3. Simplify \[ \frac{x^{2}+x-12}{4 x-2} \div \frac{3 x^{2}+18 x+24}{2 x-1} \]

To simplify the expression \[ \frac{x^{2}+x-12}{4 x-2} \div \frac{3 x^{2}+18 x+24}{2 x-1}, \] we start by rewriting the division as multiplication by the reciprocal: \[ \frac{x^{2}+x-12}{4 x-2} \times \frac{2 x-1}{3 x^{2}+18 x+24}. \] Next, we factor each polynomial. 1. Factoring \(x^{2}+x-12\): \[ x^{2}+x-12 = (x+4)(x-3). \] 2. Factoring \(4x-2\): \[ 4x-2 = 2(2x-1). \] 3. For \(3x^{2}+18x+24\), we can factor out a 3: \[ 3x^{2}+18x+24 = 3(x^{2}+6x+8). \] Now, we factor \(x^{2}+6x+8\): \[ x^{2}+6x+8 = (x+2)(x+4). \] Therefore, \[ 3x^{2}+18x+24 = 3(x+2)(x+4). \] Now we substitute back into our expression: \[ \frac{(x+4)(x-3)}{2(2x-1)} \times \frac{2x-1}{3(x+2)(x+4)}. \] Next, we can cancel common factors: - The \(2x-1\) terms cancel. - The \(x+4\) terms also cancel. Now, we are left with: \[ \frac{(x-3)}{2 \cdot 3(x+2)} = \frac{x-3}{6(x+2)}. \] Therefore, the simplified expression is \[ \frac{x-3}{6(x+2)}. \]