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zerlinav

  • 3 years ago

What are the possible number of positive, negative, and complex zeros of f(x) = 3x4 - 5x3 - x2 - 8x + 4

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  1. richyw
    • 3 years ago
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    Ok well first of all it's a quadratic function, so the total will be 4. Use Descartes’ Rule of Signs. Your function is \[p(x)=3x^4-5x^3-x^2-8x+4\] So count the number of sign changes. A "sign change" is when the sign the coefficient of the \(a^n\) is different from the sign of the coefficient of \(a^{n-1}\). So here you have one sign changes. You do the same thing for \(p(-x)\) and also find one sign change (don't include the sign change for the 4! only the ones that multiply by x matter. Since there must be 4 total you know that it is one positive, one negative and two complex (the complex ones have to be an even number!)

  2. richyw
    • 3 years ago
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    sorry not quadratic. haha. 4th degree polynomial

  3. richyw
    • 3 years ago
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    or "quartic (I just had to google that, lol)

  4. cruffo
    • 3 years ago
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    I thought there were two sign changes. \[ p(x)=3x^4-5x^3-x^2-8x+4 \rightarrow \color{red}{+}\color{red}{-}- \color{red}- \color{red}+\]

  5. cruffo
    • 3 years ago
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    # of possible positive zeros is the number of sigh changes of the polynomial, or less than that number by an even integer. So the number of possible positive zeros is 2 or 0

  6. richyw
    • 3 years ago
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    there is only one sign change though. That last one doesn't matter.

  7. cruffo
    • 3 years ago
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    ??? http://www.purplemath.com/modules/drofsign.htm

  8. richyw
    • 3 years ago
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    weird, I might be wrong. But I don't think so.

  9. cruffo
    • 3 years ago
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    Take a look at the graph: http://www.wolframalpha.com/input/?i=p%28x%29%3D3x^4-5x^3-x^2-8x%2B4+from+-3+to+3

  10. richyw
    • 3 years ago
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    oh yeah you are totally correct!

  11. richyw
    • 3 years ago
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    so in this case using the rule of signs all that you can know is one of four possible combinations then right? Also I have been doing this wrong for like 5 years haha. good thing we have calculators...

  12. cruffo
    • 3 years ago
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    : ) Sounds right: degree 4 means you will find 4 solutions to p(x) = 0 For this polynomial, just looking at the signs, we can tell that there will be 2 or 0 positive roots, 2 or 0 negative roots. For the complex roots - they always come in pairs. so... case 1: 2 + real zeros, 0 - real zeros, and 2 complex zeros case 2: 0 + real zeros, 2 - real zeros, and 2 complex zeros case 3: 2 + real zeros, 2 - real zeros, and 0 complex zeros case 4: 0 + real zeros, 0 - real zeros, and 4 complex zeros

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