Multiply and simplify: \( \frac{x^{2}-3 x-4}{x^{2}+2 x-8} \div \frac{x+1}{x+4} \)

To solve the expression \( \frac{x^{2}-3 x-4}{x^{2}+2 x-8} \div \frac{x+1}{x+4} \), we can start by rewriting the division as multiplication by the reciprocal: \[ \frac{x^{2}-3 x-4}{x^{2}+2 x-8} \times \frac{x+4}{x+1} \] Now, we need to factor the polynomials in the numerator and denominator. **1. Factor \( x^2 - 3x - 4 \):** To factor \( x^2 - 3x - 4 \), we look for two numbers that multiply to -4 and add up to -3. These numbers are -4 and 1. Thus, we can factor it as: \[ x^2 - 3x - 4 = (x - 4)(x + 1) \] **2. Factor \( x^2 + 2x - 8 \):** Next, we will factor \( x^2 + 2x - 8 \). We need two numbers that multiply to -8 and add up to 2. These numbers are 4 and -2. So, we can factor it as: \[ x^2 + 2x - 8 = (x + 4)(x - 2) \] Now we can substitute these factorizations back into our expression: \[ \frac{(x - 4)(x + 1)}{(x + 4)(x - 2)} \times \frac{x + 4}{x + 1} \] **3. Simplify the expression:** When we multiply, we combine the fractions: \[ \frac{(x - 4)(x + 1)(x + 4)}{(x + 4)(x - 2)(x + 1)} \] Now, we can cancel the common factors \( (x + 1) \) in the numerator and denominator, as well as \( (x + 4) \): \[ \frac{(x - 4)}{(x - 2)} \] Thus, the simplified expression is: \[ \frac{x - 4}{x - 2} \] So the final answer is: \[ \boxed{\frac{x - 4}{x - 2}} \]