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anonymous

  • 5 years ago

Hi again, could anyone help with x3-3x2-4x-12=0 the 3 and 2 after the x's rep. cube and square. Thanks

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  1. bahrom7893
    • 5 years ago
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    take a look at this: http://home.scarlet.be/~ping1339/cubic.htm I will try to get to this question later...

  2. anonymous
    • 5 years ago
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    This one's actually easier then it might look at first glance. The key is to realize you can group the first two terms and the last two terms in such a way that you'll get a common polynomial factor. \[ \begin{align} x^3-3x^2-4x-12&=0\\ (x^3-3x^2) - (4x-12) &=0\\ x^2(x-3) - 4(x-3) &=0\\ (x^2-4)(x-3)&=0\\ (x+2)(x-2)(x-3)&=0 \end{align} \]

  3. anonymous
    • 5 years ago
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    thank you, ALOT

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