If pqrs is a square solve for x what do I do?
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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.
<em>Therefore, the possible numbers of zeros are:</em>
<em>2 positive, 1 negative and 0 complex</em>
<em>0 positive, 1 negative and 2 complex.</em>
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).
<h3>Hence, the number of complex zeros for the given polynomial is zero.</h3>