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

Below are one acceptable set of complete factorizations for the given expressions. ────────────────────────────── 2.1 Factorize a²x² – 3a²xy Step 1. Notice that a²x is common to both terms.   a²x² – 3a²xy = a²x(x – 3y) ────────────────────────────── 2.2 Factorize 6a² – 11ab – 10b² Step 1. Look for two numbers whose product is (6 · –10 = –60) and whose sum is –11. The numbers –15 and 4 work since (–15) + 4 = –11. Step 2. Rewrite the middle term:   6a² – 15ab + 4ab – 10b² Step 3. Factor by grouping:   = (6a² – 15ab) + (4ab – 10b²)   = 3a(2a – 5b) + 2b(2a – 5b) Step 4. Factor out the common binomial:   = (2a – 5b)(3a + 2b) ────────────────────────────── 2.3 Factorize 216p⁶ + 1 Step 1. Recognize that 216p⁶ = (6p²)³ and 1 = 1³. This is a sum of cubes. Step 2. Apply the sum of cubes formula: a³ + b³ = (a + b)(a² – ab + b²)   Let a = 6p² and b = 1, so:   216p⁶ + 1 = (6p² + 1)[(6p²)² – 6p²·1 + 1²]     = (6p² + 1)(36p⁴ – 6p² + 1) ────────────────────────────── 2.4 Factorize c² – b² – 2ab – 2ac Step 1. It is helpful to group the terms. First, rewrite the expression by grouping the c² and –b² together and factoring the –2a out of the remaining terms:   = (c² – b²) – 2a(b + c) Step 2. Recognize that c² – b² is a difference of squares:   c² – b² = (c – b)(c + b) Step 3. Now the expression is:   = (c – b)(c + b) – 2a(b + c)   Notice that (c + b) = (b + c). Factor (b + c) out:   = (b + c)[(c – b) – 2a]   = (b + c)(c – b – 2a) ────────────────────────────── 2.5 Factorize ½ – 8p¹⁶ Step 1. Factor out ½:   ½ – 8p¹⁶ = ½ (1 – 16p¹⁶) Step 2. Recognize that 1 – 16p¹⁶ is a difference of squares since 16p¹⁶ = (4p⁸)²:   1 – 16p¹⁶ = (1 – 4p⁸)(1 + 4p⁸) Step 3. Notice further that 1 – 4p⁸ is again a difference of squares because 4p⁸ = (2p⁴)²:   1 – 4p⁸ = (1 – 2p⁴)(1 + 2p⁴) Step 4. So the full factorization is:   ½ – 8p¹⁶ = ½ (1 – 2p⁴)(1 + 2p⁴)(1 + 4p⁸) ────────────────────────────── Final Answers: 2.1 a²x² – 3a²xy = a²x(x – 3y) 2.2 6a² – 11ab – 10b² = (3a + 2b)(2a – 5b) 2.3 216p⁶ + 1 = (6p² + 1)(36p⁴ – 6p² + 1) 2.4 c² – b² – 2ab – 2ac = (b + c)(c – b – 2a) 2.5 ½ – 8p¹⁶ = ½ (1 – 2p⁴)(1 + 2p⁴)(1 + 4p⁸)