Question

Use Descartes' rule of signs to determine the possible numbers of positive real zeros and negative real zeros for
P(x) = 2x3 - 4x2 + 2x + 7

Positive roots: ?

Negative roots: ?

Answer 1

Help me! please!

Answer 2

how many sign changes do you count?

Answer 3

2? I don't know what Descartes's rule, I would like to lean :)

Answer 4

learn*

Answer 5

ohh it call Zeros of Polynomials

Answer 6

2 is correct; which means there are "at most" 2 positive roots

Answer 7

P(-x) = -2x3 - 4x2 -2x + 7

with a negtive "x" how many sign changes are there?

Answer 8

1?

Answer 9

ummmm 3?

Answer 10

yep, which tells us there is at most "1" negative root

Answer 11

not 3 lol

Answer 12

ohhh -_-

Answer 13

just 1 sign change

Answer 14

P(-x) = -2x3 - 4x2 -2x + 7
------------->-->
1 0

Answer 15

AWWW !

Answer 16

at most so if at most is 1 positive that means 0 positive is also possible?

Answer 17

P(x) = 2x3 - 4x2 + 2x + 7
--->----->----->
1 1 0

2 positive roots possible

Answer 18

0 positive is possible yes; you count off in 2s since there cane be comlex multiples

Answer 19

oh okay

Answer 20

AWWW i get thank u!!! : D

Answer 21

;)

Answer 22

thanks amistre

Answer 23

there are two sign changes so there are 2 possible solutions of the given equation.Also checking i case of f(-x) there is only one sign change so it has only one negative root. http://www.purplemath.com/modules/drofsign.htm