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

The options are
1. Positive: 4, 2, or 0; Negative: 2 or 0; Complex: 6, 4, 2, or 0

2. Positive: 3 or 1; Negative: 3 or 1; Complex: 4, 2, or 0

3. Positive: 3 or 1; Negative: 2 or 0; Complex: 4, 2, or 0

4. Positive: 1; Negative: 5, 3, or 1; Complex: 4, 2, or 0

Answer 2

I think it's #4

Answer 3

Let P = max number of positive zeros, N = max number of negative zeros and n - (P+N) the minimum number of complex. n is the degree of the polynomial

n=6
Signs of coefs from highest to lowest: {-,-,-,-,-,0,+}.

Positive zeros (P) :
Number of sign changes: 1

Now negate the odd terms: {-,+,-,+,-,0,+}

Negative zeros (N):
Number of sign changes: 5,3,1

Complex zeros (min):
6 - (1+5) = 0
6 - (1+3) = 2
6 - (1+1) = 4

$$
\href{https:///en.wikipedia.org/wiki/Descartes%27_rule_of_signs}{\text{Descarte's Rule}}
$$

Does this make sense?

Answer 4

Yes!! Okay thank's!