Here's the question you clicked on:
Tati_Lee
What are the possible number of positive, negative, and complex zeros of f(x) = –x6 – x5– x4 – 4x3 – 12x2 + 12 ?
Positive: 4, 2, or 0; Negative: 2 or 0; Complex: 6, 4, 2, or 0 Positive: 3 or 1; Negative: 3 or 1; Complex: 4, 2, or 0 Positive: 3 or 1; Negative: 2 or 0; Complex: 4, 2, or 0 Positive: 1; Negative: 5, 3, or 1; Complex: 4, 2, or 0
there are 1 or 3 negative real roots. So there are 4 or 2 complex roots.
is that what your looking for Tati?
Yea Can you show me how you got it? I gotta do my DBA on the questions i got wrong on my exam review -_-
I learned that "This is a polynomial of order 6, so there are 6 roots, so there is only one sign change for the coefficients, so there is exactly 1 positive real root. there are 3 sign changes, so there are at most 3 negative real roots." And that's how I got the answer....not sure if that makes any sense..
yes it does.. thank you!