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anonymous

  • one year ago

What are the possible numbers of positive, negative, and complex zeros of f(x) = −x^6 + x^5 − x^4 + 4x^3 − 12x^2 + 12? Positive: 4, 2, or 0; negative: 2 or 0; complex: 6, 4, 2, or 0 Positive: 3 or 1; negative: 3 or 1; complex: 4, 2, or 0 Positive: 2 or 0; negative: 2 or 0; complex: 6, 4, or 2 Positive: 5, 3, or 1; negative: 1; complex: 4, 2, or 0

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  1. anonymous
    • one year ago
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    I know that you need to use the Descartes Rule or something like that, but I dont know how.

  2. welshfella
    • one year ago
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    yeah you look for changes of signs

  3. anonymous
    • one year ago
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    Can you help me find the answer?

  4. welshfella
    • one year ago
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    here's a good website http://www.purplemath.com/modules/drofsign.htm

  5. anonymous
    • one year ago
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    So 5 sign changes, now what can I do with that information?

  6. mathstudent55
    • one year ago
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    How many changes of signs do you see? |dw:1443383107699:dw|

  7. anonymous
    • one year ago
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    Already posted it ;D

  8. mathstudent55
    • one year ago
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    There maybe as many as 5 real roots. Since the polynomial has real coefficients, if it has complex roots, the complex roots must come in pairs of complex conjugate roots.

  9. anonymous
    • one year ago
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    Wait never mind, Im guessing D.

  10. mathstudent55
    • one year ago
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    That means if it does have complex roots, it may have 2 (1 pair of complex conjugate roots) or 4 complex roots (2 pairs of complex conjugate roots).

  11. mathstudent55
    • one year ago
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    That was the "positive:" case. Now replace x with -x. (x) = −x^6 - x^5 − x^4 - 4x^3 − 12x^2 + 12 Now there is only 1 sign change. There is only 1 negative root.

  12. mathstudent55
    • one year ago
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    You are correct. D is the answer.

  13. anonymous
    • one year ago
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    Thank you!

  14. mathstudent55
    • one year ago
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    yw

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