Consider the following polynomial function and use Decartes' Rule of Signs to determine the possible number of positive real zeros and negative real zeros.
\[
p(x)=x^{3}+9 x^{2}+x-1
\]
Determine the different possibilities for the number of positive zeros of $f$. Enter the values using a commaseparated list.
Determine the different possibilities for the number of negative zeros of $f$. Enter the values using a comma-separated list.
Determine the different possibilities for the number of complex zeros of $f$. Enter the values using a comma-separated list.
Answer
So, the possible number of positive real zeros is \(\boxed{1}\), the possible number of negative real zeros is \(\boxed{2}\) or \(\boxed{0}\), and the possible number of complex zeros is \(\boxed{0}\) or \(\boxed{2}\).
Step 1 ：The given polynomial is \(p(x)=x^{3}+9 x^{2}+x-1\). The coefficients of the polynomial are 1, 9, 1, and -1. The signs of the coefficients are +, +, +, and -. There is 1 sign change. Therefore, the number of positive real zeros is 1.
Step 2 ：Replace x by -x in the polynomial to get \(p(-x)=(-x)^{3}+9 (-x)^{2}+(-x)-1=-x^{3}+9 x^{2}-x-1\). The coefficients of the polynomial are -1, 9, -1, and -1. The signs of the coefficients are -, +, -, and -. There are 2 sign changes. Therefore, the number of negative real zeros is 2 or 0.
Step 3 ：The total number of zeros of a polynomial is equal to its degree. The degree of the given polynomial is 3. If there is 1 positive real zero and 2 negative real zeros, then there are 0 complex zeros. If there is 1 positive real zero and 0 negative real zeros, then there are 2 complex zeros.
Step 4 ：So, the possible number of positive real zeros is \(\boxed{1}\), the possible number of negative real zeros is \(\boxed{2}\) or \(\boxed{0}\), and the possible number of complex zeros is \(\boxed{0}\) or \(\boxed{2}\).
