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.

remainder theoremLet :

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

 

 

 

 

 

 

 

 

 

Dan

Dan

Blogger and a Math enthusiast. Has no interest in Mathematics until MMC came. Aside from doing math, he also loves to travel and watch movies.
Dan

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