QUESTION 1 \( .1 \quad \) Factorise the following expressions fully: \( \begin{array}{l}1.1 .1 \quad x^{4}-81 \\ 112\end{array} 6 x^{2} v-10 x v+15 x-25 \)

Certainly! Let's factorise each expression step by step. --- ### **1.1. Factorise \( x^{4} - 81 \)** **Step 1:** Recognize that \( x^{4} - 81 \) is a **difference of squares**, since \( x^4 = (x^2)^2 \) and \( 81 = 9^2 \). \[ x^{4} - 81 = (x^{2})^{2} - 9^{2} = (x^{2} - 9)(x^{2} + 9) \] **Step 2:** Notice that \( x^{2} - 9 \) is also a **difference of squares**. \[ x^{2} - 9 = (x - 3)(x + 3) \] **Step 3:** Combine the factors. \[ x^{4} - 81 = (x - 3)(x + 3)(x^{2} + 9) \] **Final Factorised Form:** \[ (x - 3)(x + 3)(x^{2} + 9) \] --- ### **1.2. Factorise \( 6x^{2}v - 10xv + 15x - 25 \)** **Step 1:** Group the terms to facilitate factoring. \[ 6x^{2}v - 10xv + 15x - 25 = (6x^{2}v - 10xv) + (15x - 25) \] **Step 2:** Factor out the greatest common factor (GCF) from each group. - From the first group \( (6x^{2}v - 10xv) \), the GCF is \( 2xv \): \[ 2xv(3x - 5) \] - From the second group \( (15x - 25) \), the GCF is \( 5 \): \[ 5(3x - 5) \] **Step 3:** Now, the expression looks like: \[ 2xv(3x - 5) + 5(3x - 5) \] **Step 4:** Notice that \( (3x - 5) \) is a common factor. \[ (3x - 5)(2xv + 5) \] **Final Factorised Form:** \[ (3x - 5)(2xv + 5) \] --- **Summary of Factorisations:** 1. **\( x^4 - 81 \)** \( = (x - 3)(x + 3)(x^2 + 9) \) 2. **\( 6x^{2}v - 10xv + 15x - 25 \)** \( = (3x - 5)(2xv + 5) \)