solve the equation x4+4x3+5x2+2x-2=0 of which one root is -1+root-i
I'm guessing that the root mentioned is supposed to be -1 + i, where i = root(-1).
Complex roots always come in pairs, as complex conjugates.
If one complex root is a + ib, then the other complex root is a - ib.
Since we have -1 + i, then we also have -1 - i. That means that (x - (-1+i) and (x - (-1-i)) are factors of the original 4th order equation.
(x - (-1+i))*(x - (-1-i)) = x^2 + 2x + 2
Since (x^2 + 2x + 2) is a factor of the original eqn, then (x^2 + 2x + 2) divides into the original eqn.
(x^4 + 4x^3 + 5x^2 + 2x - 2) / (x^2 + 2x + 2) = x^2 + 2x - 1.
Using the quadratic formula on x^2 + 2x - 1,
x = (-2 +/- rt(2^2 - 4*1*(-1))/(2*1)
x = (-2 +/- rt(4 + 4)/(2)
x = (-2 +/- 2rt(2)/(2)
x = -1 +/- rt(2)
The 4 roots are: (-1 + i), (-1 - i), (-1 + rt(2)), (-1 - rt(2))