Question

What are the possible number of positive, negative, and complex zeros of f(x) = –x6 + x5– x4 + 4x3 – 12x2 + 12 ?

Answer 1

use descarte's rule of signs

f(x) = (-) (+)( -) (+) (-) (+)
5 change in signs so 5 possible positive roots

f(-x) = (-) (-) (-) (-) (-) (+)
1 change in sign so 1 possible negative root

5 positive roots + 1 negative root = 6 roots

since 6 is the degree of this equation there are only 6 possible roots and since they're already occupied there are no complex roots

does that help?

Answer 2

All the numbers at x are sqaured

Answer 3

what do you mean?

Answer 4

–x^6 + x^5– x^4 + 4x^3 – 12x^2 + 12 ?

Answer 5

yeah...exactly...

Answer 6

...you have no idea what the descartes' rule of signs is...?

Answer 7

I'm trying to understand it.

Answer 8

as the name suggests...you're going to look at the signs...then you search for change in signs

Answer 9

for every change in sign in f(x) counts for 1 possible positive root

every change in sign in f(-x) counts for 1 possible negative root

Answer 10

Okay

Answer 11

just to be clear...

f(-x) = -(-x)^6 + (-x)^5 - (-x)^4 + 4(-x)^3 - 12(-x)^2+12

f(-x) = -(x^6) + (-x^5) -(x^4) + 4(-x^3) -12(x^2) + 12

f(-x) = -x^6 - x^5 - x^4 - 4x^3 -12x^2 + 12

Answer 12

i'll let you understand it now :) tag me if you have questions

Answer 13

@lgbasallote bravo