Please help with precalc problem Use Descartes Rule of Signs

Please help with precalc problem

Use Descartes\' Rule of Signs to determine how many positive and how many negative real zeros the polynomial can have. Then determine the possible total number of real zeros. (Enter your answers as comma-separated lists.) number of positive zeros possible number of negative zeros possible number of real zeros possible

Solution

P(x) = x^4 + x^3 +x^2 +x+12

Decarte\'s Rule:

There are no sign changes in P(x) , so zero number of+ve ) zeros

Now( P(-x) = x^4 -x^3 +x^2 -x +12

There are 4 sign changes so, there can maximum 4 -ve zeros, or 2 -ve zeros( with 2 complex zeros) , 0 number -ve real zeros

Number of real zeros: 4 , 2, 0