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?
Best Answer
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 $b=1$.