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Polynomial division

  1. Mar 22, 2007 #1


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    I just cant remember how to do this!!! Ive been to several sites suggesting synthetic division and other guides on polynomial division but i cant get it into my head and its driving me wild. :surprised


    synthetic division doesn't work here because the denominator is larger than the numerator..... right?

    Heres my work ive done w/ that P/Q method:

    (x^2+3x+2) = (x+2)(x+1)

    therefore use the format:

    8x-8 = A/(x+2) + B(x+1)
    and solve for A and B

    This is where im stuck. I went further and solved A = 16 and B = -8 and pluged it in to get

    16/(x+2) - 8/(x+1) but this doesnt equal (8x-8) / (x^2+3x+2) !!!!!!!

    Im stuck!!! can someone please enlighten me? Thanks a ton!

    OR: If anyone has a good helpful website that they know of that shows polynomial division when the Denominator is larger than the numerator id appreciate it. I can only find the sites where the numerator is larger than the denominator
    Last edited: Mar 22, 2007
  2. jcsd
  3. Mar 22, 2007 #2


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    Try solving for A and B again! I thin you made a mistake somewhere. :smile:
  4. Mar 22, 2007 #3


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    What I think you're trying to do is called partial fractions. Polynomial division in this case is very easy: the quotient is 0 and the remainder is 8x-8. :wink:

    Your goal, presumably, is to write the quotient

    \frac{8x - 8}{x^2 + 3x + 2}

    in the form

    [tex]\frac{A}{x+2} + \frac{B}{x+1}.[/tex]

    So you want those to be equal, right? ...

    (clearly 8x-8 = A/(x+2) + B(x+1) was a typo -- what did you really mean by that?)
    Last edited: Mar 22, 2007
  5. Mar 23, 2007 #4


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    my goal was to simplify it enough so i can integrate it easially. im going to work on it some more right now and ill report back.
  6. Mar 23, 2007 #5


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    Ok heres my new work:

    8x-8 = A(x+2) + B(x+1)

    Sub in -2 for x to get B
    8(-2)-8 = A(-2+2) + B(-2+1)

    Simplify to
    B = 24

    then i subed in -1 for x to solve for A
    8(-1)-8 = A(-1+2) + B(-1+1)

    simplify to
    A = 16 <------------ Thats where my mistake was!!! its -16 not 16!!!!



    But this brings up another question. I tried to do a problem with a perfect square in the denominator and the method doesnt work... :-(

    This is the one i made right now

    (8x+12) / (x+1)^2

    [(x+1) (x+1)^2 (8x+12)] / (x+1)^2 = [A/(x+1) + B/ (x+1)^2 (x+1)(x+1)^2]

    simplifies to

    (x+1) (8x+12) = A(x+1)^2 +B(x+1)

    when i try to solve for A and B, no matter what number i sub in for X they both end up zeroing out and im left with nothing!

    Any insite this time? Thanks so much so far
    Last edited: Mar 23, 2007
  7. Mar 23, 2007 #6


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    Ok in my attempt i multiplied both sides by both terms. I tried it by multiplying both sides by one term at a time and i think its working... ill report back
  8. Mar 23, 2007 #7


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    Nope doesnt work. im stuck again :-( I ended getting A = 0 and B = 4. A = 0 doesnt really make a whole lotta sense.
  9. Mar 23, 2007 #8


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    Actually I just figured it out. I learned that my alegebra needs some work.

    A = 8
    B = 4
    THANKS ill be back later with more questions!
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