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Algebraic Fraction

  1. Sep 7, 2005 #1
    hey,
    I am having trouble with the following...

    (6x^2+5x-6)/(6x^2+13x+6) TIMES (3x^2-4x-4)/(3x^2-8x+4)

    i can factorise them but i dont know what to do after that...

    Thanks for your help
     
  2. jcsd
  3. Sep 8, 2005 #2

    Fermat

    User Avatar
    Homework Helper

    You're meant to simplify the expression, yes?

    Just cancel out the common factors. I don't think it can be simplified any further
     
  4. Sep 8, 2005 #3
    yer, but what is left is nasty haha and its wrong
     
  5. Sep 8, 2005 #4
    it simplifys all the way too 1 eventually doesn't it?
     
  6. Sep 8, 2005 #5
    ummm nar.....

    this is what i can get to...
    {(3x+3)(2x-2)/(2x+3)(3x+2)} + {(3x+2)(x-2)/(3x-2)(x-2)}

    and the answer is...

    {2(9x^2+4)}/(3x-2)(3x+2)

    i just dont know how to get there haha
     
  7. Sep 8, 2005 #6
    nice...it does....how about it with a plus instead of a multiply between the two factions
     
  8. Sep 8, 2005 #7
    Hmm, I seem to get
    [tex]\left( {\frac{{6x^2 + 5x - 6}}
    {{6x^2 + 13x + 6}}} \right)\left( {\frac{{3x^2 - 4x - 4}}
    {{3x^2 - 8x + 4}}} \right) = \left[ {\frac{{\left( {3x + 2} \right)\left( {2x - 3} \right)}}{{\left( {3x + 2} \right)\left( {2x + 3} \right)}}} \right]\left[ {\frac{{\left( {3x + 2} \right)\left( {x - 2} \right)}}{{\left( {3x - 2} \right)\left( {x - 2} \right)}}} \right] = \frac{{\left( {3x + 2} \right)\left( {2x - 3} \right)}}{{\left( {3x - 2} \right)\left( {2x + 3} \right)}} [/tex]

    Then I suppose?

    [tex]\frac{{6x^2 + 5x - 6}}{{6x^2 + 13x + 6}} + \frac{{3x^2 - 4x - 4}}
    {{3x^2 - 8x + 4}} = \frac{{\left( {3x + 2} \right)\left( {2x - 3} \right)}}{{\left( {3x + 2} \right)\left( {2x + 3} \right)}} + \frac{{\left( {3x + 2} \right)\left( {x - 2} \right)}}{{\left( {3x - 2} \right)\left( {x - 2} \right)}} = \frac{{\left( {2x - 3} \right)\left( {3x - 2} \right) + \left( {3x + 2} \right)\left( {2x + 3} \right)}}{{\left( {2x + 3} \right)\left( {3x - 2} \right)}} = [/tex]
    [tex] \frac{{2x - 3}}{{2x + 3}} + \frac{{3x + 2}}{{3x - 2}} = \frac{{12\left( {x^2 + 1} \right)}}{{\left( {2x + 3} \right)\left( {3x - 2} \right)}} = \frac{4}{{3x - 2}} - \frac{6}{{2x + 3}} + 2 [/tex]
     
    Last edited: Sep 8, 2005
  9. Sep 8, 2005 #8
    is that with a plus in between the fractions?
     
  10. Sep 8, 2005 #9
    Yes, LaTex doesn't clearly space out the signs :rolleyes:
     
  11. Sep 8, 2005 #10

    TD

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    I believe it's a but simpler, perhaps you made some slight mistakes.

    [tex]\frac{{6x^2 + 5x - 6}}
    {{6x^2 + 13x + 6}} \cdot \frac{{3x^2 - 4x - 4}}
    {{3x^2 - 8x + 4}} = \frac{{\left( {2x + 3} \right)\left( {3x - 2} \right)}}
    {{\left( {2x + 3} \right)\left( {3x + 2} \right)}} \cdot \frac{{\left( {x - 2} \right)\left( {3x + 2} \right)}}
    {{\left( {x - 2} \right)\left( {3x - 2} \right)}} = 1[/tex]

    Cancels out nicely :smile:
     
  12. Sep 8, 2005 #11
    hmmm, maybe the answer is wrong.... it says {2(9x^2+4)}/(3x-2)(3x+2)
     
  13. Sep 8, 2005 #12
    or for the multiplication one you can expand the polynomial and the top clearly equals the bottom with reduced thinking needed!
     
    Last edited: Sep 8, 2005
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