Use Descartes' Rule of Signs to determine the possible number of positive and negative zeros of f(x)= 6x³ - 5x² + 6x - 6. A)3 positive reals or 1 positive real;3 negative reals or 1 negative real B)no positive reals;no negative reals C)3 positive reals or 1 positive real;1 negative real D)1 positive real;3 negative reals or 1 negative real E)3 positive reals or 1 positive real;no negative reals

Question

Quizplus · Accepted Answer

Descartes' Rule of Signs states that the number of positive real zeros of a polynomial function is either equal to the number of sign changes between consecutive coefficients or is less than that by an even number. For f(x) = 6x^13 - 5x^12 + 6x - 6, there are 2 sign changes (from 6 to -5, and from -5 to +6), so there could be 2 or 0 positive real zeros. However, since the question format and options provided do not match this exact outcome, it seems there might be a misunderstanding in interpreting the function's expression or the options provided. For negative zeros, we evaluate f(-x) = -6x^13 - 5x^12 - 6x - 6, which has no sign changes, indicating no negative real zeros. The closest matching option to the analysis is E, suggesting a possible misinterpretation in the question or answer choices.

Use Descartes' Rule of Signs to Determine the Possible Number