1. **State the problem:** We want to find the partial fraction decomposition of the rational expression $$\frac{x^4 - 2x^3 + 8}{x^2(x - 2)}$$. 2. **Recall the formula and rules:** For a rational expression where the denominator factors into powers of linear terms, the partial fraction decomposition takes the form: $$\frac{P(x)}{(x)^2(x-2)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x-2}$$ where $A$, $B$, and $C$ are constants to be determined. 3. **Set up the equation:** $$x^4 - 2x^3 + 8 = A x (x - 2) + B (x - 2) + C x^2$$ Multiply both sides by the denominator $x^2(x-2)$ to clear fractions. 4. **Expand the right side:** $$A x (x - 2) = A (x^2 - 2x)$$ $$B (x - 2) = B x - 2B$$ $$C x^2 = C x^2$$ So, $$A (x^2 - 2x) + B x - 2B + C x^2 = (A + C) x^2 + (B - 2A) x - 2B$$ 5. **Equate coefficients with the left side:** The left side is $$x^4 - 2x^3 + 8$$, which has terms of degree 4 and 3, but the right side only has up to degree 2. This means the original expression is an improper fraction and must be simplified by polynomial division first. 6. **Perform polynomial division:** Divide $$x^4 - 2x^3 + 8$$ by $$x^2 (x - 2) = x^3 - 2x^2$$. - First term: $$x^4 / x^3 = x$$ - Multiply divisor by $$x$$: $$x(x^3 - 2x^2) = x^4 - 2x^3$$ - Subtract: $$(x^4 - 2x^3 + 8) - (x^4 - 2x^3) = 8$$ Remainder is 8, so: $$\frac{x^4 - 2x^3 + 8}{x^3 - 2x^2} = x + \frac{8}{x^3 - 2x^2} = x + \frac{8}{x^2 (x - 2)}$$ 7. **Now decompose the proper fraction:** $$\frac{8}{x^2 (x - 2)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x - 2}$$ 8. **Multiply both sides by denominator:** $$8 = A x (x - 2) + B (x - 2) + C x^2$$ 9. **Expand right side:** $$A (x^2 - 2x) + B x - 2B + C x^2 = (A + C) x^2 + (B - 2A) x - 2B$$ 10. **Equate coefficients:** Since left side is constant 8, coefficients of $x^2$ and $x$ must be zero: - Coefficient of $x^2$: $$A + C = 0$$ - Coefficient of $x$: $$B - 2A = 0$$ - Constant term: $$-2B = 8$$ 11. **Solve the system:** From constant term: $$-2B = 8 \implies B = -4$$ From $x$ term: $$B - 2A = 0 \implies -4 - 2A = 0 \implies -2A = 4 \implies A = -2$$ From $x^2$ term: $$A + C = 0 \implies -2 + C = 0 \implies C = 2$$ 12. **Write the final decomposition:** $$\frac{8}{x^2 (x - 2)} = \frac{-2}{x} + \frac{-4}{x^2} + \frac{2}{x - 2}$$ 13. **Include the polynomial part from division:** $$\frac{x^4 - 2x^3 + 8}{x^2 (x - 2)} = x - \frac{2}{x} - \frac{4}{x^2} + \frac{2}{x - 2}$$