Question

Help please

Answer 1

rational root theorem

Answer 2

first list all the factors of the constant term (p = 15) technically -15 but it's easier to just consider the positive

then list all the factors of the leading coefficient (q = 2)

then list all possible combinations of +/- p/q factors

Answer 3

5,3,15

Answer 4

1,2 for 2

Answer 5

and i don't know what you mean by "list all possible combinations of +/- p/q factors

Answer 6

for 15 also include 1, so 1, 3, 5, 15
2 is just 1 and 2

so we start by listing all possible combinations of the p-factors (1,3,5,15) divided by q factors (1,2)

so +/- 1/1, +/- 1/2, etc. (take a look at the first example I posted if it's a bit unclear)

Answer 7

so would 15 be q and 2 p?

Answer 8

p is 15 and q is 2

Answer 9

then keep going for every possible combination of p/q factor

Answer 10

+-1/1,+-1/2,+-3/1,+-3/2,+-5/1,+-5/2,+-15/1,+-15/2

Answer 11

good, that's it (I would recommend simplifying when possible, for example 1/1 is just 1

Answer 12

Oh so all the other numbers over 1 is just that number

Answer 13

ok*

Answer 14

yes

Answer 15

so 8 possible zeros

Answer 16

yes

Answer 17

it is asking to list the possible choices so it would be the entire list you wrote not just 8

Answer 18

yup got that

Answer 19

for descartes rule of signs (#14) you will need to determine the # of sign changes for f(x) and f(-x)

for f(x) you can simply look at the function as it is already written and count the number of sign changes

Answer 20

Well i am getting 19

Answer 21

4 sign changes?

Answer 22

it only changes from + to - once so only 1 sign change

Answer 23

so you got that by apply -1 right?

Answer 24

the first term is positive
the second term is positive (no change yet, still positive)
the third term is negative (it changed from + to - so 1 sign change)
the fourth term is negative (no change, still negative)

so 1 sign change

Answer 25

got it so the final answer is one sign change

Answer 26

yup so we have at most 1 positive root

now we repeat this process with f(-x)

simply substitute (-x) for x into the function and repeat the sign change count

Answer 27

okay so 1 sign change for this?

Answer 28

check your signs carefully

I get -2x^3 + 5x^2 + 31x - 15

so how many sign changes is this?

Answer 29

from the original ?

Answer 30

2?

Answer 31

good, so at most 2 negative roots

so we have 1 possible positive roots and 2 possible negative roots

we must also account for the possibility of complex roots (they always come in pairs, so the function must have an even number of complex roots, or 0 complex roots)

positive + negative + complex roots must equal 3 for all combinations (since the degree is 3)

Answer 32

our possibilities are:

1 positive 2 negative 0 complex
1 positive 0 negative 2 complex

that's it (since we can't have 2 negative + 1 complex b/c complex roots come in pairs)

Answer 33

so the answer is 1 positive 2 negative 0 complex
1 positive 0 negative 2 complex

Answer 34

yes