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Math Help - I got stuck...mmm...I keep thinking how to get out here

  1. #1
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    I got stuck...mmm...I keep thinking how to get out here

    See attached file, please
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  2. #2
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    (6a 10) (a + 2) (10a 2) (a 2)
    (2a + 4) (2a 4) (3a 5) (9a2 + 15a + 25)
    2

    must cancel (2a + 4) not (2a - 4). Be careful.
    -O
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  3. #3
    A riddle wrapped in an enigma
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    Quote Originally Posted by Alienis Back View Post
    See attached file, please
    Hi Alienis,

    \frac{6a^2+2a+20}{4a^2-16}\cdot \frac{10a^2-22a+4}{27a^3-125}

    I'm going to approach this from a different angle. First factor out any common factors in your numerators and denominators.

    \frac{2(3a^2+a+10}{4(a^2-4)}\cdot \frac{2(5a^2-11a+2)}{27a^3-125}

    Next, factor what you have left.

    \frac{2(3a-5)(a+2)}{4(a-2)(a+2)}\cdot \frac{2(5a-1)(a-2)}{(3a-5)(9a^2+15a+25)}

    Now, simplify all you can

    \frac{\rlap{---}2(\rlap{---------}3a-5) \rlap{----------}(a+2)}{\rlap{---}4\rlap{----------}(a-2) \rlap{----------}(a+2)}\cdot \frac{\rlap{---}2(5a-1) \rlap{----------}(a-2)}{\rlap{----------}(3a-5)(9a^2+15a+25)}

    \frac{5a-1}{9a^2+15a+25}
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