$29 \cdot 43$ If $a, b$ are integers, and $x^{2}-x-1$ is a factor of $a x^{3}+b x^{2}+1$, then $b$ equals
(A) -2 .
(B) -1 .
(C) 0 .
(D) 1 .
(E) 2 .
(39th American High School Mathematics Examination, 1988)
［Solution］Using synthetic division, we have
$$
\begin{array}{l}
\left.x^{2}-x-1\right) \frac{a x+(a+b)}{a x^{3}+b x^{2}} \ldots+1 \\
\begin{array}{lll}
a x^{3}-a x^{2} & -a x & \\
\hline(b+a) x^{2} & +a x & +1
\end{array} \\
\frac{(a+b) x^{2}-(a+b) x-(a+b)}{(2 a+b) x+1-(a+b)} \\
\because \quad\left(x^{2}-x-1\right) \mid\left(a x^{3}+b x^{2}+1\right) \text {, } \\
\therefore \quad\left\{\begin{array} { l } 
{ 1 + ( a + b ) = 0 , } \\
{ 2 a + b = 0 . }
\end{array} \quad \text { Solving, we get } \quad \left\{\begin{array}{l}
a=1, \\
b=-2 .
\end{array}\right.\right. \\
\end{array}
$$

Therefore, the answer is (A).