1. **Problem statement:** Factorize the polynomial $$x^5 - 10x^4 + 35x^3 - 50x^2 + 24x$$ over the rationals using the rational root theorem. 2. **Rational Root Theorem:** Possible rational roots are factors of the constant term divided by factors of the leading coefficient. Here, the polynomial is $$x^5 - 10x^4 + 35x^3 - 50x^2 + 24x$$. 3. **Note:** The polynomial has no constant term (constant term is 0), so $$x=0$$ is a root. 4. **Factor out $$x$$:** $$x^5 - 10x^4 + 35x^3 - 50x^2 + 24x = x(x^4 - 10x^3 + 35x^2 - 50x + 24)$$ 5. **Apply Rational Root Theorem to quartic:** Possible roots are factors of 24: $$\pm1, \pm2, \pm3, \pm4, \pm6, \pm8, \pm12, \pm24$$. 6. **Test roots:** - For $$x=1$$: $$1 - 10 + 35 - 50 + 24 = 0$$, so $$x=1$$ is a root. 7. **Divide quartic by $$(x-1)$$:** Using synthetic division: $$\begin{array}{c|ccccc} 1 & 1 & -10 & 35 & -50 & 24 \\ & & 1 & -9 & 26 & -24 \\ \hline & 1 & -9 & 26 & -24 & 0 \\ \end{array}$$ Quotient: $$x^3 - 9x^2 + 26x - 24$$ 8. **Test roots on cubic:** - For $$x=1$$: $$1 - 9 + 26 - 24 = -6 \neq 0$$ - For $$x=2$$: $$8 - 36 + 52 - 24 = 0$$, so $$x=2$$ is a root. 9. **Divide cubic by $$(x-2)$$:** $$\begin{array}{c|cccc} 2 & 1 & -9 & 26 & -24 \\ & & 2 & -14 & 24 \\ \hline & 1 & -7 & 12 & 0 \\ \end{array}$$ Quotient: $$x^2 - 7x + 12$$ 10. **Factor quadratic:** $$x^2 - 7x + 12 = (x - 3)(x - 4)$$ 11. **Complete factorization:** $$x(x - 1)(x - 2)(x - 3)(x - 4)$$ --- 12. **Partial fraction decomposition:** Decompose $$\frac{2x^3 + 3x^2 - x + 1}{(x - 1)^2 (x + 2)}$$. 13. **Set form:** $$\frac{2x^3 + 3x^2 - x + 1}{(x - 1)^2 (x + 2)} = \frac{A}{x - 1} + \frac{B}{(x - 1)^2} + \frac{C}{x + 2}$$ 14. **Multiply both sides by denominator:** $$2x^3 + 3x^2 - x + 1 = A(x - 1)(x + 2) + B(x + 2) + C(x - 1)^2$$ 15. **Expand right side:** - $$A(x - 1)(x + 2) = A(x^2 + x - 2) = A x^2 + A x - 2A$$ - $$B(x + 2) = B x + 2B$$ - $$C(x - 1)^2 = C(x^2 - 2x + 1) = C x^2 - 2C x + C$$ Sum: $$A x^2 + A x - 2A + B x + 2B + C x^2 - 2C x + C = (A + C) x^2 + (A + B - 2C) x + (-2A + 2B + C)$$ 16. **Match coefficients with left side:** Left side is $$2x^3 + 3x^2 - x + 1$$. Note: Right side has no $$x^3$$ term, so coefficient of $$x^3$$ on right must be 0. 17. **Set up system:** - Coefficient of $$x^3$$: $$2 = 0$$ (contradiction) This means the assumed form is missing a term for $$x^3$$. 18. **Since numerator degree (3) equals denominator degree (3), perform polynomial division first:** Divide numerator by denominator: $$\frac{2x^3 + 3x^2 - x + 1}{(x - 1)^2 (x + 2)}$$ 19. **Calculate denominator expanded:** $$(x - 1)^2 (x + 2) = (x^2 - 2x + 1)(x + 2) = x^3 + 2x^2 - 2x^2 - 4x + x + 2 = x^3 - x - 2x + 2 = x^3 - 3x + 2$$ 20. **Divide numerator by denominator:** $$\frac{2x^3 + 3x^2 - x + 1}{x^3 - 3x + 2}$$ Leading term division: $$\frac{2x^3}{x^3} = 2$$ Multiply denominator by 2: $$2(x^3 - 3x + 2) = 2x^3 - 6x + 4$$ Subtract: $$(2x^3 + 3x^2 - x + 1) - (2x^3 - 6x + 4) = 3x^2 + 5x - 3$$ 21. **Rewrite:** $$\frac{2x^3 + 3x^2 - x + 1}{(x - 1)^2 (x + 2)} = 2 + \frac{3x^2 + 5x - 3}{(x - 1)^2 (x + 2)}$$ 22. **Now decompose:** $$\frac{3x^2 + 5x - 3}{(x - 1)^2 (x + 2)} = \frac{A}{x - 1} + \frac{B}{(x - 1)^2} + \frac{C}{x + 2}$$ 23. **Multiply both sides by denominator:** $$3x^2 + 5x - 3 = A(x - 1)(x + 2) + B(x + 2) + C(x - 1)^2$$ 24. **Expand right side:** $$A(x^2 + x - 2) + B(x + 2) + C(x^2 - 2x + 1) = (A + C) x^2 + (A + B - 2C) x + (-2A + 2B + C)$$ 25. **Match coefficients:** - $$x^2$$: $$3 = A + C$$ - $$x$$: $$5 = A + B - 2C$$ - Constant: $$-3 = -2A + 2B + C$$ 26. **Solve system:** From first: $$C = 3 - A$$ Substitute into second: $$5 = A + B - 2(3 - A) = A + B - 6 + 2A = 3A + B - 6$$ $$B = 11 - 3A$$ Substitute into third: $$-3 = -2A + 2(11 - 3A) + (3 - A) = -2A + 22 - 6A + 3 - A = 25 - 9A$$ $$-3 - 25 = -9A$$ $$-28 = -9A$$ $$A = \frac{28}{9}$$ Then: $$C = 3 - \frac{28}{9} = \frac{27}{9} - \frac{28}{9} = -\frac{1}{9}$$ $$B = 11 - 3 \times \frac{28}{9} = 11 - \frac{84}{9} = \frac{99}{9} - \frac{84}{9} = \frac{15}{9} = \frac{5}{3}$$ 27. **Final partial fraction decomposition:** $$\frac{2x^3 + 3x^2 - x + 1}{(x - 1)^2 (x + 2)} = 2 + \frac{\frac{28}{9}}{x - 1} + \frac{\frac{5}{3}}{(x - 1)^2} + \frac{-\frac{1}{9}}{x + 2}$$ Or simplified: $$= 2 + \frac{28/9}{x - 1} + \frac{5/3}{(x - 1)^2} - \frac{1/9}{x + 2}$$