Partial Fractions 43C011

1. **State the problem:** Decompose the rational expressions into partial fractions using three methods and verify the same decomposition is obtained. 2. **Problem (a):** Decompose \( \frac{2x+1}{x^2+3x-10} \). 3. **Factor the denominator:** \( x^2+3x-10 = (x+5)(x-2) \). 4. **Set up partial fractions:** \( \frac{2x+1}{(x+5)(x-2)} = \frac{A}{x+5} + \frac{B}{x-2} \). 5. **Multiply both sides by denominator:** $$ 2x+1 = A(x-2) + B(x+5) $$ --- ### Method (i): Equate coefficients 6. Expand right side: $$ 2x+1 = A x - 2A + B x + 5B = (A+B)x + (-2A + 5B) $$ 7. Equate coefficients of \(x\) and constants: $$ 2 = A + B $$ $$ 1 = -2A + 5B $$ 8. Solve system: From first: \( B = 2 - A \) Substitute into second: $$ 1 = -2A + 5(2 - A) = -2A + 10 - 5A = 10 - 7A $$ $$ 7A = 9 \Rightarrow A = \frac{9}{7} $$ $$ B = 2 - \frac{9}{7} = \frac{14}{7} - \frac{9}{7} = \frac{5}{7} $$ --- ### Method (ii): Substitute values of \(x\) 9. Substitute \( x = 2 \): $$ 2(2) + 1 = A(2-2) + B(2+5) \Rightarrow 5 = 0 + 7B \Rightarrow B = \frac{5}{7} $$ 10. Substitute \( x = -5 \): $$ 2(-5) + 1 = A(-5-2) + B(-5+5) \Rightarrow -9 = -7A + 0 \Rightarrow A = \frac{9}{7} $$ --- ### Method (iii): Using roots and formula 11. Roots are \( \alpha_1 = -5 \), \( \alpha_2 = 2 \). 12. For \( A \): $$ A = \frac{g(\alpha_1)}{h'(\alpha_1)} $$ where \( g(x) = 2x+1 \), \( h(x) = (x+5)(x-2) \), $$ h'(x) = (x+5)'(x-2) + (x+5)(x-2)' = 1 \cdot (x-2) + (x+5) \cdot 1 = 2x + 3 $$ 13. Evaluate: $$ g(-5) = 2(-5) + 1 = -9 $$ $$ h'(-5) = 2(-5) + 3 = -10 + 3 = -7 $$ $$ A = \frac{-9}{-7} = \frac{9}{7} $$ 14. For \( B \): $$ g(2) = 2(2) + 1 = 5 $$ $$ h'(2) = 2(2) + 3 = 4 + 3 = 7 $$ $$ B = \frac{5}{7} $$ --- 15. **Final decomposition for (a):** $$ \frac{2x+1}{x^2+3x-10} = \frac{9/7}{x+5} + \frac{5/7}{x-2} $$ --- 16. **Problem (b):** Decompose \( \frac{4}{x^2 - 3x} \). 17. Factor denominator: $$ x^2 - 3x = x(x-3) $$ 18. Set up partial fractions: $$ \frac{4}{x(x-3)} = \frac{C}{x} + \frac{D}{x-3} $$ 19. Multiply both sides by denominator: $$ 4 = C(x-3) + D x = C x - 3C + D x = (C + D) x - 3C $$ --- ### Method (i): Equate coefficients 20. Equate coefficients: $$ 0 = C + D $$ $$ 4 = -3C $$ 21. Solve: $$ C = -\frac{4}{3} $$ $$ D = -C = \frac{4}{3} $$ --- ### Method (ii): Substitute values 22. Substitute \( x = 0 \): $$ 4 = C(0-3) + D(0) = -3C \Rightarrow C = -\frac{4}{3} $$ 23. Substitute \( x = 3 \): $$ 4 = C(3-3) + D(3) = 3D \Rightarrow D = \frac{4}{3} $$ --- ### Method (iii): Using roots and formula 24. Roots: \( \alpha_1 = 0 \), \( \alpha_2 = 3 \). 25. Derivative of denominator: $$ h(x) = x(x-3) \Rightarrow h'(x) = (x)'(x-3) + x(x-3)' = 1 \cdot (x-3) + x \cdot 1 = 2x - 3 $$ 26. Evaluate for \( C \): $$ g(0) = 4 $$ $$ h'(0) = 2(0) - 3 = -3 $$ $$ C = \frac{4}{-3} = -\frac{4}{3} $$ 27. Evaluate for \( D \): $$ g(3) = 4 $$ $$ h'(3) = 2(3) - 3 = 6 - 3 = 3 $$ $$ D = \frac{4}{3} $$ --- 28. **Final decomposition for (b):** $$ \frac{4}{x^2 - 3x} = \frac{-4/3}{x} + \frac{4/3}{x-3} $$ --- **Summary:** All three methods yield the same partial fraction decompositions for both problems.