What is the remainder when x^{2006}-x^{2005}+(x+1)^2 is divided by x^2-1?
Source: NCTM Mathematics Teacher, November 2006

If a polynomial P(x) is divided by a polynomial D(x)\neq 0, there exists a quotient polynomial Q(x) and a remainder polynomial r(x) such that
P(x)=D(x)Q(x)+r(x) with r(x)<D(x).
P(x)= x^{2006}-x^{2005}+(x+1)^2
Though tedious it is not hard to do the long division by (x^2-1). A first few steps reveal that the quotient equals x^{2004}-x^{2003}+x^{2002}-x^{2001}+x^{2000}-x^{1999}+\cdots. Since we are after the remainder and not the quotient, let’s focus our attention on the last stage of the operation. It looks like the following
.\qquad\quad x^2
.\qquad\quad x^4 -x^2

.\qquad\quad x^2-x
.\qquad\quad x^4 -x^2
.\qquad\quad\textrm{-}x^3 +x

.\qquad\quad x^2-x+2
.\qquad\quad x^4 -x^2
.\qquad\quad\textrm{-}x^3 +x
.\qquad\quad 2x^2 -2
The remainder equals x+3.

Answer: x+3

Alternative Solution
The divisor x^2-1=(x-1)(x+1). We are going to make two divisions, first by (x-1) then by (x+1). We will get two remainders from which we calculate the final remainder.
The first division (either long or synthetic) by (x-1) yields a remainder of 4
The second division of the above expression by (x+1) yields a remainder of 1.
The rational expressions of the two remainders are \dfrac{4}{x^2-1} and \dfrac{1}{x+1}.
The remainder equals x+3.


About mvtrinh

Retired high school math teacher.
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