If √(2x+2) is a factor then 2x+2 must be a factor of (2x3-√2)2
(2x3-√2)2=4x6-4x3√2+2=2(2x6-2x3√2+1); 2x+2=2(x+1).
Use synthetic division to divide x+1 into 2x6-2x3√2+√2:
-1 | 2 0 0 -2√2 0 0 1
2 -2 2 -2 2+2√2 -2-2√2 | 2+2√2
2 -2 2 -2-2√2 2+2√2 -2-2√2 | 3+2√2 = 2x5-2x4+2x3-(2+2√2)x2+(2+2√2)x-2-2√2 rem 3+2√2.
For general x, then, √(2x+1) cannot be a factor of 2x3-√2, because there is a remainder 3+2√2.
The actual quotient would be √(2x5-2x4+2x3-(2+2√2)x2+(2+2√2)x-2-2√2).