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

Positive Real: 3 or 1
Negative Real: 1
Complex: 2 or 0

Positive Real: 3 or 1
Negative Real: 2 or 0
Complex: 1

Positive Real: 1
Negative Real: 3 or 1
Complex: 2 or 0

Positive Real: 4, 2 or 0
Negative Real: 1
Complex: 0 or 1 or 3

@DebbieG :P I don't understand 100% how to solve these and get these types of answers

Answer 2

Oh, this is that theorem about the number of sign changes, right? Does that sound familiar?

Answer 3

The signs changing.... I know two theories about the signs changing

Answer 4

Descartes rule....

Answer 5

Well, yeah

Answer 6

I know of it but it's difficult, imo, to work with

Answer 7

First, look at f(x), and count the number of times that the SIGN of the COEFFICIENTS changes.

\(\large f(x) = \color{red}-7x^4 \color{red}- 12x^3 \color{blue}+ 9x^2 \color{red}- 17x \color{blue}+ 3?\)

Answer 8

There are 3 sign changes.... so this is the "positive case". So there are a max of 3 real, positive roots.

Answer 9

Oh, that's what that means? I thought that had something to do with solving the problem..

Answer 10

You don't have solve it, just have to find out how many real positive, real negative, and complex roots there could be.

So 4 max positive reals, but you "count down" by 2's: so there could be 4, 2, or 0 real positive roots.

Answer 11

be right back...

Answer 12

Is it 3 or 4? o.o I have an answer choice of both.. (I see what you're getting at, though. Thanks for helping me understand better)

Answer 13

Oh, sorry - there are 3 sign changes, so we count down by 2's: either 3 or 1 real, positive roots.

Answer 14

Now to find max number of real negative roots, we look at the same thing, but for f(-x):

\(\large f(-x) = \color{}-7(-x)^4 \color{}- 12(-x)^3 \color{}+ 9(-x)^2 \color{}- 17(-x) \color{}+ 3
\\\large=\color{red}-7x^4 \color{blue}+ 12x^3 \color{blue}+ 9x^2 \color{blue}+ 17x \color{blue}+ 3\)

So only 1 sign change, which means a maximum of 1 real, negative root

Answer 15

so 3 or 1 real positive roots, and 1 real negative root.

Keeping in mind that complex roots come in pairs (do you understand what I mean by that?), and that this is a 4th degree polynomial so has 4 total roots (including repeated roots). Soooooo putting it all together means that we have:

3 positive real and 1 negative real (and 0 complex)
OR
1 positive real, 1 negative real and 2 complex

those are the only possibilities.

Answer 16

Here is what's WRONG with all of the other options:

Positive Real: 3 or 1
Negative Real: 2 or 0
Complex: 1
Can't have 1 complex root - they come in pairs!

Positive Real: 1
Negative Real: 3 or 1
Complex: 2 or 0
Max of 1 negative real root, so 3 negative real is not possible

Positive Real: 4, 2 or 0
Negative Real: 1
Complex: 0 or 1 or 3
Max of 3 positive real roots, so can't have 4; and also can't have 1 or 3 complex - they come in pairs!

:)