What are the possible number of positive real, negative real, and complex zeros of . f(x) = 6x3 – 3x2 + 5x + 9? . . 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.

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

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

Now (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

106 emergencies

step-by-step explanation:

-12 > a

step-by-step explanation:

-11 > 1 + a

subtract 1 from each side

-11-1 > 1-1+a

-12 > a