How do you find all the real and complex roots and use Descartes Rule of Signs to analyze the zeros of #P(x) = x^5 - 4x^4 + 3x^3 + 2x - 6#?

First, count the number of sign changes in the polynomial, from term to term.

Using P(ositive) and N(egative), it goes: P N P P N

Thus, the sign changes #3# different times.

This means that there are #3# positive roots. There could also be #1#--I'll return to this idea later.

To find the negative roots, count the sign changes in #f(-x)#:

#f(-x)=-x^5-4x^4-3x^3-2x-6#

The signs go: N N N N N

The sign never changes so there are no negative roots.

So, we know that there are a possible #3# positive roots. Since there are no negative roots, the other #2# roots (we know there are #5# since the degree of the polynomial is #5#) are complex roots.

However, since complex roots always come in pairs, the #3# positive roots may actually be expressed as #1# positive root and #2# complex roots. Again, there are the two additional complex roots not a part of the #3#, so the polynomial could have #1# positive root and #4# complex ones.

We now know that we only have to try to find the positive roots. The potential roots are #1,2,3,6#. Synthetically divide or do polynomial long division to find that #(x-3)# is a root.

#(x^5 - 4x^4 + 3x^3 + 2x - 6)/(x-3)=x^4-x^3+2#

The four imaginary roots are remarkably complex (this is just one): #x = 1/4-1/2 sqrt(1/4+(9+i sqrt(1455))^(1/3)/3^(2/3)+8/(3 (9+i sqrt(1455)))^(1/3))-1/2 sqrt(1/2-(9+i sqrt(1455))^(1/3)/3^(2/3)-8/(3 (9+i sqrt(1455)))^(1/3)-1/(4 sqrt(1/4+(9+i sqrt(1455))^(1/3)/3^(2/3)+8/(3 (9+i sqrt(1455)))^(1/3))))#

There is a method to find the roots to a quartic equation, but it is very convoluted.

Check a graph:

graph{x^5 - 4x^4 + 3x^3 + 2x - 6 [-29.71, 28.03, -18.82, 10.05]}