1. Stating the problem: Simplify the expression $$\frac{-x^3 + 2x^2 - 3x - 6}{2x + 1}$$. 2. Formula and rules: To simplify a rational expression, we try to factor the numerator and denominator and then cancel common factors. 3. Factor the numerator: We look for roots of the cubic polynomial $$-x^3 + 2x^2 - 3x - 6$$ by testing possible rational roots using the Rational Root Theorem. 4. Test $$x = -1$$: $$-(-1)^3 + 2(-1)^2 - 3(-1) - 6 = 1 + 2 + 3 - 6 = 0$$, so $$x = -1$$ is a root. 5. Use synthetic division or polynomial division to divide the numerator by $$x + 1$$: Dividing $$-x^3 + 2x^2 - 3x - 6$$ by $$x + 1$$ gives quotient $$-x^2 + 3x - 6$$. 6. So, $$-x^3 + 2x^2 - 3x - 6 = (x + 1)(-x^2 + 3x - 6)$$. 7. Factor the quadratic $$-x^2 + 3x - 6$$ if possible. The discriminant is $$\Delta = 3^2 - 4(-1)(-6) = 9 - 24 = -15 < 0$$, so it cannot be factored further over the reals. 8. The denominator is $$2x + 1$$, which cannot be factored further. 9. Since there are no common factors between numerator and denominator, the simplified form is: $$\frac{(x + 1)(-x^2 + 3x - 6)}{2x + 1}$$. Final answer: $$\frac{(x + 1)(-x^2 + 3x - 6)}{2x + 1}$$