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anonymous

  • 5 years ago

Factor f(x) = x^4 + 5x^2 - 14 completely.

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  1. anonymous
    • 5 years ago
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    let y = x^2 ; it won't factor any further than two quadratics in y.

  2. anonymous
    • 5 years ago
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    The answer is supposed to look something like (this probably isnt the answer but for example) : \[(x - \sqrt{7})(x - \sqrt{7})(x + \sqrt{2}i)(x- \sqrt{2}i)\]

  3. anonymous
    • 5 years ago
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    Im not sure how to solve for it this way :(

  4. anonymous
    • 5 years ago
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    If you go that far, then so be-it. HOWEVER, if you want it in that form (I normally don't) who am I to stop you. First, factor f(x) into two quadratics: Let y = x^2 => f(x) = y^2 + 5y - 14 = (y+7)(y-2) => f(x) = (x^2+7)(x^2-2) Then you can factorise these (well, if you call using imaginary numbers factorising :@)

  5. anonymous
    • 5 years ago
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    I got \[(x + \sqrt{7}i)(x-\sqrt{7}i)(x+\sqrt{2})(x-\sqrt{2})\] does that seem right you?

  6. anonymous
    • 5 years ago
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    I got \[(x + \sqrt{7}i)(x-\sqrt{7}i)(x+\sqrt{2})(x-\sqrt{2})\] does that seem right you?

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