1. **Problem statement:** Find the partial fraction decomposition of the following functions: (i) $ \frac{2x^3 + 7x^2 - 2x - 27}{(x - 1)(x + 4)} $ (ii) $ \frac{6x^3 + 5x^2 + 4x + 3}{(x^2 + x + 1)(x^2 - 1)} $ 2. **Formula and rules:** For partial fraction decomposition, express the rational function as a sum of simpler fractions whose denominators are factors of the original denominator. - For linear factors like $ (x - a) $, use terms like $ \frac{A}{x - a} $. - For irreducible quadratic factors like $ (x^2 + x + 1) $, use terms like $ \frac{Bx + C}{x^2 + x + 1} $. 3. **Part (i):** Denominator factors: $ (x - 1)(x + 4) $ are linear. Set: $$ \frac{2x^3 + 7x^2 - 2x - 27}{(x - 1)(x + 4)} = \frac{A}{x - 1} + \frac{B}{x + 4} $$ Multiply both sides by $ (x - 1)(x + 4) $: $$ 2x^3 + 7x^2 - 2x - 27 = A(x + 4) + B(x - 1) $$ 4. **Solve for A and B:** Expand right side: $$ A x + 4A + B x - B = (A + B) x + (4A - B) $$ Equate coefficients: - Coefficient of $ x^3 $: Left side has 2, right side 0, so this indicates improper fraction; perform polynomial division first. 5. **Polynomial division for (i):** Divide numerator by denominator: Denominator: $ (x - 1)(x + 4) = x^2 + 3x - 4 $ Divide $ 2x^3 + 7x^2 - 2x - 27 $ by $ x^2 + 3x - 4 $: - First term: $ \frac{2x^3}{x^2} = 2x $ Multiply back: $ 2x(x^2 + 3x - 4) = 2x^3 + 6x^2 - 8x $ Subtract: $$ (2x^3 + 7x^2 - 2x - 27) - (2x^3 + 6x^2 - 8x) = x^2 + 6x - 27 $$ - Next term: $ \frac{x^2}{x^2} = 1 $ Multiply back: $ 1(x^2 + 3x - 4) = x^2 + 3x - 4 $ Subtract: $$ (x^2 + 6x - 27) - (x^2 + 3x - 4) = 3x - 23 $$ 6. **Rewrite original fraction:** $$ \frac{2x^3 + 7x^2 - 2x - 27}{(x - 1)(x + 4)} = 2x + 1 + \frac{3x - 23}{(x - 1)(x + 4)} $$ Now decompose the proper fraction: $$ \frac{3x - 23}{(x - 1)(x + 4)} = \frac{A}{x - 1} + \frac{B}{x + 4} $$ Multiply both sides by denominator: $$ 3x - 23 = A(x + 4) + B(x - 1) = (A + B) x + (4A - B) $$ Equate coefficients: - Coefficient of $ x $: $ 3 = A + B $ - Constant term: $ -23 = 4A - B $ 7. **Solve system:** From $ 3 = A + B $, $ B = 3 - A $ Substitute into second: $$ -23 = 4A - (3 - A) = 4A - 3 + A = 5A - 3 $$ $$ 5A = -20 \Rightarrow A = -4 $$ Then $ B = 3 - (-4) = 7 $ 8. **Final decomposition (i):** $$ \frac{2x^3 + 7x^2 - 2x - 27}{(x - 1)(x + 4)} = 2x + 1 + \frac{-4}{x - 1} + \frac{7}{x + 4} $$ 9. **Part (ii):** Denominator: $ (x^2 + x + 1)(x^2 - 1) = (x^2 + x + 1)(x - 1)(x + 1) $ Set: $$ \frac{6x^3 + 5x^2 + 4x + 3}{(x^2 + x + 1)(x - 1)(x + 1)} = \frac{Ax + B}{x^2 + x + 1} + \frac{C}{x - 1} + \frac{D}{x + 1} $$ Multiply both sides by denominator: $$ 6x^3 + 5x^2 + 4x + 3 = (Ax + B)(x - 1)(x + 1) + C(x^2 + x + 1)(x + 1) + D(x^2 + x + 1)(x - 1) $$ 10. **Simplify terms:** Note $ (x - 1)(x + 1) = x^2 - 1 $ So: $$ (Ax + B)(x^2 - 1) + C(x^2 + x + 1)(x + 1) + D(x^2 + x + 1)(x - 1) $$ Expand: - $ (Ax + B)(x^2 - 1) = A x^3 + B x^2 - A x - B $ - $ C(x^2 + x + 1)(x + 1) = C(x^3 + 2x^2 + 2x + 1) $ - $ D(x^2 + x + 1)(x - 1) = D(x^3 - 1) $ Sum all: $$ (A + C + D) x^3 + (B + 2C) x^2 + (-A + 2C) x + (-B + C - D) $$ 11. **Equate coefficients:** From left side $ 6x^3 + 5x^2 + 4x + 3 $, equate: - $ x^3: 6 = A + C + D $ - $ x^2: 5 = B + 2C $ - $ x^1: 4 = -A + 2C $ - Constant: 3 = -B + C - D 12. **Solve system:** From $ 4 = -A + 2C $, get $ A = 2C - 4 $ Substitute into $ 6 = A + C + D $: $$ 6 = (2C - 4) + C + D = 3C + D - 4 \Rightarrow D = 10 - 3C $$ From $ 5 = B + 2C $, get $ B = 5 - 2C $ From constant term: $$ 3 = -B + C - D = -(5 - 2C) + C - (10 - 3C) = -5 + 2C + C - 10 + 3C = 6C - 15 $$ $$ 6C = 18 \Rightarrow C = 3 $$ Then: $$ A = 2(3) - 4 = 6 - 4 = 2 $$ $$ D = 10 - 3(3) = 10 - 9 = 1 $$ $$ B = 5 - 2(3) = 5 - 6 = -1 $$ 13. **Final decomposition (ii):** $$ \frac{6x^3 + 5x^2 + 4x + 3}{(x^2 + x + 1)(x^2 - 1)} = \frac{2x - 1}{x^2 + x + 1} + \frac{3}{x - 1} + \frac{1}{x + 1} $$