problem 4 & 5 is wrong

x^100 =(x^2 -3x +2)(q(x)) + (ax+b)

x^100 =(x-1)(x-2)(q(x)) + (ax+b)

At x=1 ,

1^100=0*q(1) +a(1)+b

1=a+b (1)

At x=2 ,

2^100=0*q(2) +a(2)+b

2^100=2a+b (2)

Solve (1) and (2) by elimination

2^100 -1= a

1=2^100 -1 +b

b=2-2^100

ax+b=(2^100 -1)x + (2-2^100) which equals to (2^100-1)X-(2^99-1)×2

I started dividing powers of $x$ by $x^2-3x+2,$ starting with $x$ and then progressing to higher powers and noticed a pattern. For $x^n$ divided by $x^2-3x+2$ , the following always held: $R(x)=(2^n-1)x-2(2^{n-1}-1).$ So, let's prove this by induction.

Base case: For $x$ divided by $x^2-3x+2,$ the remainder is $x,$ which equals $(2^1-1)x-2(2^{1-1}-1).$

Suppose that for some $n,$ the remainder, when $x^n$ is divided by $x^2-3x+2$ , is $R(x)=(2^n-1)x-2(2^{n-1}-1).$

So, there is some polynomial $p(x)$ such that $x^n = (x^2-3x+2)p(x) + (2^n-1)x-2(2^{n-1}-1).$

$\Rightarrow x^{n+1} = (x^2-3x+2)xp(x) + (2^n-1)x^2-2(2^{n-1}-1)x$

$=(x^2-3x+2)(xp(x)+2^n-1)-(2^n-1)x^2+3(2^n-1)x-2(2^n-1) +(2^n-1)x^2-2(2^{n-1}-1)x$

$=(x^2-3x+2)(xp(x)+2^n-1)+[3(2^n-1)-2(2^{n-1}-1)]x-2(2^n-1)$

$=(x^2-3x+2)(xp(x)+2^n-1) + [2^{n+1}-1]x-2(2^n-1)$

We see that this property of the remainder holds for $n+1.$ Therefore, by the principle of mathematical induction, this formula for $R(x)$ holds for all $n\geq 1.$

The answer to this problem follows.