How do I show that x^5+x^2+1=0 has 4 imaginary roots?

Answer 1

There are no sign changes in the polynomial fro term to term, therefore there ar no real positive solutions

f(-x) = -x^5 +x^2 +1 =0
Therefore there is only one real negative root.
Thergore the remaining four roots must be imaginary.
This Descarte's law of signs.