LilyMQ one year ago 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

1. LilyMQ

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

2. LilyMQ

I think it's #4

3. ybarrap

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?

4. LilyMQ

Yes!! Okay thank's!