$29 \cdot 58$ When $x^{100}$ is divided by $x^{2}-3 x+2$, the remainder $R$ is a polynomial of degree less than 2, so $R$ can be written as
(A) $2^{100}-1$.
(B) $2^{100}(x-1)-(x-2)$.
(C) $2^{100}(x-3)$.
(D) $x\left(2^{100}-1\right)+2\left(2^{99}-1\right)$.
(E) $2^{100}(x+1)-(x+2)$.
(20th American High School Mathematics Examination, 1969)
[Solution] Let $x^{100}=Q(x)\left(x^{2}-3 x+2\right)+R(x)$, where $Q(x)$ is a polynomial of degree 98, and the degree of $R(x)$ is less than 2, let $R(x)=a x+b$. Then
$$
x^{100}=Q(x)(x-2)(x-1)+(a x+b)
$$

Let $x=1,2$ to get $\quad 1=a+b, 2^{100}=2 a+b$.
Thus, $a=2^{100}-1, b=1-a=2-2^{100}$.
Therefore, $\quad R(x)=2^{100}(x-1)-(x-2)$.
Hence, the answer is $(B)$.