2015/106) Factorize: 2x^2-5xy-3y^2+3x+19y-20

The factor is of the form $(ax+by+c)(dx+ey+f)$

if we ignore c and f and multiply we get

$(ax+by)(dx+ey)= adx^2 + (ae+bd)xy + bey^2$

so we can first factor $2x^2-5x-3y^2$ and then evaluate the other parts

$2x^2-5xy-3y^2= (2x+y)(x-3y)$ factored by quadratic method

so

 $2x^2-5xy-3y^2+3x+19y-20= (2x+y+e)(x-3y+f)$

      $= 2x^2-3y^2-5xy+(2f+e)x + (f-3e)y + ef$

Comparing coefficients of x , y and constant separately we get
$2f+e = 3, f- 3e = 19$ and $ef = - 20$
solving 1^st 2 equations we get $e = -5$ and $f = 3$ and 3^rd equation also meets criteria
so we get 

$2x^2-5xy-3y^2+3x+19y-20= (2x+y-5)(x-3y+4)$