## State possible number of imaginary zeros... help!

Ok, a multiple choice question wants me to:

"State the possible number of imaginary zeros of g(x)=x^4+3x^3+7x^2-6x-13."

(A) 3 or 1
(B) 2, 4, or 0
(C) Exactly 1
(D) Exactly 3

Using Descartes Rule of Sings I get:

Exactly 1 positive zero, 3 or 1 negative zeros, and 0 or 2 Imaginary zeros. They only want imaginary but I gave you the pos. and neg. because I used them to figure out the imaginary ones with a chart, like below.

|P_|_N_|_I_|
| 1 | 3 | 0 | 1+3+0=4
| 1 | 1 | 2 | 1+1+2=4

I've done similar questions given in my book and I've gotten them all right but for this question there is no option for what I found as you can see.

Am I right and the test question is just wrong? Did I do something wrong? I have gone over the book's example a thousand times and I don't see any other way.

Thanks,
Alan
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 I've graphed this poly with a program and from what I can see there would be 2 imaginaries, 1 positive, and 1 negative. I really think their answers are all wrong.
 The complex roots of a polynomial with all real coefficients must come in pairs, a complex number adn its conjugate. That should help you.

## State possible number of imaginary zeros... help!

Well... I said 2 imaginary which would make sense since they come in pairs. Sooo... am I right?

 Quote by alancj Well... I said 2 imaginary which would make sense since they come in pairs. Sooo... am I right?
I don't know.. maybe. The questions asked for the number of "possible" zeros not the actual number so knowing what I said in my previous post which is the correct answer?
 None of them! The closest thing (B: 2, 4, or 0) would be wrong since 4 imaginaries aren't possible. If I had 4 imaginaries then I would also still have my one positive and negative zero. That would add up to 6 zero’s which is more than the 4 zero maximum.

 Quote by alancj None of them! The closest thing (B: 2, 4, or 0) would be wrong since 4 imaginaries aren't possible. If I had 4 imaginaries then I would also still have my one positive and negative zero. That would add up to 6 zero’s which is more than the 4 zero maximum.
Without doing anything, any math or test at all. There is a possibility that this polynomial has 4 complex zeros. And what does OR mean?
 It means that there are possibly 2, 4, or 0 imaginary zeros.

 Quote by alancj It means that there are possibly 2, 4, or 0 imaginary zeros.
So then is that true or false?
 They ask “State the possible number of imaginary zeros of …” and the “possible number” is 0 or 2 imaginary zeros. 4 can’t be in the list because it is not a possibility. “OR” doesn’t excuse 4 for being there, the list of possibilities needs to be completely the same or it is completely wrong. Is “2, 4, or 0” the same as “2 or 0”? No it’s not. If the question asked "which choice includes the possible number or imaginary zeros?" then I could select (B).

 Quote by alancj They ask “State the possible number of imaginary zeros of …” and the “possible number” is 0 or 2 imaginary zeros. 4 can’t be in the list because it is not a possibility. “OR” doesn’t excuse 4 for being there, the list of possibilities needs to be completely the same or it is completely wrong. Is “2, 4, or 0” the same as “2 or 0”? No it’s not. If the question asked "which choice includes the possible number or imaginary zeros?" then I could select (B).
Explain to me why 4 is not a possibility? This is a polynomial of degree 4 which means that in the domain of complex numbers there must be 4 roots including multiplicities. Since all the coefficients are real numbers any complex roots must come in conjugate pairs meaning that if there are complex roots there must be an even number of them. Four complex roots is certainly a possibility, that doesn't mean that there are 4 complex roots it just means that there may be 4 complex roots. I have a strong feeling that this is all the problem wanted you to do, not tests for how many positive or negative roots there could be but just look at the degree of the polynomial and use what you know to be true about a polynomial with real coefficients.
 Hi D leet: Because x^4+3x^3+7x^2-6x-13 has one sign change (++ - -; + to -), Descartes guarantees existence of exactly one positive real root. Further, because f(-x) changes sign three times, the function has 3 or 1 negative real roots. Hence, no real roots (a necessary condition of the "four imaginary root" scenario) is not possible. It therefore follows that possible imaginary roots are numbered at zero or two, concurring with Alan's previous statement. Regards, Rich B.

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 Quote by d_leet I have a strong feeling that this is all the problem wanted you to do, not tests for how many positive or negative roots there could be but just look at the degree of the polynomial and use what you know to be true about a polynomial with real coefficients.
I think this is exactly what the problem intended. More advanced techniques can narrow down the answer even more, but given the choices they aren't looking for a tighter range on the number of imaginary zeros. It may have been better worded as "the number of imaginary zeros is in which set: A. {1,3}, B.{0,2,4}, C.{1}, D.{3}"
 Recognitions: Gold Member Science Advisor Staff Emeritus If I know a number has to be 0 or 2, I also know that number has to be 0, 2, or 4.

 Quote by d_leet Explain to me why 4 is not a possibility? This is a polynomial of degree 4 which means that in the domain of complex numbers there must be 4 roots including multiplicities. Since all the coefficients are real numbers any complex roots must come in conjugate pairs meaning that if there are complex roots there must be an even number of them. Four complex roots is certainly a possibility, that doesn't mean that there are 4 complex roots it just means that there may be 4 complex roots. I have a strong feeling that this is all the problem wanted you to do, not tests for how many positive or negative roots there could be but just look at the degree of the polynomial and use what you know to be true about a polynomial with real coefficients.
If that's what they intended then the textbook certainly didn't bother to make it more obvious. First of all I have no idea what the “domain of complex numbers” is, what “multiplicities” is, and I wouldn’t know what “complex roots” are if it wasn’t for my searching on the internet. I would never be able to find that out from my text book. And if I did know those things I would not have logically concluded that 4 would be a possibility, because the ONLY method given to find out possible imaginaries is to find out the positive and negatives first. I simply found the example in the book that sounded like my question and did what it said. Why would anyone think outside of that box? No one would unless they read math books for fun.

So, example 2 in my textbook is “find numbers of positive and negative zeros.” The question in the example is “State the possible number of positive real zeros, negative real zeros, and imaginary zeros of p(x)=…” When I, or anybody else, would read that they would think “yippy, here is how I do this problem.” So I found the pos. and neg. zeros and made the chart and found the imaginaries… 4 was not one of them. The example question was virtually the same as the test question only I just needed to give them the imaginary part of the answer for the test and not the rest of the possibilities. I would have no reason to believe that the 4 is a possibility because no one going through the algebra 2 textbook for the first time would have concluded or thought along the lines that you did. It isn’t till the next page that they tell me what the “Complex Conjugates Theorem” is…. which sounded like it didn’t have a thing to do with my question. I still don’t know what the hell a conjugate is, even thought I looked it up in the index.

This is a high school correspondence course and it wouldn't be the first time these people have totally screwed a test question up. The last time was with a science class and I had to write them a letter proving that half dozen questions I did were right and that they needed to change their test questions.

I have wasted so much time and energy on this stupid question it's not even funny. I'll still probably get a B on this test .

-Alan

 Quote by Hurkyl If I know a number has to be 0 or 2, I also know that number has to be 0, 2, or 4.
Not if all you know is what the book tells you. Then you would be an idiot to think that 1+1+4=4. I would like to give the person who wrote this damn question a knuckle sandwich.

 Quote by Rich B. Hi D leet: Because x^4+3x^3+7x^2-6x-13 has one sign change (++ - -; + to -), Descartes guarantees existence of exactly one positive real root. Further, because f(-x) changes sign three times, the function has 3 or 1 negative real roots. Hence, no real roots (a necessary condition of the "four imaginary root" scenario) is not possible. It therefore follows that possible imaginary roots are numbered at zero or two, concurring with Alan's previous statement. Regards, Rich B.
At least somebody understands me...