What are the possible number of positive real, negative real, and complex zeros of . f(x) = 6x3 – 3x2 + 5x + 9?. Answer:. . A)Positive Real: 1. Negative Real: 0. Complex: 2. . B)Positive Real: 2 or 0. Negative Real: 2 or 0. Complex: 1. . C)Positive Real: 2 or 0. Negative Real: 1. Complex: 2 or 0. . D)Positive Real: 1. Negative Real: 2 or 0. Complex: 2 or 0.

The correct answer between all the choices given is the third choice or letter C. I am hoping that this answer has satisfied your query and it will be able to help you in your endeavor, and if you would like, feel free to ask another question.

JeanaShupp JeanaShupp · Answer 2 · 2018-11-01T12:43:44+01:00

Answer:C) The possible number of roots are

Positive Real root : 2 or 0.

Negative Real root: 1.

Complex root : 2 or 0

Step-by-step explanation:

Given :A cubic polynomial arranged in descending order [tex]f(x)=6x^3-3x^2+5x+9[/tex]

Therefore by Descartes rules of signs . It has 2 sign change therefore it has 2 or zero positive real roots.

Now [tex]f(-x)=-6x^3-3x^2-5x+9[/tex] (change signs of coefficients of odd terms)

Now it has 1 sign change i.e. f(-x) has exactly 1 positive real roots. That means f(x) has exactly 1 negative real roots.

For complex roots , we know that it is a cubic polynomial i.e. it has exactly 3 roots and complex roots always occur in pair, Therefore it will have zero or 2 complex roots.

Therefore the possible number of roots are

Positive Real root : 2 or 0.

Negative Real root: 1.

Complex root : 2 or 0

Therefore option C is the correct answer .