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

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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!)
sorry not quadratic. haha. 4th degree polynomial
or "quartic (I just had to google that, lol)

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I thought there were two sign changes. \[ p(x)=3x^4-5x^3-x^2-8x+4 \rightarrow \color{red}{+}\color{red}{-}- \color{red}- \color{red}+\]
# 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
there is only one sign change though. That last one doesn't matter.
weird, I might be wrong. But I don't think so.
Take a look at the graph:^4-5x^3-x^2-8x%2B4+from+-3+to+3
oh yeah you are totally correct!
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...
: ) 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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