Question

What are the possible number of positive, negative, and complex zeros of f(x) = –x^6 + x^5– x^4 + 4x^3 – 12x^2 + 12 ?

Answer 1

descartes rule of sign for this one
a good simple explanation with example here
http://www.purplemath.com/modules/drofsign.htm

Answer 2

\[f(x) = –x^6 + x^5– x^4 + 4x^3 – 12x^2 + 12
\]
has 5 " changes in sign "
a change of sign for every coefficient
so there are either 5, 3 or 1 positive real zeros

Answer 3

now try
\[f(-x) = –x^6 - x^5– x^4 - 4x^3 – 12x^2 + 12
\]
which has only one change is sign
so one negative real zero

Answer 4

When using the rule of sign, where do the 3 and 1 come from? I see there is five sign changes but how do you get 5, 3, and 1?

Answer 5

since there are 5 changes in sign you know it has at most 5 positive real zeros
then you count down by twos

Answer 6

i.e. it could have 5, but it might have 3 or 1
it cannot have 4 or 2 or none

Answer 7

and since there is one change of sign for \(f(-x)\) there must be one negative real zero, because you count down by twos and obviously there cannot be minus one negative real zero

Answer 8

so there must be one negative real zeros
there are 6 zeros in total
so either
1 negative 5 positive
1 negative 3 positive 2 complex
1 negative 1 positive 4 complex
the complex ones come in conjugate pairs
in fact for this one there is one negative zero, one positive zero, and 4 complex zeros

Answer 9

Alright, thank you.

Answer 10

this is how i know
http://www.wolframalpha.com/input/?i=%E2%80%93x^6+%2B+x^5%E2%80%93+x^4+%2B+4x^3+%E2%80%93+12x^2+%2B+12

Answer 11

yw