Hello! I don't know how to foil and add/subtract an imaginary numbers.
$\displaystyle x^2 (x+2i) (x-2i) (x - (3-i)) (x-(3+i)) $
thanks...
is that so?
$\displaystyle f(2i) = (2i)^6 - 6(2i)^5 + 14(2i)^4 - 24(2i)^3 + 40(2i)^2
$
$\displaystyle f(2i) = 64i^6 - 192i^5 + 224i^4 - 192i^3 + 160i^2$
$\displaystyle f(2i) = -64 - 192i + 224 + 192i - 160 = 0$
... and I guarantee you that $\displaystyle f(3+i) = 0$ ... complex zeros always come in conjugate pairs.