4. \( \left(x^{2}-2 x-8\right)(x+1) \)

To solve the expression \( (x^{2}-2x-8)(x+1) \), we first need to factor the quadratic part, \( x^{2}-2x-8 \). This can be factored into \( (x-4)(x+2) \), since \( -4 \) and \( +2 \) are the numbers that multiply to \( -8 \) and add to \( -2 \). Thus, we can rewrite the expression as \( (x-4)(x+2)(x+1) \). Next, we can expand this product if needed. Distributing \( (x-4) \) and \( (x+2) \) will involve multiplying: 1. \( (x-4)(x+2) = x^{2} + 2x - 4x - 8 = x^{2} - 2x - 8 \). 2. Now multiply this by \( (x+1) \): \( (x^{2}-2x-8)(x+1) = x^{3} + x^{2} - 2x^{2} - 2x - 8x - 8 = x^{3} - x^{2} - 10x - 8 \). So the final expansion of \( (x^{2}-2x-8)(x+1) \) is \( x^{3} - x^{2} - 10x - 8 \).