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Math Help - divide

  1. #1
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    Talking divide

    divide the first expression by the second.
    (12x^3-11x^y-12xy-25y^3)/(3x-5y)

    i got lost when divisor has two variables.... thanks

    yeah there is an error here.
    (12x^3-11x^2y-12xy-25y^3)/(3x-5y)

    and it is really 12xy there.
    thanks
    Last edited by princess_21; January 16th 2009 at 05:50 AM. Reason: error
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  2. #2
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    well there are a few ways to do this
    first is synthetic division which they tought you in school but I don't quite remember
    the next is just simply doing it one step at a time, it comes out to be the synthetic method I believe but this way you can see what you are doing

    (12x^3-11x^y-12xy-25y^3)/(3x-5y)
    so start with the first term
    12x^3, it is 4x^2 times the first term in the bottom
    so replace 12x^3 with 4x^2(3x - 5y) + 20yx^2 , is it clear why I did that?
    anyways now you have
    (4x^2) + (20yx^2-11x^y-12xy-25y^3)/(3x-5y)
    and you can keep going

    hopefully this helps, I just woke up so I'm still not entirely lucid
    Last edited by jbpellerin; January 15th 2009 at 09:00 AM. Reason: typo
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  3. #3
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    Hello, princess_21!

    I will assume there are typos in your problem . . .


    (12x^3-11x^{{\color{red}2}}y-12xy^{{\color{red}2}}-25y^3) \div (3x-5y)

    i got lost when divisor has two variables . . . really?


    . . \begin{array}{ccccccc}<br />
& & & 4x^2 & +3xy & +y^2 \\<br />
& & --- & --- & --- & --- \\<br />
3x-5y & ) & 12x^3 & -11x^2y & -12xy^2 & -25y^3 \\<br />
& & 12x^3 & -20x^2y \\<br />
& & --- & --- \\<br />
& & & 9x^2y & -12xy^2 \\<br />
& & & 9x^2y & -15xy^2 \\<br />
& & & -- & --- \\<br />
& & & & 3xy^2 & -25y^3 \\<br />
& & & & 3xy^2 & -5y^3 \\<br />
& & & & --- & --- \end{array}

    . . . . . . . . . . . . . . . . . . . . . . . . . . -20y^3


    Answer: . 4x^2 + 3xy + y^2 - \frac{20y^3}{3x-5y}

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  4. #4
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    Quote Originally Posted by princess_21 View Post
    divide the first expression by the second.
    (12x^3-11x^y-12xy-25y^3)/(3x-5y)

    i got lost when divisor has two variables.... thanks
    \frac{12x^3-11x^2y-12xy^2-25y^3}{3x-5y} =y^2\left(\frac{12a^3-11a^2-12a-25}{3a-5}\right) ,where a = \frac{x}{y} Now you need to factorise in one variable... Hope you wont get lost
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  5. #5
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    Cool

    yeah there is an error here.
    (12x^3-11x^2y-12xy-25y^3)/(3x-5y)

    and it is really 12xy there.
    thanks
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