Divide and simplify. \[ \frac{2 x^{2}+15 x-38}{4 x^{2}+11 x-3} \div \frac{3 x^{2}-4 x-4}{3 x^{2}+11 x+6} \]

To solve the expression \[ \frac{2 x^{2}+15 x-38}{4 x^{2}+11 x-3} \div \frac{3 x^{2}-4 x-4}{3 x^{2}+11 x+6}, \] we first rewrite the division as a multiplication by the reciprocal: \[ \frac{2 x^{2}+15 x-38}{4 x^{2}+11 x-3} \times \frac{3 x^{2}+11 x+6}{3 x^{2}-4 x-4}. \] Next, we will factor each polynomial. 1. **Factoring \(2x^{2}+15x-38\)**: - We need two numbers that multiply to \(2 \times -38 = -76\) and add to \(15\). These numbers are \(19\) and \(-4\). - The expression can be factored into \(2x^2 + 19x - 4x - 38 = (2x - 4)(x + 19)\). - Thus, \(2x^2 + 15x - 38 = (2x - 4)(x + 19) = 2(x - 2)(x + 19)\). 2. **Factoring \(4x^{2}+11x-3\)**: - We need numbers that multiply to \(4 \times -3 = -12\) and add up to \(11\). The numbers \(12\) and \(-1\) work. - Factoring gives us \(4x^2 + 12x - x - 3 = (4x - 1)(x + 3)\). 3. **Factoring \(3x^{2}+11x+6\)**: - We need numbers that multiply to \(3 \times 6 = 18\) and add to \(11\). The numbers \(9\) and \(2\) work. - Thus, \(3x^2 + 9x + 2x + 6 = (3x + 2)(x + 3)\). 4. **Factoring \(3x^{2}-4x-4\)**: - We need numbers that multiply to \(3 \times -4 = -12\) and add to \(-4\). The numbers \(-6\) and \(2\) work. - We can factor as \(3x^2 - 6x + 2x - 4 = (3x + 2)(x - 2)\). Now substituting back the factored forms into our expression, we have: \[ \frac{2(x - 2)(x + 19)}{(4x - 1)(x + 3)} \times \frac{(3x + 2)(x + 3)}{(3x + 2)(x - 2)}. \] Now we can cancel common factors: - The \((x - 2)\) from the numerator and denominator. - The \((3x + 2)\) from the numerator and denominator. - The \((x + 3)\) from the numerator and denominator. After canceling, we are left with: \[ \frac{2}{4x - 1}. \] Thus, the simplified expression is: \[ \frac{2}{4x - 1}. \]