Finding big remainder in a small way (Tomato objective 288 )

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

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