Consider the following polynomial.
\[
f(x)=x^{3}-9 x^{2}+20 x-12
\]

Step 1 of 3 : Use Descartes' Rule of Signs to determine the possible number of positive and negative real zeros. Be sure to include all possibilities

Step 1 ：Consider the polynomial \(f(x)=x^{3}-9 x^{2}+20 x-12\).

Step 2 ：Use Descartes' Rule of Signs to determine the possible number of positive and negative real zeros. Descartes' Rule of Signs states that the number of positive real roots of a polynomial is either equal to the number of sign changes between consecutive coefficients, or less than that by an even number. The number of negative real roots is found by applying the rule to the polynomial obtained by replacing x by -x.

Step 3 ：In the given polynomial, the coefficients are 1, -9, 20, and -12. The signs of the coefficients are +, -, +, and -, respectively. There are 3 sign changes, so there could be 3 or 1 positive real roots.

Step 4 ：To find the number of negative real roots, we replace x by -x in the polynomial, which gives us \(f(-x)=(-x)^{3}-9 (-x)^{2}+20 (-x)-12 = -x^{3}-9 x^{2}-20 x-12\). The signs of the coefficients are -, -, -, and -, respectively. There are no sign changes, so there are no negative real roots.

Step 5 ：Final Answer: The possible number of positive real zeros is 3 or 1, and the possible number of negative real zeros is 0. Therefore, the answer is \(\boxed{3 \text{ or } 1 \text{ positive real zeros and } 0 \text{ negative real zeros}}\).