# Remainder Theorem With Second Degree Divisor

What is the remainder when x2013 is divided by x2– 1? This problem was taken from 2009 MTAP-Dep-Ed Math Challenge Finals. The problem looks very easy but if you don’t know the basic principle how to deal with this problem you can never answer this in just a minute. This article will teach you how to deal with problem like this.

Let :

$P(x)$     be a polynomial with degree n.

$D(x)$     be the divisor in second degree.

$Q(x)$    be the quotient when $P(x)$ is divided by $D(x)$

$R(x)$     be the remainder when $P(x)$     is divided by $D(x)$

If $D(x)$ is a  1st degree expression, the remainder is constant.

If $D(x)$ is a second degree expression and  the remainder is a in 1st degree expression in the form of $Ax+B$.

We can say that $P(x) = D(x)(Q(x))+R(x)$

Worked Problem 1:

What is the remainder when $x^2013$ is divided by $x^2- 1?$

Given:

$P(x)= x^2013$                      $D(x)= x^2- 1 or (x-1)(x+1)$       Required: $R(x)$

Solution:

$P(x) = D(x)(Q(x))+R(x)$ $x^2013= (Q(x))(x-1)(x+1)+Ax +B$

Let $x=1$ to simplify things out.

$(1)^{2013}= (Q(1))(1-1)(1+1)+A(1) +B$

$1=A+B$,   let’s call this equation 1.

Let $x=-1$

$(-1)^{2013}= (Q(11))(-1-1)(-1+1)+A(-1) +B$

$-1=-A+B$ let’s call this equation 2.

Solve equation 1 and 2 simultaneously.

$A=1 , B=0$

Therefore the remainder is $1(x)+0$ or $x$.

Worked Problems 2:

What is the remainder when $x^{2014}-2x^{2013}+1$ is divided by $x^2-2x$?

Solution:

Given:

$P(x)= x^{2014}-2x^{2013}+1$        $x^2-2x$

$P(x) = D(x)(Q(x))+R(x)$ $x^{2014}-2x^{2013}+1= Q(x)(x)(x-2)+Ax+B$

Let x=2

$(2)^{2014}-2(2)^{2013}+1= Q(2)(2)(2-2)+A(2)+B$ $2^{2014}-2^{2014}+1= 0+2A+B$

$2A+B=1$,  equation 1

Let x=0

$(0)^{2014}-2(0)^{2013}+1= Q(0)(0)(0-2)+A(0)+B$ $1=B$

From equation 1,

$2A+B=1$

$2A+1=1$,

$A=0$

The remainder is $(0)x+1$ or simply $1$