Assuming we can factor just using integers, there are four possibilities to begin with, just based on the coefficient 2 in 2y^2.

(2y + u) (y + v)

(2y + u) (y - v)

(2y - u) (y + v)

(2y - u) (y - v)

Seeing the + in front of the three, we can narrow down to

(2y + u) (y + v)

(2y - u) (y - v)

Seeing the - in front of the 7y, we can narrow down to

(2y - u) (y - v)

possible (u,v) are (1,3), (3,1), giving

(2y - 1) (y - 3)

(2y - 3) (y - 1)

A little examination reveals

(2y - 1) (y - 3)

In practice you can do most of the steps in your head, I would just write

and then fill in the blanks.Code:(2y - )(y - )