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

Question

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?

Answer 1

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 $b=1$.

Answer 2

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,

$(x - 1)^{100} + (x - 2)^{200} \equiv 1\;\mod (x^2-3x+2)$.

Answer 3

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 $(x-1)(x-2)$.

Answer 4

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.

Answer 5

$P(x)=(x-1)^{100}+(x-2)^{200}=Q(x)×(x^2-3x+2)+ax+b$

$P(1)=(-1)^{200}=a+b, a+b=1$

$P(2)=(1)^{100}=2a+b, 2a+b=1$

$a=0 , b=1$

Hence remainder :- $R(x)=1$