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

**5.1. Factorise \( 5ab+2a \)** Factor out the common factor \( a \): \[ 5ab+2a = a(5b+2). \] **5.2. Factorise \( 6x(2a-b)+5y(2a-b) \)** Both terms have the common factor \( (2a-b) \). Factor it out: \[ 6x(2a-b)+5y(2a-b) = (2a-b)(6x+5y). \] **5.3. Factorise \( 8x^{2}-2 \)** First, factor out the common factor \( 2 \): \[ 8x^{2}-2 = 2(4x^{2}-1). \] Recognize that \( 4x^2-1 \) is a difference of two squares: \[ 4x^2-1 = (2x-1)(2x+1). \] Thus, \[ 8x^{2}-2 = 2(2x-1)(2x+1). \] **5.4. Factorise \( x^{2}-x-12 \)** We look for two numbers that multiply to \(-12\) and add to \(-1\). Those numbers are \(-4\) and \(3\): \[ x^{2}-x-12 = (x-4)(x+3). \] **5.5. Factorise \( x^{2}-16+a x-4a \)** First, rearrange the terms to group like terms: \[ x^{2}+ ax - 4a - 16. \] Group the terms: \[ (x^{2}+ ax) - (4a+16). \] Factor out common factors from each group: \[ x(x+a) - 4(a+4). \] Notice that \( a+4 \) is common (since \( x+a \) can be written as \( (a+ x) \)) if we rearrange: \[ x(x+a) - 4(a+4) = (x-4)(x+a+4). \] Thus, the factorised form is: \[ x^{2}-16+a x-4a = (x-4)(x+a+4). \]