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anonymous

  • 5 years ago

What are the possible number of positive real, negative real, and complex zeros of f(x) = –7x4 – 12x3 + 9x2 – 17x + 3? Answer: A)Positive Real: 1 Negative Real: 3 or 1 Complex: 2 or 0 B)Positive Real: 3 or 1 Negative Real: 2 or 0 Complex: 1 C)Positive Real: 3 or 1 Negative Real: 1 Complex: 2 or 0 D)Positive Real: 4, 2 or 0 Negative Real: 1 Complex: 0 or 1 or 3 This the last question like this, help me please:)

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  1. anonymous
    • 5 years ago
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    descartes rule of sign for this one http://www.purplemath.com/modules/drofsign.htm

  2. anonymous
    • 5 years ago
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    three changes of sign in your example, \[–7x^4 – 12x^3 + 9x^2 – 17x + 3\] \[–7x^4 – 12x^3_{\text{ here }} + 9x^2_{\text{ here }} – 17x_{\text{ here }} + 3\]

  3. anonymous
    • 5 years ago
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    so possible positive roots: 3 or 1

  4. anonymous
    • 5 years ago
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    \[f(-x) = –7(-x)^4 – 12(-x)^3 + 9(-x)^2 – 17(-x) + 3\] \[f(-x) = –7x^4 + 12x^3 + 9x^2 + 17x + 3\] this has one change in sign

  5. anonymous
    • 5 years ago
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    so one possible negative root. degree is 4, so there are 4 roots in total Positive Real: 3 or 1 Negative Real: 1 Complex: 2 or 0

  6. anonymous
    • 5 years ago
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    Okay thank you very much!:))

  7. anonymous
    • 5 years ago
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    yw btw the link i sent has a brief clear explanation with examples, if you have to take a test on this it is easy enough with a little practice

  8. anonymous
    • 5 years ago
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    okay thank you, i am understanding bits and pieces of this lesson:)

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