Let us factorise the following algebraic expressions:

a² – 9b² + 4c² – 25d² – 4ac + 30bd

Given, a² – 9b² + 4c² – 25d² – 4ac + 30bd

⇒ a² – 9b² + 4c² – 25d² – 4ac + 30bd

⇒ a² – 4ac + 4c² – 9b² – 25d² + 30bd

⇒ a² – 2(2ac) + (2c)² – (3b)² – (5d)² + 2(3b)(5d)

⇒ (a – 2c)² – (3b – 5d)²

[Since, from the identity III we know that, (x² – y²) = (x – y)(x + y)]

⇒ ((a – 2c) – (3b – 5d))(a – 2c – 3b + 5d)

⇒ (a – 2c – 3b – 5d)(a – 2c – 3b + 5d)