Remainder

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

SOLUTION
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
D(x)=x^2-1
x^{2006}-x^{2005}+(x+1)^2=x^{2006}-x^{2005}+x^2+2x+1
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
x^2-1|\overline{x^4-x^3+x^2+2x+1}
.\qquad\quad x^4 -x^2
.\qquad\quad\overline{\textrm{-}x^3+2x^2}+2x+1

.\qquad\quad x^2-x
x^2-1|\overline{x^4-x^3+x^2+2x+1}
.\qquad\quad x^4 -x^2
.\qquad\quad\overline{\textrm{-}x^3+2x^2}+2x+1
.\qquad\quad\textrm{-}x^3 +x
.\qquad\quad\overline{2x^2+x+}1

.\qquad\quad x^2-x+2
x^2-1|\overline{x^4-x^3+x^2+2x+1}
.\qquad\quad x^4 -x^2
.\qquad\quad\overline{\textrm{-}x^3+2x^2}+2x+1
.\qquad\quad\textrm{-}x^3 +x
.\qquad\quad\overline{2x^2+x+}1
.\qquad\quad 2x^2 -2
.\qquad\quad\overline{x+3}
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
\dfrac{x^{2005}(x-1)+x^2+2x+1}{x-1}=x^{2005}+\dfrac{x^2+2x+1}{x-1}
=x^{2005}+x+3+\dfrac{4}{x-1}
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}.
\dfrac{4}{x^2-1}+\dfrac{1}{x+1}=\dfrac{4+x-1}{x^2-1}
=\dfrac{x+3}{x^2-1}
The remainder equals x+3.

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About mvtrinh

Retired high school math teacher.
This entry was posted in Problem solving and tagged , , , , , , . Bookmark the permalink.

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