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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 $(2^{100}-1)x-(2^{99}-1)$