Math question algebra 2, fundamental theorem of algebra, stste the number of complex roots and the possible number of real and imaginary roots for each equation. then find all roots. one root has been given. x^6 - 3x^5 + 2x^4 - 6x^3 - 15x^2 + 45x = 0; 3 (if possible provide work)

By Descartes' Rule of Signs:
The signs from the original equation are: + - + - - +. 3 sign changes mean that there are either 3 or 1 positive roots.
If we change the signs of the odd-powered terms: + + + + - -. This 1 sign change means that there is exactly 1 negative root.

All roots:
We can immediately factor out x from the equation:
x (x^5 - 3x^4 + 2x^3 - 6x^2 - 15x + 45) = 0
Factor out (x-3) since x = 3 is a root:
x (x - 3) (x^4 + 2x^2 - 15) = 0
Factor the last term further:
x (x - 3) (x^2 - 3) (x^2 + 5) = 0
This allows us to determine the rest of the x-values:
x = 0, 3, sqrt3, -sqrt3, i*sqrt5, i*-sqrt5