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Division of Unknown Polynomials


Date: 3/18/96 at 15:6:36
From: Anonymous
Subject: Algebra problem

I am an eighth grade student at the Willmar Junior High School in 
Willmar, Minnesota.  I have a math question that I need help on.  

When a polynomial P(x) is divided by x-1, the remainder is 3.
When P(x)is divided by x-2, the remainder is 5.  Find the 
remainder when P(x) is divided by x^2-3x+2.  

Thank you,
Jacob


Date: 3/22/96 at 19:28:51
From: Doctor Steven
Subject: Re: Algebra problem

This is a tough problem for an eighth grade class!

First let's look at what we know:

    1. We know P(x) when divided by (x-1) gives a remainder of 3.
       So P(x) = f(x)*(x-1) + 3.  (We don't care what f(x) is)

    2. We know P(x) when divided by (x-2) gives a remainder of 5.
       So P(x) = g(x)*(x-2) + 5, also.(We don't care what g(x) is 
       either!)

    So P(x) = f(x)*(x-1) + 3 = g(x)*(x-2) + 5.

    Subtract 5 from every part of this three part equation to get:
    P(x) - 5 = f(x)*(x-1) - 2 = g(x)*(x-2) + 0.             

    Well, we also know that (x-1) = 1*(x-2) + 1. So the remainder of
    x-1 when divided by x-2 is +1.

    Now check this out for some higher mathematics (really though it's
    pretty easy):
                  ____                                    ___
        Look at 4| 5   its remainder is 1.  Now look at 4| 3  its 
        remainder is 3.
                      ____
        Now look at 4| 15  its remainder is 3, or 3*1 the product               
        of the remainders of 5/4 and 3/4.  This works for anything 
        that is multiplied together.

    So f(x)*(x-1) - 2 better not have a remainder when divided by 
    (x-2) (since it equals the far righthand side, which definitely 
    doesn't have a remainder when divided by (x-2) ).

    This tells us that f(x) better have a remainder of 2 when 
    divided by (x-2) (since rem(2)*rem(1) - rem(2) = rem(0) ).

    So f(x) = m(x)*(x-2) + 2. (We don't care what m(x) is)

    Now we have:

        P(x) - 5 = [m(x)*(x-2) + 2](x-1) - 2 = g(x)(x-2) + 0.

    But all we need to look at now is:

        P(x) - 5 = [m(x)*(x-2) + 2](x-1) - 2.

    Add 5 to both sides to get:

        P(x) = [m(x)*(x-2) + 2](x-1) + 3.

    Then simplify on the right to get:

        P(x) = m(x)*(x-2)*(x-1) + (2x - 2) + 3.

    Simplify some more to get:

        P(x) = m(x)*(x^2 - 3x + 2) + 2x + 1.

    So the remainder of P(x) when divided by x^2 - 3x + 2 is 
        (2x + 1) 

Whew! That was a toughy. ;)

-Doctor Steven,  The Math Forum

    
Associated Topics:
High School Polynomials

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