Let $f(x) = x^{100}$ .

We note that $x^2-3x+2 = (x-1)(x-2)$ .

By remainder theorem, the remainder when:

• $f(x)$ divided by $(x-1)$ is $r(1) = f(1) = 1$
• $f(x)$ divided by $(x-2)$ is $r(2) = f(2) = 2^{100}$

Since $f(x)$ is rational, we can assume $r(x)$ to be:

$\begin{aligned} r(x) & = a(x-2) + b(x-1) = \begin{cases} \text{when } x = 1 & \Rightarrow a(1-2) + b(1-1) = 1 \\ & \Rightarrow a = -1 \\ \text{when } x = 2 & \Rightarrow a(2-2) + b(2-1) = 2^{100} \\ & \Rightarrow b = 2^{100} \end{cases}\\ \Rightarrow r(x) & = -(x-2) + 2^{100}(x-1) \\ & = \left( 2^{100} - 1\right)x - 2^{100} + 2 \\ & = \boxed{\left( 2^{100} - 1\right)x - 2 \left(2^{99} - 1\right)} \end{aligned}$

I used a very very simple concept.

let remainder be mx+c . It will be of first degree or less as divisor is of second degree .

DIVIDEND-REMAINDER=QUOTIENT*DIVISOR

P(x)=x^100-mx-c must me divisible by (x-1)(x-2). Hence we get the values of m and c by P(1)=0 and P(2)=0.