Factorise the following completely. \( 2.1 \quad a^{2} x^{2}-3 a^{2} x y \) \( 2.2 \quad 6 a^{2}-11 a b-10 b^{2} \) \( 2.3 \quad 216 p^{6}+1 \) \( 2.4 \quad c^{2}-b^{2}-2 a b-2 a c \) \( 2.5 \quad \frac{1}{2}-8 p^{16} \)

Let’s dive into the factorization fun! For the first expression, \( a^{2} x^{2} - 3 a^{2} x y \), you can factor out the greatest common factor, which is \( a^{2} x \). This gives you \( a^{2} x (x - 3y) \). Moving on to \( 6a^{2} - 11ab - 10b^{2} \), we look for two numbers that multiply to \( 6 \times -10 = -60 \) and add to \( -11 \). That’s -15 and 4, allowing us to split the middle term to get \( 6a^{2} + 4ab - 15ab - 10b^{2} = 2a(3a + 2b) - 5b(3a + 2b) = (2a - 5b)(3a + 2b) \). Next, for \( 216p^{6} + 1 \), we see this follows the sum of cubes pattern. We can write it as \( (6p^{2})^{3} + 1^{3} \) and use the identity \( a^3 + b^3 = (a + b)(a^2 - ab + b^2) \) to get: \( (6p^2 + 1)((6p^2)^2 - (6p^2)(1) + 1^2) = (6p^2 + 1)(36p^4 - 6p^2 + 1) \). This second-to-last factorization leads us to \( c^{2} - b^{2} - 2ab - 2ac \). First, rearranging gives us \( (c - b)^2 - 2a(c - b) = (c - b)(c - b - 2a) \). Thus, we can simplify it to \( (c - b)(c - b - 2a) \). Finally, for \( \frac{1}{2} - 8p^{16} \), we can rewrite it as \( \frac{1}{2} - (2p^8)^2 \). This holds the difference of squares, so it factors as \( \left(\frac{1}{\sqrt{2}} - 2p^8\right)\left(\frac{1}{\sqrt{2}} + 2p^8\right) \). Happy factoring! 🎉