Enter the number of complex zeros for the polynomial function in the box. f(x)=x^3−96x^2+400

Given polynomial function:

.

We need to apply Descartes' rule of sign to identify the number of complex roots.

The given polynomial is f(x)=x^3-96x^2+400

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

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.

Let us see the graph:

In the graph, we can see that the graph cuts the x axis at three points (2 positive, 1 negative points).

ca = ec

cpctc makes it so that corresponding parts of the triangle you proved to be equal are equal, the other options aren't part of the part of the triangles you proved to be equal.