Solution:
There is no (zero) complex zeros for the given polynomial.
Explanation:
We apply Descartes' rule of sign to identify the number of complex roots.
The given polynomial is ![f(x)=x^3-96x^2+400](https://tex.z-dn.net/?f=f%28x%29%3Dx%5E3-96x%5E2%2B400)
Let us see the number of sign changes in f(x)
+,-,+
There are 2 sign changes in f(x). One from plus to minus and second from plus to minus. Hence, there 2 or 0 positive roots.
Now, let us see the number of sign changes in f(-x)
![f(-x)=-x^3-96x^2+400](https://tex.z-dn.net/?f=f%28-x%29%3D-x%5E3-96x%5E2%2B400)
-,-,+
There are only one sign change. Hence, there will be 1 negative roots.
The degree of the polynomial is 3. Hence, there will be exactly 3 zeros.
Therefore, the possible numbers of zeros are
2 positive, 1 negative and 0 complex
0 positive, 1 negative and 2 complex.
Hence, possible number of complex zeros are 0 and 2.
Now, we graph the function in xy plane and see that the graph cuts the x axis at three points. It means all the zeros are real , which is the case of first possibility (2 positive, 1 negative and 0 complex).
Hence, the number of complex zeros for the given polynomial is zero.