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Math Help - Need help factoring

  1. #1
    Newbie
    Joined
    Nov 2009
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    Need help factoring

    Hi
    I was wondering if anyone could help me with this equation i have, im trying to factor this in my ti92+ calculator and im not seeing the textbook result my book shows. Im told to factor this

    1/9*y^2 - 1/6*y*x^2 + 1/16*x^4

    the answer the book gives is

    (1/3*y - 1/4*x^2)^2

    is there anything special i need to do to factor this equation.
    The calculator gives me this as the answer

    (3*x^2 - 8)^2 / 144

    I know if i factor it by hand i can get the proper answer but i have alot of these to do and im not sure what im doing wrong. Ive also tried this in the sage math application and it didnt show the proper form either.


    Any help would be appreciated thanks.
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  2. #2
    MHF Contributor Amer's Avatar
    Joined
    May 2009
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    Jordan
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    1,093
    Quote Originally Posted by newbie43 View Post
    Hi
    I was wondering if anyone could help me with this equation i have, im trying to factor this in my ti92+ calculator and im not seeing the textbook result my book shows. Im told to factor this

    1/9*y^2 - 1/6*y*x^2 + 1/16*x^4

    the answer the book gives is

    (1/3*y - 1/4*x^2)^2

    is there anything special i need to do to factor this equation.
    The calculator gives me this as the answer

    (3*x^2 - 8)^2 / 144

    I know if i factor it by hand i can get the proper answer but i have alot of these to do and im not sure what im doing wrong. Ive also tried this in the sage math application and it didnt show the proper form either.


    Any help would be appreciated thanks.
    1/9*y^2 - 1/6*y*x^2 + 1/16*x^4

    \frac{y^2}{9} - \frac{yx^2}{6} + \frac{x^4}{16}

    \left(\frac{1}{9}\right)\left(y^2 - \frac{yx^2 \cdot 9}{6} + \frac{9x^4}{16}\right)

    you can consider that x is a constant and apply the general law (I do not know what is the mathematics name)

    y = \frac{-b \mp \sqrt{b^2 -4ac}}{2a}

    you have
    a=1 , b=\frac{-9x^2}{6} , c=\frac{9x^4}{16}

    y= \frac{\frac{9x^2}{6} \mp \sqrt{\left(\frac{9x^2}{6}\right)^2-4(1)\left(\frac{9x^4}{16}\right)}}{2}

    y=\frac{\frac{9x^2}{6}\mp 0 }{2} = \frac{\frac{3x^2}{2}}{2}

    y=\frac{3x^2}{4}

    so

    y^2 - \frac{yx^2 \cdot 9}{6} + \frac{9x^4}{16} = \left(y - \frac{3x^2}{4} \right)^2

    but

    \left(\frac{1}{9}\right)\left(y^2 - \frac{yx^2 \cdot 9}{6} + \frac{9x^4}{16}\right) = \left(\frac{1}{9} \right)\left(y-\frac{3x^3}{4}\right)

    \left(\frac{1}{9}\right)\left(y^2 - \frac{yx^2 \cdot 9}{6} + \frac{9x^4}{16}\right) = \left(\frac{y}{3} - \frac{x^2}{4}\right)^2
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  3. #3
    Newbie
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    Thanks very much for the help.
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