1. **Express the following as partial fractions:** $$\frac{9x^2 + 20x - 10}{(x + 2)(3x - 1)}$$ 2. **Determine the partial fraction decomposition:** $$\frac{4x^2 - 22x + 7}{(2x + 3)(x - 2)^2}$$ 3. **Solve the equation:** $$3 \coth^2 x - 8 \cosech x + 1 = 0$$ --- ### Step 1: Partial fraction decomposition of $$\frac{9x^2 + 20x - 10}{(x + 2)(3x - 1)}$$ 1. The denominator factors are linear and distinct: $(x+2)$ and $(3x-1)$. 2. Set up the decomposition: $$\frac{9x^2 + 20x - 10}{(x + 2)(3x - 1)} = \frac{A}{x + 2} + \frac{B}{3x - 1}$$ 3. Multiply both sides by the denominator: $$9x^2 + 20x - 10 = A(3x - 1) + B(x + 2)$$ 4. Expand the right side: $$9x^2 + 20x - 10 = 3Ax - A + Bx + 2B = (3A + B)x + (-A + 2B)$$ 5. Equate coefficients: - Coefficient of $x^2$: Left side has $9$, right side has $0$ (no $x^2$ term) $ o$ contradiction. This means the numerator degree is equal to denominator degree, so first perform polynomial division. 6. Divide numerator by denominator: Denominator expanded: $(x+2)(3x-1) = 3x^2 + 5x - 2$ Divide $9x^2 + 20x - 10$ by $3x^2 + 5x - 2$: - Leading term division: $9x^2 / 3x^2 = 3$ - Multiply denominator by 3: $3(3x^2 + 5x - 2) = 9x^2 + 15x - 6$ - Subtract: $(9x^2 + 20x - 10) - (9x^2 + 15x - 6) = 5x - 4$ 7. So: $$\frac{9x^2 + 20x - 10}{(x + 2)(3x - 1)} = 3 + \frac{5x - 4}{(x + 2)(3x - 1)}$$ 8. Now decompose the proper fraction: $$\frac{5x - 4}{(x + 2)(3x - 1)} = \frac{A}{x + 2} + \frac{B}{3x - 1}$$ 9. Multiply both sides by denominator: $$5x - 4 = A(3x - 1) + B(x + 2) = 3Ax - A + Bx + 2B = (3A + B)x + (-A + 2B)$$ 10. Equate coefficients: - For $x$: $5 = 3A + B$ - Constant: $-4 = -A + 2B$ 11. Solve system: From first: $B = 5 - 3A$ Substitute into second: $-4 = -A + 2(5 - 3A) = -A + 10 - 6A = 10 - 7A$ $-4 - 10 = -7A o -14 = -7A o A = 2$ Then $B = 5 - 3(2) = 5 - 6 = -1$ 12. Final decomposition: $$\frac{9x^2 + 20x - 10}{(x + 2)(3x - 1)} = 3 + \frac{2}{x + 2} - \frac{1}{3x - 1}$$ --- ### Step 2: Partial fraction decomposition of $$\frac{4x^2 - 22x + 7}{(2x + 3)(x - 2)^2}$$ 1. The denominator has a linear factor $(2x + 3)$ and a repeated linear factor $(x - 2)^2$. 2. Set up decomposition: $$\frac{4x^2 - 22x + 7}{(2x + 3)(x - 2)^2} = \frac{A}{2x + 3} + \frac{B}{x - 2} + \frac{C}{(x - 2)^2}$$ 3. Multiply both sides by denominator: $$4x^2 - 22x + 7 = A(x - 2)^2 + B(2x + 3)(x - 2) + C(2x + 3)$$ 4. Expand terms: - $(x - 2)^2 = x^2 - 4x + 4$ - $(2x + 3)(x - 2) = 2x^2 - 4x + 3x - 6 = 2x^2 - x - 6$ 5. Substitute: $$4x^2 - 22x + 7 = A(x^2 - 4x + 4) + B(2x^2 - x - 6) + C(2x + 3)$$ 6. Expand right side: $$= A x^2 - 4A x + 4A + 2B x^2 - B x - 6B + 2C x + 3C$$ Group like terms: - $x^2$: $A + 2B$ - $x$: $-4A - B + 2C$ - Constant: $4A - 6B + 3C$ 7. Equate coefficients: - $x^2$: $4 = A + 2B$ - $x$: $-22 = -4A - B + 2C$ - Constant: $7 = 4A - 6B + 3C$ 8. Solve system: From first: $A = 4 - 2B$ Substitute into second: $-22 = -4(4 - 2B) - B + 2C = -16 + 8B - B + 2C = -16 + 7B + 2C$ $-22 + 16 = 7B + 2C o -6 = 7B + 2C$ Substitute $A$ into third: $7 = 4(4 - 2B) - 6B + 3C = 16 - 8B - 6B + 3C = 16 - 14B + 3C$ $7 - 16 = -14B + 3C o -9 = -14B + 3C$ 9. System now: $$\begin{cases} -6 = 7B + 2C \\ -9 = -14B + 3C \end{cases}$$ Multiply first by 3: $-18 = 21B + 6C$ Multiply second by 2: $-18 = -28B + 6C$ Subtract second from first: $0 = 49B o B = 0$ 10. Substitute $B=0$ into $-6 = 7B + 2C$: $-6 = 0 + 2C o C = -3$ 11. Substitute $B=0$ into $A = 4 - 2B$: $A = 4 - 0 = 4$ 12. Final decomposition: $$\frac{4x^2 - 22x + 7}{(2x + 3)(x - 2)^2} = \frac{4}{2x + 3} + \frac{0}{x - 2} + \frac{-3}{(x - 2)^2} = \frac{4}{2x + 3} - \frac{3}{(x - 2)^2}$$ --- ### Step 3: Solve $$3 \coth^2 x - 8 \cosech x + 1 = 0$$ 1. Recall identities: $$\coth^2 x - \cosech^2 x = 1$$ 2. Express $\coth^2 x$ as $1 + \cosech^2 x$: $$3(1 + \cosech^2 x) - 8 \cosech x + 1 = 0$$ 3. Expand: $$3 + 3 \cosech^2 x - 8 \cosech x + 1 = 0$$ $$3 \cosech^2 x - 8 \cosech x + 4 = 0$$ 4. Let $y = \cosech x$, then: $$3 y^2 - 8 y + 4 = 0$$ 5. Solve quadratic: $$y = \frac{8 \pm \sqrt{(-8)^2 - 4 \times 3 \times 4}}{2 \times 3} = \frac{8 \pm \sqrt{64 - 48}}{6} = \frac{8 \pm 4}{6}$$ 6. Two solutions: - $y = \frac{8 + 4}{6} = 2$ - $y = \frac{8 - 4}{6} = \frac{2}{3}$ 7. Recall $y = \cosech x = \frac{1}{\sinh x}$, so: $$\sinh x = \frac{1}{y}$$ 8. For $y=2$: $$\sinh x = \frac{1}{2}$$ $$x = \sinh^{-1} \left( \frac{1}{2} \right) = \ln \left( \frac{1}{2} + \sqrt{\left( \frac{1}{2} \right)^2 + 1} \right) = \ln \left( \frac{1}{2} + \sqrt{\frac{1}{4} + 1} \right) = \ln \left( \frac{1}{2} + \frac{\sqrt{5}}{2} \right) = \ln \left( \frac{1 + \sqrt{5}}{2} \right)$$ 9. For $y=\frac{2}{3}$: $$\sinh x = \frac{3}{2}$$ $$x = \sinh^{-1} \left( \frac{3}{2} \right) = \ln \left( \frac{3}{2} + \sqrt{\left( \frac{3}{2} \right)^2 + 1} \right) = \ln \left( \frac{3}{2} + \sqrt{\frac{9}{4} + 1} \right) = \ln \left( \frac{3}{2} + \frac{\sqrt{13}}{2} \right) = \ln \left( \frac{3 + \sqrt{13}}{2} \right)$$ --- ### **Final answers:** 1. $$\frac{9x^2 + 20x - 10}{(x + 2)(3x - 1)} = 3 + \frac{2}{x + 2} - \frac{1}{3x - 1}$$ 2. $$\frac{4x^2 - 22x + 7}{(2x + 3)(x - 2)^2} = \frac{4}{2x + 3} - \frac{3}{(x - 2)^2}$$ 3. $$x = \ln \left( \frac{1 + \sqrt{5}}{2} \right) \quad \text{or} \quad x = \ln \left( \frac{3 + \sqrt{13}}{2} \right)$$