D_homes wrote:
You could use the remainder/factor theorem.
x is a factor (solution) if f(x) = 0
To find the values for x where f(x) could possibly equal zero, divide all the factors of 'a' by all the factors of 'c', and plug those values into the equation.
He want's to find the number of solutions, not necessarily what they are. Also, we're talking real numbers here, trial and error is not a good idea. For this same reason, factoring is also not possible.
For example, consider x^2 + 0*x - 5. This is a simple polynomial (I'm even using integers), yet trying to get a computer to find the exact values of the solution by trial and error is no walk in the park.
There are easy ways to know how many real solutions exist for a polynomial of at most degree 2.