Question

what are the number of possible negative real zeros for f(x) = x^3 - 4x^2 - 3x -9

Answer 1

you need to apply Descarte's Law of Signs. I'm a bit rusty at this to be honest.

Answer 2

that's what I was thinking but the degree of the polynomial will only give number of possible roots, correct? Not the number of negative real roots.

Answer 3

right

Answer 4

there is only 1 change of sign - from + x^3 to -4x^2 which I think means there is 1 or 3 positive roots
you find the possible number of negative roots by changing the signs and counting the number of changes

Answer 5

ok...thanks...

Answer 6

if we multiply each term by -1 and count the number of sign changes its 1
so that means there is 0 or 1 number of negative roots

Answer 7

let x=-1 and count the number of sign changes that occur.

Answer 8

in effect (-1)^even = 1 and (-1)^odd = -1 so all odd degrees reverse in sign if you can remember that trick

Answer 9

so in this problem, the negative roots would match

x^3 - 4x^2 - 3x -9
to
-x^3 - 4x^2 + 3x -9

Answer 10

@amistre64 - is that a typo?

Answer 11

i cant see a typo ... what are you refering to?

Answer 12

shouldn't the result after multiplying by -1 be

- x^3 + 4x^2 +3x + 9 ?

Answer 13

no, we dont multiply thru by -1

we assume that the value of x is negative,

f(-x) = (-x)^3 - 4(-x)^2 - 3(-x) -9

f(-x) = -x^3 - 4x^2 + 3x -9

Answer 14

oh - I see - i was under a misapprehension - think id better look at Descartes rule again!

Answer 15

- so thats 2 changes of sign right?

Answer 16

now count that chang in signs

and by the way, we cant have 0 or 1 as options

complex roots come in pairs, so

say we have 3 changes, we have 3, or 1 possible neg roots
say we have 1 changes, we have 1 possible neg root
say we have 4 changes, we have 4, or 2, or 0 neg roots

Answer 17

in this case, we have 2 sign changes, yes; so account for complex solutions to find possible number of neg roots

Answer 18

so we could have 1 or zero negative roots?

Answer 19

we cant have 1 or 0 a the same time

complex roots eat at our other roots, they eat away at them in 2s

Answer 20

no - must be zero

Answer 21

we can either have: 2 neg and 0 complex
or 0 neg and 2 complex

so the 'possible' number of neg roots is 2 or 0

Answer 22

yes - complex roots exist as conjugates

Answer 23

yes - i see now

Answer 24

spose we counted 3 sign changes

we can have 3 neg and 0 complex
or 1 neg and 2 complex

possible number of neg roots would then be 3 or 1

Answer 25

right

Answer 26

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