The possible number of the real positive root is 2 and the negative roots are 2.
Given that
Function; [tex]\rm p(x)= -3x^3 + 11x^2 + 12x - 8[/tex]
We have to determine
The possible number of positive real zeros and negative real zeros for each polynomial function is given by Descartes's rule of signs.
According to the question
Descartes's rule of signs tells us that we then have exactly 3 real positive zeros or less but an odd number of zeros.
Descartes's rule says that the polynomial will have the number of zeroes equal to the number of times the sign-off the coefficient changes.
Function; [tex]\rm p(x)= -3x^3 + 11x^2 + 12x - 8[/tex]
For the positive real roots are -3, 11, 12, and 8.
There will be 2 positive real roots.
For the real negative roots substitute P(x) = P(-x)
[tex]\rm P(x)= P(-x)= -3(-x)^3 + 11(-x)^2 + 12(-x) - 8\\\\P(x)= P(-x)= +3(x)^3 + 11(x)^2 - 12(x) - 8\\[/tex]
There are 2 negative real roots.
Hence, the possible number of the real positive root is 2 and the negative roots are 2.
To know more about Descartes's rule click the link given below.
https://brainly.com/question/2599579
