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Math Help - Remainders of a Unknown Polynomial.

  1. #1
    Senior Member MacstersUndead's Avatar
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    Remainders of a Unknown Polynomial.

    When P(x) is divided by x-1 the remainder is 1, and when divided by (x-2)(x-3), the remainder is 5. Find the remainder when P(x) is divided by (x-1)(x-2)(x-3).

    I'm using the formula where
    P(x)=(x-1)(x-2)(x-3)Q(x)+ax^2+bx+c
    --

    hints / steps to solve this will be greatly appreciated. thanks in advance.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by MacstersUndead View Post
    When P(x) is divided by x-1 the remainder is 1, and when divided by (x-2)(x-3), the remainder is 5. Find the remainder when P(x) is divided by (x-1)(x-2)(x-3).

    I'm using the formula where
    P(x)=(x-1)(x-2)(x-3)Q(x)+ax^2+bx+c
    --

    hints / steps to solve this will be greatly appreciated. thanks in advance.
    Put R(x)=ax^2+bx+c then R(x)=(x-1)(ax+(a+b)) + (a+b+c), but we know that the remainder when $$ p(x) is divided by $$ x-1 is $$ 1 so we have a+b+c=1.

    A similar process for the other remainder will give you two more equations in $$ a,\ b and $$ c which you then solve.

    CB
    Last edited by CaptainBlack; June 28th 2010 at 10:41 PM. Reason: removed plural from remainders
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    Senior Member MacstersUndead's Avatar
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    Thank you, CB! It makes sense now. I just couldn't figure it out trying to help the one who showed me this problem. Although, he might have to re-iterate it to make sure it's clear.

    From my understanding of the problem, however. there are only two remainders given. one when dividing by (x-1) and the other by (x-2)(x-3), but in that case you would have one dependent variable. and therefore many remainders.
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by MacstersUndead View Post
    Thank you, CB! It makes sense now. I just couldn't figure it out trying to help the one who showed me this problem. Although, he might have to re-iterate it to make sure it's clear.

    From my understanding of the problem, however. there are only two remainders given. one when dividing by (x-1) and the other by (x-2)(x-3), but in that case you would have one dependent variable. and therefore many remainders.
    For the other remainder you have:

    ax^2+bx+c=a(x-2)(x-3)+5

    Expand and equate coefficients and you will find that you have two equations.

    CB
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  5. #5
    Senior Member MacstersUndead's Avatar
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    <br />
R(x)=ax^2+bx+c<br />
    <br />
R(x)=(x-2)(x-3)a +(b+5a)x+(c-6a)<br />
    if my polynomial long division is correct. I'm a bit rusty. so that would mean
    b+5a = 0 and c-6a = 5?

    thanks again for all your help.
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