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Remainder Theorem

  1. Apr 23, 2009 #1
    1. The problem statement, all variables and given/known data

    Find the remainder when (x^80 - 8x^30 + 9x^24 + 5x + 6) is divided by (x+1)

    2. Relevant equations

    3. The attempt at a solution

    So I'm not really sure where to start. I tried starting by doing long polynomial division, but I get stuck. How do I start this?
     
  2. jcsd
  3. Apr 23, 2009 #2

    rock.freak667

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    What does the remainder theorem say?
     
  4. Apr 23, 2009 #3

    Dick

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    Yeah, you would get stuck doing the division. It's a long haul. But look, suppose you did do the division f(x)=(x^80 - 8x^30 + 9x^24 + 5x + 6) by (x+1)? Then you would get f(x)=q(x)*(x+1)+r, right? Where q(x) is the quotient and r is the remainder. What happens if you put x=(-1) into that?
     
  5. Apr 23, 2009 #4
    Remainder Theorem:
    If p(x) / (x – a) = q(x) with remainder r(x),

    then p(x) = (x – a) q(x) + r(x).
     
  6. Apr 23, 2009 #5
    You get r ?
     
  7. Apr 23, 2009 #6

    Dick

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    Well, yes. You get f(-1)=r. That's the remainder theorem. So what is r?
     
  8. Apr 23, 2009 #7
    Do you get:

    (1 - 8 + 9 -5 + 6) = 3 ? So r = 3?
     
  9. Apr 23, 2009 #8

    Dick

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    Sure. If you don't believe it make up a simpler example where you can actually do the long division and check that it works. It's good for you.
     
  10. Apr 23, 2009 #9
    Thanks. How did you know to use -1?
     
  11. Apr 23, 2009 #10

    Dick

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    Look back at the problem. I'll give you three guesses. The first one had better be right.
     
  12. Apr 23, 2009 #11
    Because a = x + 1, so x = -1?
     
  13. Apr 23, 2009 #12

    Dick

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    What's a???? If a=x+1 then x=a-1. You are onto the second guess.
     
    Last edited: Apr 23, 2009
  14. Apr 23, 2009 #13
    Well heck, I'm not sure. :O
    I assume we're using -1 because of something to do with (x+1) = 0 or something?
     
  15. Apr 23, 2009 #14

    Dick

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    Yes, if you had spent all day figuring out the q(x) in f(x)=q(x)*(x+1)+r by doing the horrible division, at the end of it all you could realize that you didn't need to find q(x) at all because if you put x=(-1) the q(x) disappears. That's the remainder theorem.
     
  16. Apr 23, 2009 #15
    Sweets.
    So if for example I was dividing by (x-4), I would use 4 instead of -1?
     
  17. Apr 23, 2009 #16

    Dick

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    f(x)=q(x)*(x-4)+r. Sure, f(4)=r. You don't need to find q(x) before you know the remainder.
     
  18. Apr 23, 2009 #17
    How cool is that.
     
  19. Apr 23, 2009 #18

    Dick

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    Way cool.
     
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