Short-hand? Do you mean synthetic division?
If 2i is a root, then both it and its complex conjugate should be roots.
(x + 2i) and (x - 2i) should both work.
You can foil these two to get a single binomial.
Then you could use that binomial to use as a divisor into the polynomial. If all works out with no remainder, then you've shown it's a root.