x-2 is to be a factor of 2x³-6x²+5x+a

We divide either by long division or by synthetic division:
and set the remainder = 0

By long division:           By synthetic division

        2x²-2x+1            2|2 -6  5  a
x-2)2x³-6x²+5x+a             |   4 -4  2
    2x³-4x²                   2 -2  1  a+2
       -2x²+5x
       -2x²+4x
             x+a
             x-2
               a+2

Either way, the remainder is a+2.
We want the remainder to be 0.
So we set a+2 = 0
            a = -2

Edwin