x = 4 ; x = 3 + i ; x = 3 - i
(If you get a zero that is adding or subtracting, you always need to write it twice but change the sign do they cancel out)
f(x) = (x-4)(x-3-i)(x-3+i)
Distributing the last two parenthesis first is always the best way to start off
(x-3-i)(x-3+i) has (x-3) in common so it can be separated to
(x-3)^2 + (-i)(+i)
(x^2 - 6x + 9) ; (-i)(+i) is always +1
(x^2 - 6x + 9) + 1
(x^2 - 6x + 10)
Now multiply this with (x-4)
x^3 - 6x^2 + 10x
- 4x^2 + 24x - 40
x^3 - 10x^2 + 34x - 40 = f(x)
