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Math Help - Simplifying and Factorising Algebraic Fractions

  1. #1
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    Simplifying and Factorising Algebraic Fractions

    Hi, how do I fully solve this problem?
    I've factorised both denominators, but not the numerators.
    Please help.


    Thanks
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  2. #2
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    Quote Originally Posted by BG5965 View Post
    Hi, how do I fully solve this problem?
    I've factorised both denominators, but not the numerators.
    Please help.


    Thanks
    2x^2-3xy+4x-6y = 2x^2+4x - 6y -3xy = 2x(x+2)-3y(x+2) = (x+2)(2x-3y)

    should 4x^2+6xy+9y be y^2 at the end?
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  3. #3
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    Quote Originally Posted by BG5965 View Post
    I've factorised both denominators, but not the numerators.

    You have the following:

    . . . . . \frac{2x^2\, -\, 3xy\, +\, 4x\, -\, 6y}{8x^3\, -\, 27y}\, \times\, \frac{4x^2\, +\, 6xy\, +\, 9y}{x^2\, +\, x\, -\, 2}

    You say that you have factored both of the denominators. I do not see any way of factoring the first denominator, since it is not a difference of squares or cubes, and contains no common factors, so I would be interested in seeing your work for this. I will guess that the other factorization led to the following:

    . . . . . \frac{2x^2\,  -\, 3xy\, +\, 4x\, -\, 6y}{8x^3\, -\, 27y}\, \times\, \frac{4x^2\, +\, 6xy\, +\, 9y}{(x\, +\, 2)(x\, -\, 1)}

    To learn how to factor the first numerator, try here. You should get the following:

    . . . . . \frac{(2x\, -\, 3y)(x\, +\, 2)}{8x^3\, -\, 27y}\, \times\, \frac{4x^2\, +\, 6xy\, +\, 9y}{(x\, +\, 2)(x\, -\, 1)}

    Since the last term in the second numerator is not 9y^2, I don't see any factoring you can do here, either. So you're stuck with just the one cancellation, and then multiplying the rest out.

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