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?

1 Answer
Jan 14, 2016

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]}