The polynomial p(x)=3x^3+2x^2-6x has _ local maxima and minima. also explain how to do it if you can.

1.  Find the derivative of P(x)=3x^3+2x^2-6x.  It's P'(x)=9x^2 + 4x - 6.
2.  Set this result equal to zero and solve for the critical values:
             9x^2 + 4x - 6 = 0      Using the quadratic formula, I got
              x = [-4 plus or minus sqrt(232)] / 18.  Reducing this,
               x = [-4 plus or minus 2 sqrt(58)]; thus, there are two real, unequal roots and two real, unequal critical values.
3.   One at a time, examine the two critical values:  determine whether the derivative changes from neg to pos or from pos to neg at each of these values.  Example:  If the derivative is pos to the left of the first c. v. and neg to the right, we've got a local max.

4.  Since there are only 2 critical values, you can have no more than 1 local max (corresponding to a change in the sign of the derivative from pos to neg) and one local min.  (from neg to pos).

Message me if this explanation is not sufficient to help you understand this problem thoroughly.