Question

HELP! WILL GIVE MEDAL!
--What are the possible numbers of positive, negative, and complex zeros of f(x) = -3x4 - 5x3 - x2 - 8x + 4?
(a) Positive: 2 or 0; negative: 2 or 0; complex: 4 or 2 or 0
(b) Positive: 1; negative: 3 or 1; complex: 2 or 0
(c) Positive: 3 or 1; negative: 1; complex: 2 or 0
(d) Positive: 4 or 2 or 0; negative: 2 or 0; complex: 4 or 2 or 0

Answer 1

@amistre64

Answer 2

@ganeshie8

Answer 3

@uri @jdoe0001

Answer 4

decartes has a rule of sign, can you define it?

Answer 5

Is it the one where if its a negative or something?

Answer 6

The change in signs right?

Answer 7

yes, change in sign for positive x, for negative x, and then if we have doubles we can pull out they are complexes

Answer 8

Theres one sign change. I don't know how to determine whether its positive, negative etc.

Answer 9

-3x4 - 5x3 - x2 - 8x + 4

let x = 1, its a postive number, and the signs (operations) dont change

-3 - 5 - 1 - 8 + 4
|------------|--
0 1

there is only 1 sign change, meaning that we have at best 1 postivie root and 0 complex roots



let x = -1, its a negative and the odd powers are going to change sign: (-1)^2 = 1 and (-1)^3 = -1 so we are just swapping odd degree signs since we are multiplying by -1

-3x4 - 5x3 - x2 - 8x + 4
-3x4 + 5x3 - x2 + 8x + 4
|-----|-----|---|-------
0 1 2 3

so 3 negative roots and 0 complex roots

or 1 negative root and 2 complex roots

Answer 10

since only 1 option has postive: 1 and thats all we get ... then its prolly the one we want

Answer 11

Positive: 1; negative: 3 or 1; complex: 2 or 0
yep yep yep

Answer 12

Got it! Thanks :)

Answer 13

youre welcome