Question

Does anybody know how to find the negative zeros of this function f(x)=x^3-4x^2-3x-36?

Answer 1

By hand?

Answer 2

yes. I know all you have to do is plug in -x for x in the function but I'm not sure how to find the sign changes?

Answer 3

Ah, OK you don't actually want a negative zero.

You want the possible number of them, is that right?

Answer 4

eg if you put -x in x^3 the result is -x^3 which is a sign change.

Answer 5

Yes that is correct.

Answer 6

OK, the only terms that will be affected are odd terms (squaring etc doesn't change the sign) so you get
-x^3 -4x^2+3x-36 which has 2 sign changes so
number of possible negative zeros is 2 (or 0)

Answer 7

Oh thank you now I understand :D

Answer 8

ur welcome.

Answer 9

Do you know how to find the complex zeros?

Answer 10

Usually you just take the order (3 for a cubic) and subtract the number of possible negative and positive zeros.
But you may not be able to say precisely in every case how many of every kind of root there is.

Answer 11

Ohh ohkay thank you. Do you know how to use the factor theorem for this function?

Answer 12

This function has no "easy" solution (you can follow the lengthy solution of a cubic to find the roots (1 positive real and 2 complex)

Answer 13

Umhh I don't understand.

Answer 14

In other words, you cannot use the factor theorem

Answer 15

Oh alright thanks.

Answer 16

ur welcome.