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20 February, 07:20

What are the possible numbers of positive, negative, and complex zeros of f (x) = x6 - x5 - x4 + 4x3 - 12x2 + 12?

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  1. 20 February, 07:27
    0
    Using Descartes' rule of signs.

    There are 4 changes of signs of coefficients:

    + to - (+x^6-x^5)

    - to + (-x^4+4x^3)

    + to - (+4x^3-12x^2)

    - to + (-12x^2+12)

    Therefore, there are 4,2 or 0 positive roots at most. So it must be the third answer.
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