Question

What are the possible numbers of positive real, negative real, and complex zeros of
f(x) = -7x4 - 12x3 + 9x2 - 17x + 3?

Answer 1

use the descartes rule of signs

Answer 2

So 3 negatives?

Answer 3

These are my possible choices:
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

Answer 4

There are 3 changes in the f(x) case.
so 3 is max number of positives
so that does mean it can be 3 or count down by 2
so 3 or 1 for the positives

Answer 5

now find f(-x)

Answer 6

and count those changes

Answer 7

I got 2 changes. So B?

Answer 8

what did you get for f(-x)?

Answer 9

-7x^4-9x^2+17x+3

Answer 10

i think you are missing a term

Answer 11

i forgot 12x^3

Answer 12

-7x^4+12x^3+9x^2+17x+3
is this waht you meant?

Answer 13

Yes

Answer 14

how many changes do you see?

Answer 15

in the signs

Answer 16

1. But how is -9(-x)^2 positive? Wouldn't it be -9X^2?

Answer 17

your function you gave was f(x)=-7x^4-12x^3+9x^2-17x+3

Answer 18

f(-x)=-7(-x)^4-12(-x)^3+9(-x)^2-17(-x)+3
=-7x^4+12x^3+9x^2+17x+3

Answer 19

Whoops... I copied it on my paper wrong. My bad!

Answer 20

My apologies...

Answer 21

and 1 change is right for the function you have here

Answer 22

so 1 negative

Answer 23

So we have 3 positive and 1 negative root?

Answer 24

well 3 or 1 pos and 1 neg

Answer 25

Right. Thank you so much for your help

Answer 26

np

Answer 27

1 negative, 1 positive, 2 complex or
1 negative, 3 positive

Answer 28

1 neg, 1 pos and 2 complex.
See the attachment.