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anonymous

  • 5 years ago

What are the possible number of positive, negative, and complex zeros of f(x) = 3x^4 – 5x^3 – x^2 – 8x + 4 ? Answer: A)Positive: 2 or 0; Negative: 2 or 0; Complex: 4 or 2 B)Positive: 4 or 2 or 0; Negative: 2 or 0; Complex: 4 or 2 or 0 C)Positive: 2 or 0; Negative: 2 or 0; Complex: 4 or 2 or 0 D)Positive: 3 or 1; Negative: 1; Complex: 2 or 0 help pleeeease:)

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  1. Hero
    • 5 years ago
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    I think it's C

  2. Mertsj
    • 5 years ago
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    The number of positive real zeros is equal to the number of sign changes or is less than that by an even number. There are two sign changes. So the positive roots would be 2 or 0 The number of negative real zeros is equal to the number of sign changes in f(-x) so find that first.

  3. Mertsj
    • 5 years ago
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    f(-x)=3x^4+5x^3-x^2+8x+4 So 2 sign changes implies 2 or 0 negative roots

  4. Hero
    • 5 years ago
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    So far, it's C :D

  5. Hero
    • 5 years ago
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    mertsj, how can you tell if it's A or C?

  6. Mertsj
    • 5 years ago
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    It's not A because that does not allow for the possibility that there are 0 complex roots. If there are 2 positive and 2 negative (possibly) then there would be 0 complex since it has at most 4 roots.

  7. anonymous
    • 5 years ago
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    so it's c:)

  8. Mertsj
    • 5 years ago
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    yes

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