Question

HEEEEEEEEEEEEEEEEEEEEELP how you would analyze the zeros of the polynomial function f(x) = 7x5 + 15x4 – x3 + 4x2 – 6x – 11 using Descartes’ Rule of Signs.

Answer 1

all i remember is you look at signs
+-+--

Answer 2

The function has either 3 or 1 positive real zeros, 1 negative real zero, and 1 complex zero. is this right? tthats what i got

Answer 3

i am not sure about the complex

Answer 4

descartes rule of sign tells you nothing about the actual zeros. it only tells you the number of possible zeros

Answer 5

look at the number of "changes of sign"
there are 3

Answer 6

Yeah so what are they ? was my answer right?

Answer 7

I just dont know how many complex, i got 1.

Answer 8

this means there are at most 3 POSITIVE zeros

Answer 9

there cannot be only one complex zero. they come in conjugate pairs

Answer 10

lets go slow

Answer 11

okay!

Answer 12

do you see the 3 "changes of sign"?

Answer 13

yes so that means there is 3 or 1 possible positive zeros right ?

Answer 14

\[f(x) = 7x^5 + 15x^4 – x^3 + 4x^2 – 6x – 11\]
\[f(x) = 7x^5 + 15x^4 \text {change 1} – x^3 \text{ change 2}+ 4x^2 \text{ change 3} – 6x – 11\]\]

Answer 15

this means there are AT MOST 3 positive zeros. now we count the possible negative ones. to do this replace x by -x

Answer 16

\[f(-x)=-7x^5+14x^4+x^3+4x^2+6x-11\]

Answer 17

hope it is clear what i did to replace x by -x and i hope i didn't make any typos

Answer 18

now there are only 2 changes of sign, from -7x^5 to 14x^4 and from +6x to -11 so there are
AT MOST 2 negative real zeros. that is all descartes rule of sign tells you

Answer 19

at most 3 positive real zeros
at most 2 negative ones.
it doesn't say there are 3 positive and two negative, but the complex roots must come in conjugate pairs, so there cannot be only one of them

Answer 20

as you can see from the picture, there are actually 2 negative ones and one positive one
http://www.wolframalpha.com/input/?i=f%28x%29+%3D+7x5+%2B+15x4+%E2%80%93+x3+%2B+4x2+%E2%80%93+6x+%E2%80%93+11+

Answer 21

Are there zero complex zeros?

Answer 22

no . there are 2 complex ones in this case

Answer 23

But aren't there only 5 possible zeros?

Answer 24

because from the picture you can see that there are 3 real zeros, so there must be two complex ones.
again descartes rule tells you this has
AT MOST 3 positive and
AT MOST 2 negative

Answer 25

Oh, okay so 4,2, or 0 complex zeros?? and thank you so much for all your help. you are the bestttttttttt!!!!!!!!!!!!!!!!!!! but would that be right? 4,2,0?

Answer 26

it does not say it MUST HAVE 3 positive ones, it has AT MOST 3 positive ones. can't have 4 or 5 but it could have 0, 1, 2, or 3

Answer 27

so the answer here to the question
"analyze the zeros of the polynomial function f(x) = 7x5 + 15x4 – x3 + 4x2 – 6x – 11 using Descartes’ Rule of Signs. "
is
descartes rule tells you this has
AT MOST 3 positive zeros and
AT MOST 2 negative ones

Answer 28

it may have 0, 2 or 4 complex ones yes. so in particular you know it must have
AT LEAST one real one

Answer 29

But sorry one more question, if i had 1 positive and 0 negative couldn't I have 4 complex? Oh okay, yeah.

Answer 30

no you can't have one positive, one negative and 4 complex. that is one too many

Answer 31

oh i am sorry yes, you could have one negative and 4 complex yes

Answer 32

1 positive and 4 complex

Answer 33

2 negative, one positive and two complex, which is what you actually have

Answer 34

or even one negative, two positive and two complex

Answer 35

Okay, thank you so much for all of your help. I really appreciate it :D