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anonymous

  • 5 years ago

What are the possible number of positive real, negative real, and complex zeros of f(x) = 6x3 – 3x2 + 5x + 9? Answer: A)Positive Real: 1 Negative Real: 0 Complex: 2 B)Positive Real: 2 or 0 Negative Real: 2 or 0 Complex: 1 C)Positive Real: 2 or 0 Negative Real: 1 Complex: 2 or 0 D)Positive Real: 1 Negative Real: 2 or 0 Complex: 2 or 0 Help me please:)))

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  1. amistre64
    • 5 years ago
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    +6x3 -3x2 +5x +9 1 | 1 | 0 | sign changes twice, which means there is either 2 or 0 positives swap the signs of the odd powers: -6x3 -3x2 -5x +9 1 | 0 | signs changes once; there is at elast 1 negative solution

  2. amistre64
    • 5 years ago
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    id say its "c"

  3. anonymous
    • 5 years ago
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    haha yeah i kinda got tht from you explaining it, thank you very much:)))

  4. amistre64
    • 5 years ago
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    :) youre welcome

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