Enter the number of complex zeros for the polynomial function in the box. f(x)=x3−96x2+400

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

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)

-,-,+

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.