Question

For the following function:
f(x)=2x^5-6x^3+5x^2-3x-1

i. Find the maximum number of real zeros, and
ii. Use Descartes's rule of signs to determine how many positive and how many negative zeros the function has. You do not need to find the zeros.
Please show all of your work. For the following function:
f(x)=2x^5-6x^3+5x^2-3x-1

i. Find the maximum number of real zeros, and
ii. Use Descartes's rule of signs to determine how many positive and how many negative zeros the function has. You do not need to find the zeros.
Please show all of your work. @Mathematics

Answer 1

i. Look at the degree. This will tell you how many real zeros there can be.
ii. Descartes' rule, you're going to look at the given function for the positive real zeros. Depending on the number of variations of signs will tell you the possible numbers.
There's a change in variation between 2nd and 3rd, 3rd and 4th. This means there are 2 to 1 positive real zeros.

Now look at the negative function:
f(-x) = 2(-x)^5-6(-x)^3+5(-x)^2-3(-x)-1
= -2x^5 + 6x^3 + 5x^2 + 3x - 1
How many variations are there in the signs with the f(-x). This will tell you how many negative real zeros there are.

Answer 2

hhmm

Answer 3

what are you thinking?

Answer 4

I dunno how to do this.

Answer 5

Where do the signs change in the f(-x)?

Answer 6

This was part of the other example I had assisted with.

= -2x^5 + 6x^3 + 5x^2 + 3x - 1
Where does it go from negative to positive or vice versa.

Answer 7

You can do this. :)

Answer 8

in front of the x?

Answer 9

i think it goes from negative to positive in front of the x?

Answer 10

Eh, you're getting warmer.

= -2x^5 + 6x^3 + 5x^2 + 3x - 1 change in variation occurs
^ ^

Answer 11

it starts off as a negative, then adds, then goes back to negative.

Answer 12

This is how you count the change in variations. And it means the amount of real negative zeros since it is the f(-x) = -2x^5 + 6x^3 + 5x^2 + 3x - 1

Answer 13

So as you can see there are 2 or 1 real negative zeros.

Answer 14

when you say 2 or 1 real negative zeros...are you counting the numbers in front of the x?

Answer 15

when you solve for a real zero, you would use synthetic division. Then you would use the numbers in the equation. But when I'm saying 2 or 1, that means that out of the possible "real zeros" there could be 2 or 1 of them that may come out as a zero when performing synthetic division.

Answer 16

ok so there are 2 which would be ±1, ± 1/2. Is this correct?

Answer 17

No. Possible zeros would be coming from the 2x^5 and the 3x.

(-2x^5) + 6x^3 + 5x^2 + (3x )- 1

Answer 18

ok so does it have 3 real zeros and 1 positive zero(root)
2 negative zeros(roots)??

Answer 19

Those are possibilities. You have to do synthetic division in order to see whether the possible real zeros are really real zeros.

Answer 20

hmm...ok well now i dont know.