3x(x+2)

5x(y + 2)

x(5y + 3)

2(x + 2)

Here, we have a common factor 3 in all the terms.

∴    9x + 18y + 6xy + 27 = 3[3x + 6y + 2xy +9]

  We find that      3x + 6y = 3(x + 2y) and 2xy + 9 = 1(2xy+9)

i.e. a common factor in both the groups does not exist,

 Thus, 3x + 6y + 2xy + 9 cannot be factorised.

On regrouping the terms, we have

   3x + 6y + 2xy + 9 = 3x + 9 + 2xy + 6y
                            = 3(x + 3) + 2y(x + 3)
                            = (x + 3) (3 + 2y)

Now, 3[3x + 6y + 2xy + 9] = 3[(x + 3)(3 + 2y)]

Thus, 9x + 18y + 6xy + 27 = 3(x+3) (2y+3)