smc

Let the polynomial $Q(x)$ be the quotient when $x^{100}$ is divided by $x^2-3x+2$, and let the remainder $R=ax+b$, for some real $a$ and $b$. Then we can write: $x^{100}=(x^2-3x+2)Q(x)+ax+b$. Since it is hard to deal with $Q(x)$ (it is of degree 98!), we factor $x^2-3x+2$ as $(x-2)(x-1)$ so we can eliminate $Q(x)$ by plugging in $x$ values of $2$ and $1$. $x^{100}=(x-2)(x-1)Q(x)+ax+b$, $2^{100}=(0)Q(2)+2a+b$, $2^{100}=2a+b$. Similarly, $1^{100}=a+b$. Solving this system of equations gives $a=2^{100}-1$ and $b=2-2^{100}$. Thus, $R=ax+b=(2^{100}-1)x+(2-2^{100})$. Expanding and combining terms, we see that the answer is $\boxed{2^{100}(x-1)-(x-2)}$.