Use Descartes' rule of signs to determine the different possibilities for the numbers of positive, negative, and nonreal complex zeros for the following function.
\[
f(x)=-8 x^{4}+7 x^{3}-3 x^{2}+7 x-7
\]

The possible number of positive real zeros is $4,2,0$
(Type a whole number. Use a comma to separate answers as needed.)
The possible number of negative real zeros is $\square$.
(Type a whole number. Use a comma to separate answers as needed.)

Step 1 ：Use Descartes' rule of signs to determine the different possibilities for the numbers of positive, negative, and nonreal complex zeros for the function \(f(x)=-8 x^{4}+7 x^{3}-3 x^{2}+7 x-7\).

Step 2 ：The possible number of positive real zeros is \(4,2,0\).

Step 3 ：Descartes' rule of signs states that the number of positive real roots of a polynomial is equal to the number of sign changes in the list of its coefficients, or less than that by an even number. The number of negative real roots is found similarly from the list of coefficients of the polynomial obtained by replacing x by -x.

Step 4 ：Replace x by -x in the polynomial to get \(f(-x)=-8 (-x)^{4}+7 (-x)^{3}-3 (-x)^{2}+7 (-x)-7\).

Step 5 ：Simplify the polynomial to get \(f(-x)=-8 x^{4}-7 x^{3}-3 x^{2}-7 x-7\).

Step 6 ：Count the number of sign changes in the list of coefficients of this polynomial. The coefficients are -8, -7, -3, -7, -7. There are no sign changes in this list.

Step 7 ：According to Descartes' rule of signs, the possible number of negative real zeros is 0.

Step 8 ：\(\boxed{0}\) is the possible number of negative real zeros.