$29 \cdot 62$ When the polynomial $x^{8}$ is divided by $x+\frac{1}{2}$, let $q_{1}(x)$ and $r_{1}$ represent the quotient and remainder, respectively. When $q_{1}(x)$ is again divided by $x+\frac{1}{2}$, let $q_{2}(x)$ and $r_{2}$ represent the quotient and remainder, respectively. Then $r_{2}$ equals
(A) $\frac{1}{256}$.
(B) $-\frac{1}{16}$.
(C) 1 .
(D) -16 .
(E) 256 .
(30th American High School Mathematics Examination, 1979)
[Solution] Let $a=-\frac{1}{2}$, using the Remainder Theorem we get $r_{1}=a^{8}$. From the equation
$$
x^{8}=(x-a) q_{1}(x)+r_{1}=(x-a) q_{1}(x)+a^{8} \text {, }
$$

we have
$$
q_{1}(x)=\frac{x^{8}-a^{8}}{x-a}=x^{7}+a x^{6}+\cdots+a^{7} \text {, }
$$

or
$$
q_{1}(x)=\left(x^{4}+a^{4}\right)\left(x^{2}+a^{2}\right)(x+a) .
$$

Therefore, when $q_{1}(x)$ is divided by $x-a$, using the Remainder Theorem we have
$$
r_{2}=q_{1}(a)=8 a^{7}=-\frac{1}{16} \text {. }
$$

Thus, the answer is $(B)$.