3x2 + bx + 2

Since 3 and 2 are both prime, they have only two factors
Therefore any factorization would have to be like this:

(3x   2)(x   1)

or like this:

(3x   1)(x   2)

And since the last term, 2, is positive,
the signs must be the same, so there are
four possible factorizations, each of the
above with + signs, and each of the above 
with - signs.

(3x + 2)(x + 1) = 3x2 + 5x + 2 in which case b = 5 
(3x - 2)(x - 1) = 3x2 - 5x + 2 in which case b = -5
(3x + 1)(x + 2) = 3x2 + 7x + 2 in which case b = 7
(3x - 1)(x - 2) = 3x2 - 7x + 2 in which case b = -7

Edwin