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Thread: factoring higher quadratics

  1. #1
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    factoring higher quadratics

    Solve by factoring
    $\displaystyle x^4-12x^2-64=0$
    How? -.-
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  2. #2
    Senior Member Stroodle's Avatar
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    Quote Originally Posted by m0nkeysensei View Post
    Solve by factoring
    $\displaystyle x^4-12x^2-64=0$
    How? -.-
    Let $\displaystyle x^2=y$

    Then you have:

    $\displaystyle y^2-12y-64=0$

    $\displaystyle (y+4)(y-16)=0$

    So,

    $\displaystyle (x^2+4)(x^2-16)=0$

    $\displaystyle (x^2+4)(x+4)(x-4)=0$

    $\displaystyle \therefore x=\pm 4$ and $\displaystyle x=\pm 2i$
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