Determine the maximum number of real zeros that the following polynomial function may have. Then use Descartes' Rule of Signs to determine how many positive and how many negative real zeros the polynomial function may have. Do not attempt to find the zeros.
\[
f(x)=6 x^{4}-8 x^{2}-x+9
\]
What is the maximum number of zeros that $f(x)=6 x^{4}-8 x^{2}-x+9$ can have?
4 (Type a whole number.)
How many positive real zeros may the function $f(x)=6 x^{4}-8 x^{2}-x+9$ have?
2,0 (Type a whole number. Use a comma to separate answers as needed.)
How many negative real zeros may the function $f(x)=6 x^{4}-8 x^{2}-x+9$ have?
(Type a whole number. Use a comma to separate answers as needed.)

Step 1 ：The maximum number of real zeros a polynomial can have is equal to its degree. In this case, the degree of the polynomial is 4, so the maximum number of real zeros is 4.

Step 2 ：Descartes' Rule of Signs states that the number of positive real zeros of a polynomial is equal to the number of sign changes in the polynomial's terms, or less than that by an even number. The polynomial \(f(x)=6 x^{4}-8 x^{2}-x+9\) has 2 sign changes (from positive to negative at \(-8x^2\) and from negative to positive at \(9\)), so it can have 2 or 0 positive real zeros.

Step 3 ：To find the number of negative real zeros, we substitute \(-x\) for \(x\) in the polynomial and then apply Descartes' Rule of Signs again. The polynomial becomes \(f(-x)=6 x^{4}-8 x^{2}+x+9\), which has 1 sign change (from negative to positive at \(x\)), so it can have 1 or 0 negative real zeros.

Step 4 ：Final Answer: The maximum number of zeros that \(f(x)=6 x^{4}-8 x^{2}-x+9\) can have is \(\boxed{4}\). The function \(f(x)=6 x^{4}-8 x^{2}-x+9\) may have \(\boxed{2}\) or \(\boxed{0}\) positive real zeros. The function \(f(x)=6 x^{4}-8 x^{2}-x+9\) may have \(\boxed{1}\) or \(\boxed{0}\) negative real zeros.