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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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

Mathematics
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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?
kinda
3 sign change plus 2 sign change?

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wait nvm
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|
That makes more sense, but in my answer choices it says Positive 5, 3 , or 1
Yeah, and we counted 5.
Oh, I see, my bad.
Yeah, you subtract multiples of 2 as well. So you include 5 - 2 = 3, 3 - 2 = 1
ohhh ok, that makes sense, would the answer be ~~Positive: 5, 3, or 1; negative: 1; complex: 4, 2, or 0~~
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.
yea since it consecutive with the -'s until 12
I'm sorry, I meant for complex zeros. You do not have an option that is 6, 4, 2, 0, correct?
no :o
oh wait!
yes i do
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
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.
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
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?
I believe i have but that was last semester, so i kinda forgot alittle about them lol
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.
So, lets say this is true \( 2 < 2i\)
This is just an assumption, im not saying it is or it isnt.
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!
Lol it's ok, this can come in handy if i need to remeber what a complex number is cx and thank you! xD
Lol, no problem :)

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