1. **State the problem:** Decompose the rational expression $$\frac{x^2 + x + 1}{(x-1)(x+2)}$$ into partial fractions. 2. **Formula and rules:** For partial fraction decomposition, if the denominator factors into linear terms like $(x-1)$ and $(x+2)$, we express the fraction as: $$\frac{x^2 + x + 1}{(x-1)(x+2)} = \frac{A}{x-1} + \frac{B}{x+2}$$ where $A$ and $B$ are constants to be found. 3. **Multiply both sides by the denominator:** $$x^2 + x + 1 = A(x+2) + B(x-1)$$ 4. **Expand the right side:** $$x^2 + x + 1 = A x + 2A + B x - B$$ 5. **Group like terms:** $$x^2 + x + 1 = (A + B) x + (2A - B)$$ 6. **Equate coefficients:** - Coefficient of $x^2$: Left side is 1, right side has none, so $1 = 0$ which is impossible. This means the numerator degree is not less than denominator degree, so first perform polynomial division. 7. **Polynomial division:** Divide $x^2 + x + 1$ by $(x-1)(x+2) = x^2 + x - 2$. Since degrees are equal, divide: $$\frac{x^2 + x + 1}{x^2 + x - 2} = 1 + \frac{3}{x^2 + x - 2}$$ 8. **Now decompose the remainder:** $$\frac{3}{(x-1)(x+2)} = \frac{A}{x-1} + \frac{B}{x+2}$$ Multiply both sides by denominator: $$3 = A(x+2) + B(x-1)$$ Expand: $$3 = A x + 2A + B x - B = (A + B) x + (2A - B)$$ Equate coefficients: - Coefficient of $x$: $0 = A + B$ - Constant term: $3 = 2A - B$ From $0 = A + B$, we get $B = -A$. Substitute into constant term: $$3 = 2A - (-A) = 3A \implies A = 1$$ Then $B = -1$. 9. **Final decomposition:** $$\frac{x^2 + x + 1}{(x-1)(x+2)} = 1 + \frac{1}{x-1} - \frac{1}{x+2}$$