Find the remainder when $(x - 1)^{100} + (x - 2)^{200}$ is divided by $x^2 - 3x + 2$ . 
Find the remainder when $(x - 1)^{100} + (x - 2)^{200}$ is divided by $x^2 - 3x + 2$ .

What I tried: In some step I messed up with this problem and so I think I am getting my answer wrong, so please correct me.
We have $x^2 - 3x + 2$ = $(x - 1)(x - 2)$
and I can see $(x - 1)^2 \equiv 1$ $($mod $x - 2)$ . We also have :-
$$\frac{(x - 1)^{100}}{(x - 1)(x - 2)} = \frac{(x - 1)^{99}}{(x - 2)}.$$
We have :- $(x - 1)^{98} \equiv 1$ $($mod $x - 2).$ $\rightarrow (x - 1)^{99} \equiv (x - 1)$ $($mod $x - 2)$. Now for the case of $(x - 2)^{200}$ we have :-
$$\frac{(x - 2)^{200}}{(x - 1)(x - 2)} = \frac{(x - 2)^{199}}{(x - 1)}.$$
We have :- $(x - 2) \equiv (-1)$ $($mod $x - 1)$ $\rightarrow (x - 2)^{199} \equiv (-1)$ $($mod $x - 1)$.
Adding all these up we have :- $(x - 1)^{100} + (x - 2)^{200} \equiv (x - 2)$ $($mod $x² - 3x + 2)$ .
On checking my answer with wolfram alpha , I found the remainder to be $1$, so I messed up in some step .
Can anyone help me?
 A: You are right that $(x-1)^{98}\equiv1\pmod{(x-2)}$. But that implies
$$(x-1)^{100}\equiv(x-1)^2=x(x-2)+1\equiv1\pmod{x-2}.$$
More naively, as
$$x-1\equiv1\pmod{x-2}$$
then
$$(x-1)^{100}\equiv1^{100}=1\pmod{x-2}.$$
Similarly,
$$x-2\equiv-1\pmod{x-1}$$
and
$$(x-2)^{200}\equiv(-1)^{200}=1\pmod{x-1}.$$
So $(x-1)^{100}+(x-2)^{200}$ is congruent to $1$ modulo both $x-1$ and $x-2$,
and so also modulo $(x-1)(x-2)$.
A: Since $(x - 1)^{99} \equiv (x - 1)\;\mod (x - 2)$, we get that
$(x - 1)^{100}\equiv (x-1)^2\;\mod (x-2)(x-1). \quad(*)$
Since $(x-2)^{199}\equiv -1\;\mod (x-1)$, we get that
$(x-2)^{200}\equiv -(x-2)\;\mod (x-1)(x-2). \quad(**)$
So, by adding $(*)$ and $(**)$, it follows that
$(x - 1)^{100} + (x - 2)^{200} \equiv (x-1)^2-(x-2)\\\mod (x-1)(x-2),$
that is
$(x - 1)^{100} + (x - 2)^{200}\equiv x^2-3x+3\mod (x^2-3x+2)$.
Hence,
$(x - 1)^{100} + (x - 2)^{200} \equiv 1\;\mod (x^2-3x+2)$.
A: The remainder of $(x - 1)^{100}$ divided by $(x - 1)(x - 2)$ will be $(x - 1) (2 - 1)^{99} = x - 1$. The remainder of $(x - 2)^{200}$ divided by $(x - 1)(x - 2)$ will be $(x - 2)(1 - 2)^{199} = 2 - x$ Therefore, the total remainder will be 1.
A: $P(x)=(x-1)^{100}+(x-2)^{200}=Q(x)×(x^2-3x+2)+ax+b$
$P(1)=(-1)^{200}=a+b,  a+b=1$
$P(2)=(1)^{100}=2a+b,  2a+b=1$
$a=0 , b=1$
Hence remainder :- $R(x)=1$
A: Write $$(x - 1)^{100} + (x - 2)^{200}=k(x)(x-2)(x-1)+ax+b$$
SInce this is valid for all $x$ it is valid also for $$x=1: \;\;\; 1=a+b$$ and $$x=2: \;\;\; 1=a2+b$$
So $a=0$ and $b=1$.
