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DeadShot

  • one year ago

What are the possible number of positive, negative, and complex zeros of f(x) = x^6 + x^5+ x^4 + 4x^3 - 12x^2 + 12 ?

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  1. e.cociuba
    • one year ago
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    Descartes' Rule of Signs tells you that there could be four positive roots, and two negative roots, and any pair of those positive and negative roots could be replaced by a pair of complex conjugates. In actuality, that function has two positive, two negative and two complex roots, but Descartes isn't as precise as that.

  2. DeadShot
    • one year ago
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    Ok, thanks!

  3. e.cociuba
    • one year ago
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    np:)

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