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Math Help - Another factorising algebraic fraction

  1. #1
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    Another factorising algebraic fraction

    Simplify:
    (x^2 - y^2 - 2x - 2y) X (xy-3y^2)
    (x^2 - 2xy - 3y^2) X (y-x+2)
    Guide through it will be appreciated, thanks
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  2. #2
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    \frac{(x^2 - y^2 - 2x - 2y)(xy - 3y^2)}{(x^2 - 2xy - 3y^2)(y - x + 2)}

    Take out a factor of y on the top from the last bracket, and factorise the first bracket on the bottom:

    = \frac{y(x^2 - y^2 - 2x - 2y)(x - 3y)}{(x - 3y)(x + y)(y - x + 2)}

    Cancel the (x - 3y) term which appears on both the top and bottom:

    = \frac{y(x^2 - y^2 - 2x - 2y)}{(x + y)(y - x + 2)}

    Expand the brackets on the bottom:

    = \frac{y(x^2 - y^2 - 2x - 2y)}{-x^2 + y^2 + 2x + 2y}

    Take out a factor of -1 on the bottom:

    = \frac{y(x^2 - y^2 - 2x - 2y)}{-(x^2 - y^2 - 2x - 2y)}

    Now cancel the term (x^2 - y^2 - 2x - 2y) which appears on the top and bottom of the fraction:

    = \frac{y}{-1} = -y
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  3. #3
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    Thanks!
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  4. #4
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    Hello, Supaweak!

    Another approach . . .


    Simplify: . \frac{(x^2-y^2-2x-2y)\cdot(xy-3y^2)}{(x^2-2xy-3y^2)\cdot(y-x+2)}

    Factor each expression . . .

    . . x^2-y^2 - 2x - 2y \:=\:(x-y)(x+y) - 2(x+y) \:=\:(x+y)(x-y-2)

    . . xy-3y^2 \:=\:y(x-3y)

    . . x^2-2xy - 3y^2 \:=\:(x+t)(x-3y)

    . . y - x + 2 \:=\:-(x-y-2)



    The problem becomes: . \frac{(x+y)(x-y-2)\cdot y(x-3y)}{(x+y)(x-3y) \cdot -(x-y-2)}


    Reduce: . \frac{{\color{blue}\rlap{//////}}(x+y){\color{red}\rlap{/////////}}(x-y-2)\cdot y{\color{green}\rlap{///////}}(x-3y)}{{\color{blue}\rlap{//////}}(x+y){\color{green}\rlap{///////}}(x-3y) \cdot -{\color{red}\rlap{/////////}}(x-y-2)} \;=\;\frac{y}{\text{-}1} \;=\;\boxed{-y}

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  5. #5
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    Thanks Soroban! I was gonna ask how (x-y)(x+y) - 2(x+y) = (x+y)(x-y-2) worked but I get it now, Thanks again for your help!
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