) \\
= & (x-r)(8 x+16 r-4)+24 r^{2}-8 r-42 .
\end{aligned}
$$

Thus,
$$
24 r^{2}-8 r-42=0 \text {. }
$$

Solving this, we get
$$
r=\frac{3}{2} \text { or }-\frac{7}{6} \text {. }
$$

Since $r=-\frac{7}{6}$ does not make the previous remainder zero, $r=\frac{3}{2}=1.5$. Therefore, the answer is $(D)$.