Question

The remainder on dividing x^100-x^51+1 by x^2-1 is Ax+B Then A+2B

Answer 1

Step-by-step explanation:

by reminder theorem.

23

Down vote

Accepted

This is the standard approach, especially if you know the roots of the divisor.

Let f(x)=x100−2x51+1, and f(x)=g(x)(x2−1)+ax+b be the division

Then, 0=f(1)=g(1)(12−1)+a(1)+b=a+b,

and 4=f(−1)=g(−1)((−1)2−1)+a(−1)+b=−a+b.

Hence a=−2,b=2.

Thus the remainder is −2x+2.