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anonymous

  • one year ago

HELP please :c What are the possible numbers of positive, negative, and complex zeros of f(x) = -x^6 + x^5- x4 + 4x^3 - 12x^2 + 12?.... O.o

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  1. anonymous
    • one year ago
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    So for finding the number of possible positive zeros, we simply count the amount of times the terms change signs as we look at each one. So the first term is negative and the 2nd is positive. That's one sign change. Then the next term is negative, another sign change The next term is positive, a 3rd sign change. In the end, we changed sign 5 times, so there are 5 possible positive zeros. Is that okay so far?

  2. anonymous
    • one year ago
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    kinda

  3. anonymous
    • one year ago
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    3 sign change plus 2 sign change?

  4. anonymous
    • one year ago
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    wait nvm

  5. anonymous
    • one year ago
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    Yeah, from left to right, every time a consecutive term has a different sign, I count it. However many sign changes I count is the amount of possible positive solutions. |dw:1433722328021:dw|

  6. anonymous
    • one year ago
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    That makes more sense, but in my answer choices it says Positive 5, 3 , or 1

  7. anonymous
    • one year ago
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    Yeah, and we counted 5.

  8. anonymous
    • one year ago
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    Oh, I see, my bad.

  9. anonymous
    • one year ago
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    Yeah, you subtract multiples of 2 as well. So you include 5 - 2 = 3, 3 - 2 = 1

  10. anonymous
    • one year ago
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    ohhh ok, that makes sense, would the answer be ~~Positive: 5, 3, or 1; negative: 1; complex: 4, 2, or 0~~

  11. anonymous
    • one year ago
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    Well, let's look at the negative. We have to replace every x with -x and then do the same thing, count the sign changes \[-(-x)^{6} +(-x)^{5} -(-x)^{4}+4(-x)^{3}-12(-x)^{2}+12 \] \(=-x^{6}-x^{5}-x^{4}-4x^{3}-12x^{2}+12\) So yes, negative would just be 1. As for complex zeros, it can be 0 or any multiple of 2. So 0, 2, 4, 6.

  12. anonymous
    • one year ago
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    yea since it consecutive with the -'s until 12

  13. anonymous
    • one year ago
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    I'm sorry, I meant for complex zeros. You do not have an option that is 6, 4, 2, 0, correct?

  14. anonymous
    • one year ago
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    no :o

  15. anonymous
    • one year ago
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    oh wait!

  16. anonymous
    • one year ago
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    yes i do

  17. anonymous
    • one year ago
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    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

  18. anonymous
    • one year ago
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    Ah, okay. So yeah, it would be 4, 2, 0 for complex. The reason being is from the work before, we're guaranteed to have 1 positive root and 1 negative root. There could be 3 or 5 positive, but we're guaranteed at least 1 positive, 1 negative. So that means there are only 4 roots left. So that means we could have 4, 2, or 0 complex. Complex numbers have no sense of positive or negative to them, so they are considered separate. There really is no such thing as a positive or negative complex number. That is why once we've guaranteed 1 positive and 1 negative, we cant say 6 anymore for complex, there can only be 4, 2, or 0.

  19. anonymous
    • one year ago
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    ohhh, Thank you for helping me! the complex part is kinddda mind puzzling but you made the rest make since( positive and negative)! Thank you cx

  20. anonymous
    • one year ago
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    Yeah, no matter what, complex numbers have to come in pairs, you can never have an odd amount. Have you see before how i^2 = -1 and those properties of complex numbers?

  21. anonymous
    • one year ago
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    I believe i have but that was last semester, so i kinda forgot alittle about them lol

  22. anonymous
    • one year ago
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    okay, as long as you know i^2 is -1, I can show this example. So, this is an example of why we don't consider complex numbers either positive or negative, and essentially why we wee forced to only have a max of 4 complex numbers in your problem.

  23. anonymous
    • one year ago
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    So, lets say this is true \( 2 < 2i\)

  24. anonymous
    • one year ago
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    This is just an assumption, im not saying it is or it isnt.

  25. anonymous
    • one year ago
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    So, if \(2<2i\), I should be able to multiply by some non-negative number and keep the inequality the same. So let's multiply by i \(2<2i\) \(2i < 2i*i\) \(2i < 2i^{2}\) \(2i<2(-1)\) \(2i<-2\) Well, if you think about that, it makes no sense. I started by saying 2i is bigger than 2, but now I multiplied by a value and im told 2i is less than -2. Thats not possible. So, its just a weird example of why we dont think of complex numbers as being positive or negative, theyre just different, lol. Okay, my random lecture is over, good luck!

  26. anonymous
    • one year ago
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    Lol it's ok, this can come in handy if i need to remeber what a complex number is cx and thank you! xD

  27. anonymous
    • one year ago
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    Lol, no problem :)

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