Use Descartes' Rule of Signs to determine the possible numbers of positive and negative real zeros of f(x)=x^{3}+5 x^{2}+6 x+8

Step 1 ：Use Descartes' Rule of Signs to determine the possible numbers of positive and negative real zeros of the polynomial \(f(x)=x^{3}+5 x^{2}+6 x+8\).

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 between consecutive coefficients, or less than that by an even number. The number of negative real zeros is found by applying the rule to the polynomial \(f(-x)\).

Step 3 ：For the given polynomial \(f(x)=x^{3}+5 x^{2}+6 x+8\), we first need to count the number of sign changes in the coefficients to find the possible number of positive real zeros. Then we substitute x with -x in the polynomial and again count the number of sign changes to find the possible number of negative real zeros.

Step 4 ：The coefficients of the polynomial are all positive, so there are no sign changes. This means that there are no positive real zeros. When we substitute x with -x, the coefficients remain the same, so there are also no negative real zeros.

Step 5 ：\(\boxed{\text{Final Answer: The possible numbers of positive and negative real zeros of the polynomial } f(x)=x^{3}+5 x^{2}+6 x+8 \text{ are both 0. Therefore, the polynomial has no real zeros.}}\)