Question

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

Be sure to provide the answer in your explanation.

Answer 1

how many sign changes are there in f(x)

Answer 2

3?

Answer 3

good

Answer 4

so there are at most 3 positive real roots

Answer 5

now evaluate f(-x) and tell me what you get

Answer 6

uh i dont know how.

Answer 7

start with f(x) and replace all copies of x you see with -x

then simplify

Answer 8

f(-x) = -7x5 - 15x4 + x3 - 4x2 + 6x + 11
is this what you meant?

Original function: f(x) = 7x5 + 15x4 – x3 + 4x2 – 6x – 11

Answer 9

f(x) = 7x^5 + 15x^4 – x^3 + 4x^2 – 6x – 11

f(-x) = 7(-x)^5 + 15(-x)^4 – (-x)^3 + 4(-x)^2 – 6(-x) – 11

f(-x) = -7x^5 + 15x^4 + x^3 + 4x^2 + 6x – 11

Answer 10

so i sorta did it right?

Answer 11

a bit, but you have signs off in some places

Answer 12

shouldnt 4x^2 be a negative ?

Answer 13

nope because (-x)^2 = x^2

Answer 14

so +4(-x)^2 = +4x^2

Answer 15

oh okay.

Answer 16

this is true for any even exponent

Answer 17

ohkay so like is that it?

Answer 18

now you count the sign changes in f(-x)

Answer 19

2

Answer 20

so there are at most 2 negative real roots

Answer 21

what is the degree of this function

Answer 22

of the original or new one?

Answer 23

either one, but mainly the original

Answer 24

5?

Answer 25

the degree tells you what about the total number of roots

Answer 26

so 5?

Answer 27

good, so if you have a max of 3 positive real roots and a max of 2 negative real roots

how many complex roots do you have?

Answer 28

none

Answer 29

now let's say you only had 2 positive real roots and a max of 2 negative real roots


how many complex roots do you have?

Answer 30

still none right?

Answer 31

you have 2+2 = 4 roots so far

but there are 5 total, so you have room for 1 more

Answer 32

but you can't have just 1 complex root in the form a+bi since they always come in pairs

Answer 33

make sense?

Answer 34

yea.

Answer 35

ok great

Answer 36

so you could have 5 real roots, with no imaginary roots at all

or you could have 3 real roots with 2 imaginary roots

or you could have 1 real root with 4 imaginary roots

Answer 37

and the real roots break down like this

3 max positive roots
2 max negative roots

Answer 38

okay.