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Polynomial and Remainder


Date: 02/19/98 at 15:23:36
From: George Freeman
Subject: Division of unkown polynomials

An unknown polynomial f(x) of degree 37 yields a remainder of 1 when 
divided by x-1, a remainder of 3 when divided by x-3, a remainder of 
21 when divided by x-5. Find the remainder when f(x) is divided by 
(x-1)(x-3)(x-5).


Date: 02/19/98 at 16:53:05
From: Doctor Rob
Subject: Re: Division of unkown polynomials

Okay!  There are a couple of ways of solving this.


The first method uses the Remainder Theorem:  

    f(x) = (x-a)*g(x) + f(a).

The first condition says that 
    
    f(x) = (x-1)*g(x) + 1.  

The second condition says that 

    f(x) = (x-1)*g(x) + 1 = (x-3)*h(x) + 3.  

Substituting in x = 1 tells you that 1 = -2*h(1) + 3, so h(1) = 1, so 
h(x) = (x-1)*k(x) + 1. 

Then f(x) = (x-3)*[(x-1)*k(x) + 1] + 3 = (x-1)*(x-3)*k(x) + x.  

The third condition says that f(x) = (x-5)*m(x) + 21.  Substituting 
in x = 5 tells you that 21 = (5-1)*(5-3)*k(5) + 5, so k(5) = 2, so 
k(x) = (x-5)*n(x) + 2.

Then  f(x) = (x-1)*(x-3)*[(x-5)*n(x) + 2] + x,
           = (x-1)*(x-3)*(x-5)*n(x) + 2*(x-1)*(x-3) + x,
           = (x-1)*(x-3)*(x-5)*n(x) + 2*x^2 - 7*x + 6.

Thus the remainder sought is 2*x^2 - 7*x + 6.

The second method uses the Chinese Remainder Theorem.

    f(x) = 1  (mod x-1),
    f(x) = 3  (mod x-3),
    f(x) = 21 (mod x-5).

First we must find any polynomials a(x), b(x) and c(x) such that

   a(x)*(x-3)*(x-5) + b(x)*(x-1)*(x-5) + c(x)*(x-1)*(x-3) = 1

a(x) = 1/8, b(x) = -1/4, and c(x) = 1/8 will do. Then the answer is 
given by

  f(x) = 1*a(x)*(x-3)*(x-5) + 3*b(x)*(x-1)*(x-5) + 21*c(x)*(x-1)*(x-3)
                                              (mod (x-1)*(x-3)*(x-5)),

       = 1*(1/8)*(x-3)*(x-5) + 3*(-1/4)*(x-1)*(x-5) + 
        21*(1/8)*(x-1)*(x-3)
                                              (mod (x-1)*(x-3)*(x-5)),

       = x^2*(1/8-3/4+21/8) + x*(-1+9/2-21/2) + (15/8-15/4+63/8)
                                              (mod (x-1)*(x-3)*(x-5)),
 
       = 2*x^2 - 7*x + 6 (mod (x-1)*(x-3)*(x-5)),

and the remainder is 2*x^2 - 7*x + 6.


It is comforting that the answers agree!

Note that the degree is immaterial, as long as it is at least 2.

-Doctor Rob,  The Math Forum
Check out our web site http://mathforum.org/dr.math/   
    
Associated Topics:
High School Polynomials

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