I hope you understand FOIL, i.e., FIRSTS, OUTERS, INNERS, LASTS. I'm assuming you do.

2ab - a + b + ___

Write this:

(__ ± __)(__ ± __)

Let's fix up the FIRSTS to be 2ab.  We will split 2ab up into factors 2a and b
and put them for the FIRSTS

(2a ± __)(b ± __)

Now let's fix up the OUTERS.  We notice to get the term " -a " for the OUTERS,
we will need the term on the far right to be -1/2.

(2a ± __)(b - 1/2)

Now let's fix up the INNERS.  We notice to get the term +b for the
INNERS, we will need the term +1 in the remaining blank:

(2a + 1)(b - 1/2)  <--factorization as the product of two linear expressions

The LASTS are now already fixed up. 

So we proceed to FOIL that out:

(2a)(b)+(2a)(-1/2)+(+1)(b)+(1)(-1/2)

2ab - a + b + (-1/2)

So we see that the answer is -1/2

Edwin

Write three first terms in this form

    2ab - a + b + ___ = 2*(ab - a/2 + b/2 + ___).


Look at the expression in parentheses.  

It is clear that it should be

    (ab - a/2 + b/2 - 1/4),


which is the product of linear binomials (a+1/2)*(b-1/2).


Now you have this identity

    2ab - a + b - 1/2  = 2*(a+1/2)*(b-1/2).


You can relate the factor 2 to the first or to the second binomial factor.  
It will give you two possible decompositions

    2ab - a + b - 1/2 = (2a+1)*(b-1/2)

or 

    2ab - a + b - 1/2 = (a+1/2)*(2b-1).


In any case, the blank term is -1/2.