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Dividing complex polynomials

  1. Sep 27, 2014 #1
    1. The problem statement, all variables and given/known data

    (3y-2/y+3) - (3y+1)/(y2+6y-9)
    2. Relevant equations


    3. The attempt at a solution
    Ok, I have attempted to solve in 2 steps, step 1: solve 3y-2/y+3 step 2: solve 3y+1/y2+6y-9 and then subtract the answers. This doesn't seem to work, as I get:

    3y-2/y+3 = 3 remainder 11
    and
    3y+1/y2+6y-9 = 3y+1/(y-3)(y-3)
    *y2+6y-9 simplified = (y-3)(y-3)

    so final answer I got was 3r11-[(3y+1)/(y-3)(y-3)], which doesn't make sense. I think I am approaching the problem wrong, please provide some direction...
     
  2. jcsd
  3. Sep 27, 2014 #2

    HallsofIvy

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    Actually, "dividing polynomials" is not relevant here. This is a problem in subtracting fractions- and you need to get a common denominator.
    The denominator of the first fraction is y+ 3 and the denominator of the second fraction is [itex]y^2+6y- 9[/itex]. That is NOT (y- 3)(y- 3). If you were to multiply y-3 by y- 3 you would get [itex]y^2- 6y+ 9[/itex]- the signs are wrong. In fact, if we were to "complete the square" we would get [itex]y^2+ 6y+ 9- 9- 9= (y+ 3)^2- 18= (y+3- 3\sqrt{2})(y+ 3+ 3\sqrt{2})[/itex]. In any case that has no factors in common with x+ 3 so you need to multiply the numerator and denominator of the first fraction by [itex]x^2+ 6x- 9[/itex] and numerator and denominator of the second fraction by x+ 3.
     
    Last edited: Sep 27, 2014
  4. Sep 27, 2014 #3

    mfb

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    What do you mean with "solve"? There is no equation you could solve.
    Simplify? Then I don't understand your approach. Writing the second denominator as product is a good start (but check the signs)

    Where do remainders come from? Division in the complex numbers does not have or need that concept.
     
  5. Sep 27, 2014 #4

    Ray Vickson

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    You write
    [tex] \left( 3y - \frac{2}{y} +3 \right) - \frac{3y+1}{y^2 + 6y - 9}[/tex]
    Do you mean that, or did you mean
    [tex] \frac{3y - 2}{y+3} -\frac{3y+1}{y^2 + 6y - 9} \:? [/tex]
    If you meant the latter, use parentheses, like this: (3y-2)/(y+3) - (3y+1)/(y^2 + 6y - 9).

    Anyway, ##y^2 + 6y - 9 \neq (y-3)^2##. Are you sure you have copied out the problem correctly? The version you wrote does NOT simplify at all, but just becomes messier as you do more work on it.
     
  6. Sep 27, 2014 #5
    My apologies to everyone that has attempted this so far, I transcribed the problem wrong, here is the actual problem:

    (3y-2)/(y+3) - (3y+1)/(y^2 + 6y + 9)

    In the original problem, I placed the parenthesis wrong AND put the wrong sign in one place...I am very sorry and need help with this corrected problem...

    in which simplifying (y^2+6y+9) = (y+3)^2
     
  7. Sep 27, 2014 #6
    Thanks, you caught where I copied the problem wrong, and where I misused the parenthesis, it was the latter, with +9 instead of -9!
     
  8. Sep 27, 2014 #7
    With the problem copied correctly and reworked, I came up with:

    (3y-2)/(y+3) - (3y+1)/(y+3)^2

    (y+3)^2 LCD

    (y+3)(3y-2) = 3y^2-2y+9y-6 =3y^2+7y-6

    so now the problem has been simplified to:

    (3y^2+7y-6)/(y+3)^2 - (3y+1)/(y+3)^2

    (3y^2 + 7y - 6 - 3y -1)/(y+3)^2

    (3y^2 + 4y - 7)/(y+3)^2

    (y-1)(3y+7)/(y+3)^2 :correct answer??
     
  9. Sep 27, 2014 #8

    Ray Vickson

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    Yes.
     
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