1. **State the problem:** We want to express the rational function $$\frac{10 - 17x + 14x^2}{(2 + x)(1 - 2x)^2}$$ in partial fractions of the form $$\frac{A}{2 + x} + \frac{B}{1 - 2x} + \frac{C}{(1 - 2x)^2}$$. 2. **Set up the equation:** $$\frac{10 - 17x + 14x^2}{(2 + x)(1 - 2x)^2} = \frac{A}{2 + x} + \frac{B}{1 - 2x} + \frac{C}{(1 - 2x)^2}$$ Multiply both sides by the denominator $(2 + x)(1 - 2x)^2$ to clear fractions: $$10 - 17x + 14x^2 = A(1 - 2x)^2 + B(2 + x)(1 - 2x) + C(2 + x)$$ 3. **Expand terms:** - Expand $(1 - 2x)^2 = 1 - 4x + 4x^2$ - Expand $B(2 + x)(1 - 2x)$: $$ (2 + x)(1 - 2x) = 2 - 4x + x - 2x^2 = 2 - 3x - 2x^2 $$ So, $$10 - 17x + 14x^2 = A(1 - 4x + 4x^2) + B(2 - 3x - 2x^2) + C(2 + x)$$ 4. **Distribute constants:** $$10 - 17x + 14x^2 = A - 4Ax + 4Ax^2 + 2B - 3Bx - 2Bx^2 + 2C + Cx$$ 5. **Group like terms:** Constant terms: $$A + 2B + 2C$$ $x$ terms: $$-4A x - 3B x + C x = (-4A - 3B + C) x$$ $x^2$ terms: $$4A x^2 - 2B x^2 = (4A - 2B) x^2$$ 6. **Equate coefficients:** From the original polynomial: - Constant: 10 - Coefficient of $x$: -17 - Coefficient of $x^2$: 14 Set up system: $$\begin{cases} A + 2B + 2C = 10 \\ -4A - 3B + C = -17 \\ 4A - 2B = 14 \end{cases}$$ 7. **Solve the system:** From the third equation: $$4A - 2B = 14 \implies 2A - B = 7 \implies B = 2A - 7$$ Substitute $B$ into first and second equations: - First: $$A + 2(2A - 7) + 2C = 10 \implies A + 4A - 14 + 2C = 10 \implies 5A + 2C = 24$$ - Second: $$-4A - 3(2A - 7) + C = -17 \implies -4A - 6A + 21 + C = -17 \implies -10A + C = -38$$ From second: $$C = -38 + 10A$$ Substitute $C$ into first: $$5A + 2(-38 + 10A) = 24 \implies 5A - 76 + 20A = 24 \implies 25A = 100 \implies A = 4$$ Then, $$B = 2(4) - 7 = 8 - 7 = 1$$ $$C = -38 + 10(4) = -38 + 40 = 2$$ 8. **Final answer:** $$\frac{10 - 17x + 14x^2}{(2 + x)(1 - 2x)^2} = \frac{4}{2 + x} + \frac{1}{1 - 2x} + \frac{2}{(1 - 2x)^2}$$