Let us factorise the following algebraic expressions:

(x + 1) (x + 2) (3x – 1) (3x – 4) + 12

The given expression can be rewritten as:

(x + 1)(3x – 1)(x + 2)(3x – 4) + 12

⇒ (3x² – x + 3x - 1)(3x² - 4x + 6x - 8) + 12

⇒ (3x² + 2x - 1)(3x² + 2x - 8) + 12

Let 3x² + 2x = p

⇒ (p - 1)( p – 8) + 12

⇒ p² – 9x + 8 + 12

⇒ p² – 9p + 20

⇒ p² – 5p – 4p + 20

⇒(p – 5)(p – 4)

On substituting the value of p, we get,

⇒ (3x² + 2x – 5)(3x² + 2x – 4)

⇒(3x² + 5x – 3x – 5)(3x² + 2x – 4)

⇒ (x(3x + 5) – 1(3x + 5))(3x² + 2x – 4)

⇒ (x – 1)(3x + 5) (3x² + 2x – 4)