Factorise expression?

2x^2 - 3xy - 2y^2 - 2x - 11y - 12

= (x - 2 y - 3) (2 x + y + 4)

2x² - 3xy - 2y² - 2x - 11y - 12 = (2x + y)(x - 2y) - 2x - 11y -12

2x² - 3xy - 2y² - 2x - 11y - 12 = (2x + y + a)(x -2y + b)

2x² - 3xy - 2y² - 2x - 11y - 12 = 2x² - 3xy - 2y² + (a + 2b)x + (-2a + b)y + ab

x term on each side: a + 2b = -2 …… [1]

y term on each side: -2a + b = -11 …… [2]

[1]*2 + [2]:  5b = -15  ⇒  b = -3

[1] - [2]*2:  5a = 20  ⇒  a = 4

Check: constant term on each side = ab = -12

Hence, 2x² - 3xy - 2y² - 2x - 11y - 12 = (2x + y + 4)(x -2y - 3)

The expression starts with 2x^2, so the two factors must contain 2x and x.

Then there's -2y^2, so the factors contain 2y and y, one added and the other subtracted.

There are two ways we could combine them:

 (2x ± 2y   )(x ± y   )

 (2x ± y   )(x ± 2y   )

Of these choices the only way to get -3xy is

 (2x + y   )(x - 2y   )

Finally, the constants have a product of -12. When multiplied with y and -2y, they give a sum of -11y. The only way to get an odd sum is to multiply the y by an odd factor, and the only way to get a negative sum is to multiply the -2y by the positive factor.

The only odd factors of 12 are 1 and 3, so check each of those and you'll find (y)(-3) + (-2y)(4) = -11y.

Place those into the factors:

(2x + y + 4)(x - 2y - 3)

and then verify that the multiplication of the x's with the constants produces -2x.

(x - 2y - 3)(2x + y + 4)