1.2 Factorise fully: 1.2.1 \( 6 x^{2}-7 x-20 \) 1.2 \( \quad x^{2}+x-2 x y-2 y \)

We need to factorise each expression completely. ──────────────────────── 1.2.1 Factorise 6x² – 7x – 20 Step 1. Multiply the coefficient of x² (which is 6) by the constant term (–20) to get –120. Step 2. Look for two numbers that multiply to –120 and add up to the coefficient of x (–7). Notice that –15 and 8 work because:   –15 × 8 = –120 and –15 + 8 = –7 Step 3. Rewrite the middle term –7x as –15x + 8x:   6x² – 15x + 8x – 20 Step 4. Group the terms:   (6x² – 15x) + (8x – 20) Step 5. Factor each group:   • In the first group, factor out 3x: 3x(2x – 5)   • In the second group, factor out 4: 4(2x – 5) Step 6. Now, notice that (2x – 5) is a common factor:   3x(2x – 5) + 4(2x – 5) = (2x – 5)(3x + 4) Thus, 6x² – 7x – 20 factors as (3x + 4)(2x – 5). ──────────────────────── 1.2.2 Factorise x² + x – 2xy – 2y Step 1. Group the terms:   (x² + x) + (–2xy – 2y) Step 2. Factor each group:   • From the first group, factor out x: x(x + 1)   • From the second group, factor out –2y: –2y(x + 1) Step 3. Now, (x + 1) is common:   x(x + 1) – 2y(x + 1) = (x + 1)(x – 2y) Thus, x² + x – 2xy – 2y factors as (x + 1)(x – 2y). ──────────────────────── Final Answers: 1.2.1 6x² – 7x – 20 = (3x + 4)(2x – 5) 1.2.2 x² + x – 2xy – 2y = (x + 1)(x – 2y)