12. \( 60 x^{2}+4 x-8 \)

We want to factor the expression \[ 60x^2 + 4x - 8. \] **Step 1. Factor out the greatest common factor.** All the coefficients are even, so we can factor out \(2\): \[ 60x^2 + 4x - 8 = 2(30x^2 + 2x - 4). \] **Step 2. Factor the quadratic inside the parentheses.** We focus on \[ 30x^2 + 2x - 4. \] Our goal is to factor this quadratic into the form \[ ax^2 + bx + c = (mx + n)(px + q), \] where \( m \times p = 30 \) and the constant product \( n \times q = -4 \). We begin by finding the roots using the quadratic formula. The quadratic formula for \( Ax^2 + Bx + C = 0 \) is \[ x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}. \] For our quadratic, \( A = 30\), \( B = 2\), and \( C = -4 \). First compute the discriminant: \[ \Delta = B^2 - 4AC = 2^2 - 4(30)(-4) = 4 + 480 = 484. \] Since \( \sqrt{484} = 22 \), the roots are \[ x = \frac{-2 \pm 22}{60}. \] Calculate each root: \[ x = \frac{-2 + 22}{60} = \frac{20}{60} = \frac{1}{3}, \] \[ x = \frac{-2 - 22}{60} = \frac{-24}{60} = -\frac{2}{5}. \] Thus, the quadratic factors as \[ 30x^2 + 2x - 4 = 30\left(x-\frac{1}{3}\right)\left(x+\frac{2}{5}\right). \] **Step 3. Rewrite the factors with integer coefficients.** We can eliminate the fractions by rewriting each factor. Multiply the first factor by 3 and the second factor by 5: \[ x - \frac{1}{3} = \frac{1}{3}(3x - 1), \qquad x + \frac{2}{5} = \frac{1}{5}(5x+2). \] Thus, \[ \left(x-\frac{1}{3}\right)\left(x+\frac{2}{5}\right) = \frac{1}{15}(3x-1)(5x+2). \] Substitute back into the factorization: \[ 30x^2+2x-4 = 30 \cdot \frac{1}{15} (3x-1)(5x+2) = 2(3x-1)(5x+2). \] **Step 4. Substitute back into the original expression.** Recall that \[ 60x^2 + 4x - 8 = 2(30x^2+2x-4). \] Substitute the factorization we just found: \[ 60x^2 + 4x - 8 = 2\Bigl[2(3x-1)(5x+2)\Bigr]=4(3x-1)(5x+2). \] Thus, the fully factored form is: \[ \boxed{4(3x-1)(5x+2)}. \]