Question

Part 1 [3 points]: What are the possible number of positive, negative, and complex zeros of f(x) = 2x3 – 5x2 + 6x – 4 ?
Part 2 [4 points]: Use complete sentences to explain the method used to solve this equation.

Answer 1

@jim_thompson5910

Answer 2

For part 1, you would use descartes rule of signs

http://www.purplemath.com/modules/drofsign.htm

Answer 3

ty jim

Answer 4

let me know if that helps or not

Answer 5

it does

Answer 6

ok great, glad it does

Answer 7

but i was hoping for an answer

Answer 8

cause i been working all day
and this is my last problem so i can graduate

Answer 9

well how many sign changes are there in f(x)

Answer 10

3

Answer 11

so there are at most 3 positive real roots

Answer 12

now find f(-x)

Answer 13

3 0r 1 right

Answer 14

there
is 0

Answer 15

what is f(-x)

Answer 16

negative

Answer 17

if f(x) = 2x^3 – 5x^2 + 6x – 4

then what is f(-x)

Answer 18

there is 0 possible negative roots

Answer 19

how do you know that

Answer 20

by using decartes rules of signs

Answer 21

ok what did you get for f(-x)

Answer 22

im done 4get it just did it myself and stopped being lazy

Answer 23

1.f(x) = 2x^3 - 5x^2 + 6x - 4
There are 3 or 1 possible positive zeros.

f(-x) = -2x^3 - 5x^2 - 6x - 4
There are no possible negative zeros.

There are 2 or 0 complex zeros.

2.There are 3 changes of sign in f(x).
There are no changes of sign in f(-x).

The total amount of zeros is 3-3 positive zeros -0 negative=

0 complex, and 3-1-0=2,So there are 2 or 0 complex zeros

Answer 24

just did it myself ty for your help though

Answer 25

i knew how to do it just was tired but i did it anyway ty

Answer 26

is it right though

Answer 27

yes there are at most 3 real positive roots
0 negative real roots

so, here's all the possibilities

a) there are 3 real positive roots

or

b) there is one positive real root and 2 complex roots

Answer 28

huh

Answer 29

whats not making sense