Problem: The remainder R obtained by dividing x100 by x2−3x+2 is a polynomial of degree less than 2. Then R may be written as:
Answer Choices:
A. 2100−1
B. 2100(x−1)−(x−2)
C. 2100(x−3)
D. x(2100−1)+2(299−1)
E. 2100(x+1)−(x+2)
Solution:
x100=Q(x)(x2+3x+2)+R(x)=Q(x)(x−2)(x−1)+R(x) where R(x)=ax+b.
∴R(1)=1100=1=a+b and R(2)=2100=2a+b
∴a=2100−1, b=2−2100.∴R(x)=x(2100−1)+2−2100=x⋅2100−x+2−2100.
∴R(x)=2100(x−1)−(x−2).