Since this is a cubic equation, there are 3 possible zeroes
x= 2, x = -(1-i) and x = -(1+i)
We can solve the equation by factoring the P(x)
[tex](x-2)(x^2+2x+2)[/tex]
We can get the complex factor of the second term
[tex](x-2)(x+(1+i))(x+(1-i))_{}[/tex]
Given this factor, the function P(x) will be zero if
[tex]x\text{ = 2, -1-i, and -1+i}[/tex]
