How do you find all the roots of this function?

For the complex solutions

(x-2)^2 = -1

x^2-4x+5 = 0

Division by (x^2-4x+5) gives (x^2-2x-2)

(x^2-2x-2)(x^2-4x+5) = 0

x = [2±sqrt(12)]/2

x = 1+sqrt(3), 1-sqrt(3), 2+i, 2-i

Gabriella -if (2 - i) is a solution, then by definition (2 + i) is another solution so that [x - (2 + i)]and [x - (2 - i)] are factors so multiplying them together, x^2 - 4x + 5 is a factor. To find the other two factors, there are several methods, and whichever you use is purely personal preference. One is to use Synthetic Division using the separate roots 2 + i and 2 - i, but that can be quite tedious with complex roots. You could try Long Division but that can be a pain, so if I were doing it I'd almost certainly proceed as follows: since the divisor has a term in x^2 and the dividend has x^4, the quotient must have x^2; similarly, the divisor has the constant 5, and the dividend has -10 the quotient must have - 2, so the required quadratic factor is of the form(x^2 + Ax - 2) and we have to work out the A: the term in x^3 is - 3x^3 and this is from x^2*Ax - 4x^3 = (A - 4(x^3 = 6x^3 so A - 4 = -6, so A = - 2; check with the term in x^2 and the term in x as well to be on the safe side. You now have the quadratic factor, so factorise that as normal. Hope I managed to explain that well enough.