Question

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.

Answer 1

Answer:C) The possible number of roots are

Positive Real root : 2 or 0.

Negative Real root: 1.

Complex root : 2 or 0

Explanation:

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

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

Now
`f(-x)=-6x^3-3x^2-5x+9` (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 .

Answer 2

The possible number of positive real, negative real and complex zeros of the equation f(x) = 6x3 – 3x2 + 5x + 9 are the following:

Positive Real: 2 or 0
Negative Real: 1
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.