Question

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

Answer 1

descartes has a thrm about this ....

Answer 2

count the number of times the signs change thru the equation

Answer 3

6 – x5– x4 + 4x3 – 12x2 + 12
^ ^ ^ ^

i see 4 sign changes descartes says there must be at most 4 positive zeros; and we weed out complex zeros by 2s giving us

4, or 2, or 0 positive roots

Answer 4

negating all the odd powers gives us a count for negative roots ....

Answer 5

ok

Answer 6

i have the answer thank you

Answer 7

changed changed
v v
x6 + x5– x4 - 4x3 – 12x2 + 12
^ ^ count is 2

there are either 2, or 0 negative roots

Answer 8

ok :) cause i get iffy with the complexes, i cant remember if we add up the "subtractions"

Answer 9

FOR FUTURE FLVS STUDENTS:

Answer 10

@mabitrix , didn't you read the problem? Your question had a negative symbol in front.

Answer 11

^