The factors of 12ab + 9a + 12 + 16b are
(4b+3)(3a+4)
12ab + 9a + 12 + 16b = 3a × (4b + 3) + 4 × (3 + 4b)
The factors of 12ab + 9a + 12 + 16b are (4b + 3) × (3a + 4).
So, 4b + 3 and 3a + 4 are the irreducible factors of the given algebraic expression.