Factorise \( 25 a^{2}-16+(5 a-4)(a-2) \)

Let's simplify the expression \( 25 a^{2}-16+(5 a-4)(a-2) \) step by step. First, we start by expanding \( (5 a-4)(a-2) \): \[ (5 a-4)(a-2) = 5a^2 - 10a - 4a + 8 = 5a^2 - 14a + 8. \] Now, substitute this back into our original expression: \[ 25 a^{2} - 16 + (5 a^2 - 14a + 8). \] This simplifies to: \[ 25 a^{2} - 16 + 5 a^{2} - 14 a + 8 = (25 a^{2} + 5 a^{2}) - 14 a + (-16 + 8) = 30 a^{2} - 14 a - 8. \] Next, we want to factor \( 30 a^{2} - 14 a - 8 \). To factor this quadratic expression, we can use the method of splitting the middle term. We need two numbers that multiply to \( 30 \times -8 = -240 \) and add up to \( -14 \). These numbers are \( -24 \) and \( 10 \). Now, we rewrite the middle term: \[ 30 a^{2} - 24 a + 10 a - 8. \] Group the terms: \[ (30 a^{2} - 24 a) + (10 a - 8). \] Factor each group: \[ 6a(5a - 4) + 2(5a - 4). \] Now, factor out the common factor \( (5a - 4) \): \[ (5a - 4)(6a + 2). \] Finally, we can factor \( 6a + 2 \) further as \( 2(3a + 1) \). Thus, the final factored form is: \[ \boxed{2(5a - 4)(3a + 1)}. \]