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Polynomials & divisibility

  1. Sep 24, 2009 #1
    1. The problem statement, all variables and given/known data
    A polynomial p(x) leaves the rest 3 when divided by (x+2) and the rest 8 when divided by (x-6). What's the rest r(x) when p(x) is divided by (x+2)(x-6)?


    2. Relevant equations



    3. The attempt at a solution
    I wrote the three equations:

    p(x)=q1(x+2) + 3
    p(x)=q2(x-6) + 8
    p(x)=q3(x+2)(x-6) + r(x)

    And I've tried rearranging them and I've tried to find what p(x) is and I've tried some other random things and it doesn't work out. I think the main problem is that I don't know at which angle to approach this problem and that I don't know how to reason about it..
     
  2. jcsd
  3. Sep 24, 2009 #2

    tiny-tim

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    Hi Hannisch! :smile:
    Hint: now use the fact that r(x) must be of the form Ax + B :wink:
     
  4. Sep 24, 2009 #3

    Dick

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    Multiply your first equation by (x-6) and your second equation by (x+2) and subtract them.
     
  5. Sep 26, 2009 #4
    I can get an answer using Dick's method:

    p(x)=q3(x+2)(x-6)+ (5/8)x + (34/8), where q3 = (q2-q1)/8

    But I don't know if it's correct or not? (I think it is, because I've got another problem just like this one that I do have the answers to and I applied the same method and got the correct answer.)

    And I'm not sure how to apply tiny-tim's tip? I mean, sure I can put p(x)=q3(x+2)(x-6) + r(x) = p(x)=q3(x+2)(x-6) + ax + b, and then again try to fix it, but basically I fall back into the same stupid situation I was in before..
     
  6. Sep 26, 2009 #5

    Dick

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    tiny-tim is just suggesting you use p(-2)=3=r(-2) and p(6)=8=r(6). If you set r(x)=ax+b, that's two equations in the two unknowns, a and b. And your answer is just fine. Check it with this method.
     
  7. Sep 27, 2009 #6
    Ohh, I get it now (it took me a while, I've really stared myself blind on this problem)! Anyway, I do get the same answer now (I got it wrong first, but it was a ridiculous mistake, really), so thank you very, very much!
     
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