* Problem:* The remainder R(x) obtained by dividing the polynomial \(x^{100}\) by the polynomial \(x^2-3x+2\) is

**(A)** \(2^{100}-1\)

**(B)** \((2^{100}-1)x-(2^{99}-1)\)

**(C)** \(2^{100}x-3(2^{100})\)

**(D)** \((2^{100}-1)x+(2^{99}-1)\)

**SOLUTION:****(B)**

The the divisor is a quadratic term .So, R(x) must be 1 degree less than divisor.

We know , \(f(x)=divisor.Q(x)+R(x)\)

\(=> x^{100}=(x^2-3x+2).Q(x)+ (ax+b)\) \(=> x^{100}=(x-1)(x-2).Q(x)+ (ax+b)\)when ,\(x=2\)

\(=> 2^{100}=(2-1)(2-2).Q(x)+ (2a+b)\) \(=> 2^{100}= (2a+b)……….(i)\)when ,\(x=1\)

\(=> 1^{100}=(1-1)(1-2).Q(x)+ (a+b)\) \(=> 1^{100}=(a+b)………..(ii)\)solving two equation we get, \(a=(2^{100}-1)\)

and, \(b=-(2^{99}-1)\)

The remainder R(x) is \((2^{100}-1)x-(2^{99}-1)\)