How do I solve f(x)=x^3-6x^2+16x-96 using Descartes' Rule of Signs to find the number of positive and negative real roots, along with imaginary roots, but not with a graph?

f(x) has 3 sign changes, so 3 or 1 positive roots

f(-x) has no sign changes, so no negative roots

There might be two complex roots.

To solve the equation f(x) = x^3 - 6x^2 + 16x - 96 using Descartes' Rule of Signs, we need to analyze the signs of the coefficients and count the variations in sign. This will give us information about the number of positive and negative real roots, as well as any imaginary roots.

1. Counting the number of variations in sign for the coefficients of f(x):
- Start by writing down the coefficients of the polynomial in descending order:
f(x) = x^3 - 6x^2 + 16x - 96
- Count the number of sign changes between consecutive non-zero coefficients:
The coefficients are (+), (-), (+), (-). There are two sign changes.

2. From the sign changes in step 1, determine the possible number of positive real roots:
- The number of sign changes gives the maximum number of positive real roots. In this case, there are two sign changes, so there can be at most two positive real roots.

3. Now we need to examine f(-x) to determine the number of negative real roots:
- Replace x with -x in the equation:
f(-x) = (-x)^3 - 6(-x)^2 + 16(-x) - 96
= -x^3 - 6x^2 - 16x - 96
- Count the number of sign changes between consecutive non-zero coefficients in f(-x):
The coefficients are (-), (-), (-), (-). There are no sign changes or an even number of changes.

4. From the sign changes in step 3, determine the possible number of negative real roots:
- The number of sign changes (0 or an even number) gives the maximum number of negative real roots. In this case, there can be at most zero negative real roots.

5. Finally, determine the number of imaginary roots:
- The difference between the degree of the polynomial (3 in this case) and the number of real roots gives the number of imaginary roots. In this case, since we found 2 positive real roots and 0 negative real roots, there must be 1 imaginary root.

To summarize:
- Using Descartes' Rule of Signs, we found that there can be at most 2 positive real roots, no negative real roots, and 1 imaginary root for the equation f(x) = x^3 - 6x^2 + 16x - 96.