\(2x^2-3x+2xy-3y\)
\(=\left(2x^2+2xy\right)-\left(3x+3y\right)\)
\(=2x\left(x+y\right)-3\left(x+y\right)\)
\(=\left(2x-3\right)\left(x+y\right).\)
2x2-3x+2xy-3y
= (2x2+2xy)-(3x-3y)
= 2x(x+y)-3(x+y)
= (x+y)(2x-3)
\(2x^2-3x+2xy-3y\)
=2.x.x - 3.x + 2xy - 3y
=(2x-3).x + ( 2x-3)y
=(2x-3y).(x+y)