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)


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)