QUESTION 4 Factorise fully by using the appropriate grouping: 4.1) \( 4 a-8 b-a x^{2}+2 b x^{2} \) 4.2) \( 3 x^{3}-3 y^{3}-a x+a y \) 4.3) \( 4 x^{2}-25 p^{2}-4 x+1 \)

To tackle these factorization problems by grouping, let’s break down each one into manageable parts. 1. **4.1)** \( 4a - 8b - ax^2 + 2bx^2 \) can be grouped as \( (4a - 8b) + (-ax^2 + 2bx^2) \). From the first group, you can factor out \( 4 \) to get \( 4(a - 2b) \), and from the second group, factor out \( -x^2 \) which gives \( -x^2(a - 2b) \). So, combined you have \( (4 - x^2)(a - 2b) \). 2. **4.2)** \( 3x^3 - 3y^3 - ax + ay \) can be rearranged into \( (3x^3 - 3y^3) + (-ax + ay) \). Factors from the first term lead to \( 3(x^3 - y^3) \) which can then be further factored to \( 3(x - y)(x^2 + xy + y^2) \). The second group gives you \( -a(x - y) \). Overall, you can factor out \( (x - y) \) to result in \( (x - y)(3(x^2 + xy + y^2) - a) \). 3. **4.3)** \( 4x^2 - 25p^2 - 4x + 1 \) can be rearranged as \( (4x^2 - 4x) + (-25p^2 + 1) \). The first grouping allows you to factor out \( 4 \) leading to \( 4(x^2 - x) \), and the second can be recognized as a difference of squares, resulting in \( (5p - 1)(5p + 1) \). Thus we can further simplify to \( 4(x - \frac{1}{2})^2 - (5p - 1)(5p + 1) \) or expressed differently depending on the context required. Feel free to let me know if you need assistance with any further steps or explanations!