Factorise the following algebraic expressions fully: \( \begin{array}{ll}\text { 5.1. } & 5 a b+2 a \\ \text { 5.2. } & 6 x(2 a-b)+5 y(2 a-b) \\ \text { 5.3. } & 8 x^{2}-2 \\ \text { 5.4. } & x^{2}-x-12 \\ \text { 5.5. } & x^{2}-16+a x-4 a\end{array} \)

Let’s factorise those expressions step-by-step! **5.1.** \( 5ab + 2a \) We can factor out the common term \( a \): \[ a(5b + 2) \] **5.2.** \( 6x(2a - b) + 5y(2a - b) \) Here, \( (2a - b) \) is the common factor: \[ (2a - b)(6x + 5y) \] **5.3.** \( 8x^{2} - 2 \) We factor out the common factor \( 2 \): \[ 2(4x^{2} - 1) \] Now, \( 4x^{2} - 1 \) is a difference of squares: \[ 2(2x - 1)(2x + 1) \] **5.4.** \( x^{2} - x - 12 \) We look for two numbers that multiply to \(-12\) and add to \(-1\): The numbers are \(-4\) and \(3\): \[ (x - 4)(x + 3) \] **5.5.** \( x^{2} - 16 + ax - 4a \) Rearranging it gives us \( x^{2} + ax - 20 \): Now, we group it: We want two numbers that multiply to \(-20\) and add to \(a\): Factoring gives us: \[ (x + 5)(x - 4) + a(x - 4) \] That can be factored as: \[ (x - 4)(x + 5 + a) \] And there you have it! Each expression is now fully factorised!