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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
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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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So 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?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.03 sign change plus 2 sign change?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yeah, 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0That makes more sense, but in my answer choices it says Positive 5, 3 , or 1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yeah, and we counted 5.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yeah, you subtract multiples of 2 as well. So you include 5  2 = 3, 3  2 = 1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ohhh ok, that makes sense, would the answer be ~~Positive: 5, 3, or 1; negative: 1; complex: 4, 2, or 0~~

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Well, 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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yea since it consecutive with the 's until 12

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'm sorry, I meant for complex zeros. You do not have an option that is 6, 4, 2, 0, correct?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Positive: 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ah, 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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ohhh, 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yeah, 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?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I believe i have but that was last semester, so i kinda forgot alittle about them lol

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay, 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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So, lets say this is true \( 2 < 2i\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0This is just an assumption, im not saying it is or it isnt.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So, if \(2<2i\), I should be able to multiply by some nonnegative 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!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Lol it's ok, this can come in handy if i need to remeber what a complex number is cx and thank you! xD
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