Question

What are the possible number of positive real, negative real, and complex zeros of f(x)=-7x^4-12x^3+9x^2-17x+3?

Answer 1

Why did you repost the question?

Answer 2

Do you know Descartes' rule of signs?

Answer 3

You've got the polynomial written in order of descending exponent, as you need to apply the rule of signs. How many sign changes are there?

Answer 4

Right. So that means the number of positive roots (zeros) is either 3, or 3 - a multiple of 2. In this case, the only multiple of 2 that could be subtracted is 2, so we have either 3 or 1 positive roots. Agreed?

Answer 5

Do you remember how we find out how many negative roots there can be?

Answer 6

Okay, we do the same thing, except first we multiply each term in the polynomial that has an odd exponent by -1. So x, x^3, x^5, x^7, etc. all change their signs. After doing that, we apply the rule again, counting the sign changes. What do you get when you do that?

Answer 7

2 sign changes

Answer 8

One of us made a mistake :-)

f(x)=-7x^4-12x^3+9x^2-17x+3
after multiplying odd exponent terms by -1, I get
-7x^4+12x^3+9x^2+17x+3

how many sign changes do you see there?

Answer 9

1?

Answer 10

I see 1. alternative explanation is that I messed up changing signs, but unless you see where I did so, let's go with 1.

So that means that there is 1 negative root. In general, it will be however many sign changes, possibly minus a multiple of 2, just like with the positive roots.

So, we know so far that we have 4 roots in total:

We may have 3 positive roots and 1 negative root,

or we may have 1 positive root and 1 negative root.

If the latter, how many complex roots would we need to have?

Answer 11

2 or 0. THANK YOU!! :D

Answer 12

So our final choices are either
3 pos / 1 neg / 0 complex

or

1 pos / 1 neg / 2 complex

as it turns out, this one is the latter, but I don't believe you can determine that without solving it.

Answer 13

I know I can't determine it :-)

Answer 14

It's multiple choice, that why I knew :)