Partial Fractions 87E089

1. We are asked to express each given rational function as partial fractions. 2. The general approach for partial fractions is to decompose a rational function \( \frac{P(x)}{Q(x)} \) where \(Q(x)\) factors into linear or irreducible quadratic factors. 3. For linear factors \((x-a)^n\), the partial fractions take the form \( \frac{A_1}{x-a} + \frac{A_2}{(x-a)^2} + \cdots + \frac{A_n}{(x-a)^n} \). 4. For irreducible quadratic factors \((x^2+bx+c)^m\), the partial fractions take the form \( \frac{B_1x + C_1}{x^2+bx+c} + \cdots + \frac{B_mx + C_m}{(x^2+bx+c)^m} \). --- **Problem 11 (a):** \( \frac{2x+3}{(x+1)(2x+1)(x+3)} \) 5. Set up partial fractions: $$ \frac{2x+3}{(x+1)(2x+1)(x+3)} = \frac{A}{x+1} + \frac{B}{2x+1} + \frac{C}{x+3} $$ 6. Multiply both sides by denominator: $$ 2x+3 = A(2x+1)(x+3) + B(x+1)(x+3) + C(x+1)(2x+1) $$ 7. Expand and collect terms, then equate coefficients to solve for \(A,B,C\). 8. Substituting convenient values: - For \(x=-1\): Left \(2(-1)+3=1\), Right \(A(2(-1)+1)(-1+3)=A(-1)(2)=-2A\) so \(1=-2A\) \(\Rightarrow A=-\frac{1}{2}\). - For \(x=-\frac{1}{2}\): Left \(2(-\frac{1}{2})+3=2\), Right \(B(-\frac{1}{2}+1)(-\frac{1}{2}+3)=B(\frac{1}{2})(\frac{5}{2})=\frac{5}{4}B\) so \(2=\frac{5}{4}B\) \(\Rightarrow B=\frac{8}{5}\). - For \(x=-3\): Left \(2(-3)+3=-3\), Right \(C(-3+1)(2(-3)+1)=C(-2)(-5)=10C\) so \(-3=10C\) \(\Rightarrow C=-\frac{3}{10}\). 9. Final decomposition: $$ \frac{2x+3}{(x+1)(2x+1)(x+3)} = -\frac{1}{2(x+1)} + \frac{8}{5(2x+1)} - \frac{3}{10(x+3)} $$ --- **Problem 11 (b):** \( \frac{2}{(x-1)^2(x+1)} \) 10. Partial fractions form: $$ \frac{2}{(x-1)^2(x+1)} = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{x+1} $$ 11. Multiply both sides by denominator: $$ 2 = A(x-1)(x+1) + B(x+1) + C(x-1)^2 $$ 12. Substitute values: - \(x=1\): Left \(2\), Right \(B(1+1)=2B\) so \(2=2B\) \(\Rightarrow B=1\). - \(x=-1\): Left \(2\), Right \(C(-1-1)^2 = C(-2)^2=4C\) so \(2=4C\) \(\Rightarrow C=\frac{1}{2}\). 13. To find \(A\), expand and equate coefficients: $$ 2 = A(x^2 -1) + B(x+1) + C(x^2 - 2x +1) $$ $$ 2 = A x^2 - A + B x + B + C x^2 - 2 C x + C $$ Group terms: $$ 2 = (A + C) x^2 + (B - 2C) x + (-A + B + C) $$ Since left side is constant 2, coefficients of \(x^2\) and \(x\) must be zero: - \(A + C = 0 \Rightarrow A = -C = -\frac{1}{2}\) - \(B - 2C = 0 \Rightarrow 1 - 2 \times \frac{1}{2} = 0\) (checks out) - Constant term: \(-A + B + C = 2\) Substitute \(A = -\frac{1}{2}, B=1, C=\frac{1}{2}\): $$ -(-\frac{1}{2}) + 1 + \frac{1}{2} = \frac{1}{2} + 1 + \frac{1}{2} = 2 $$ 14. Final decomposition: $$ \frac{2}{(x-1)^2(x+1)} = \frac{-\frac{1}{2}}{x-1} + \frac{1}{(x-1)^2} + \frac{\frac{1}{2}}{x+1} $$ --- **Problem 11 (c):** \( \frac{x^2 + 4x - 7}{(x^2 + 4)(x+1)} \) 15. Partial fractions form: $$ \frac{x^2 + 4x - 7}{(x^2 + 4)(x+1)} = \frac{Ax + B}{x^2 + 4} + \frac{C}{x+1} $$ 16. Multiply both sides by denominator: $$ x^2 + 4x - 7 = (Ax + B)(x+1) + C(x^2 + 4) $$ 17. Expand right side: $$ (Ax + B)(x+1) = A x^2 + A x + B x + B = A x^2 + (A + B) x + B $$ So right side: $$ A x^2 + (A + B) x + B + C x^2 + 4 C = (A + C) x^2 + (A + B) x + (B + 4 C) $$ 18. Equate coefficients: - Coefficient of \(x^2\): \(1 = A + C\) - Coefficient of \(x\): \(4 = A + B\) - Constant term: \(-7 = B + 4 C\) 19. Solve system: From first: \(C = 1 - A\) From second: \(B = 4 - A\) Substitute into third: $$ 4 - A + 4(1 - A) = -7 $$ $$ 4 - A + 4 - 4 A = -7 $$ $$ 8 - 5 A = -7 $$ $$ -5 A = -15 \Rightarrow A = 3 $$ Then: $$ C = 1 - 3 = -2 $$ $$ B = 4 - 3 = 1 $$ 20. Final decomposition: $$ \frac{x^2 + 4x - 7}{(x^2 + 4)(x+1)} = \frac{3x + 1}{x^2 + 4} - \frac{2}{x+1} $$ --- **Problem 11 (d):** \( \frac{x^3 - x^2 - 4}{x^2 - 1} \) 21. Note denominator factors as \( (x-1)(x+1) \). 22. Since numerator degree (3) \(\geq\) denominator degree (2), perform polynomial division first: Divide \(x^3 - x^2 - 4\) by \(x^2 - 1\): - Leading term: \(x^3 / x^2 = x\) - Multiply divisor by \(x\): \(x^3 - x\) - Subtract: \((x^3 - x^2 - 4) - (x^3 - x) = -x^2 + x - 4\) Next term: - Leading term: \(-x^2 / x^2 = -1\) - Multiply divisor by \(-1\): \(-x^2 + 1\) - Subtract: \((-x^2 + x - 4) - (-x^2 + 1) = x - 5\) 23. So: $$ \frac{x^3 - x^2 - 4}{x^2 - 1} = x - 1 + \frac{x - 5}{x^2 - 1} $$ 24. Decompose remainder: $$ \frac{x - 5}{(x-1)(x+1)} = \frac{A}{x-1} + \frac{B}{x+1} $$ 25. Multiply both sides: $$ x - 5 = A(x+1) + B(x-1) = (A + B) x + (A - B) $$ 26. Equate coefficients: - Coefficient of \(x\): \(1 = A + B\) - Constant term: \(-5 = A - B\) 27. Solve: Add equations: $$ 1 + (-5) = (A + B) + (A - B) = 2A \Rightarrow -4 = 2A \Rightarrow A = -2 $$ Then: $$ 1 = -2 + B \Rightarrow B = 3 $$ 28. Final decomposition: $$ \frac{x^3 - x^2 - 4}{x^2 - 1} = x - 1 - \frac{2}{x-1} + \frac{3}{x+1} $$ --- **Problem 11 (e):** \( \frac{2x^3 + 2x^2 + 2}{(x^2 + 1)(x+1)^2} \) 29. Partial fractions form: $$ \frac{2x^3 + 2x^2 + 2}{(x^2 + 1)(x+1)^2} = \frac{Ax + B}{x^2 + 1} + \frac{C}{x+1} + \frac{D}{(x+1)^2} $$ 30. Multiply both sides by denominator: $$ 2x^3 + 2x^2 + 2 = (Ax + B)(x+1)^2 + C(x^2 + 1)(x+1) + D(x^2 + 1) $$ 31. Expand \((x+1)^2 = x^2 + 2x + 1\). 32. Expand \((Ax + B)(x^2 + 2x + 1) = A x^3 + 2 A x^2 + A x + B x^2 + 2 B x + B\) 33. Expand \(C(x^2 + 1)(x+1) = C(x^3 + x^2 + x + 1) = C x^3 + C x^2 + C x + C\) 34. Add \(D(x^2 + 1) = D x^2 + D\) 35. Sum all terms: $$ (A + C) x^3 + (2A + B + C + D) x^2 + (A + 2B + C) x + (B + C + D) $$ 36. Equate coefficients with left side \(2x^3 + 2x^2 + 0 x + 2\): - \(x^3\): \(2 = A + C\) - \(x^2\): \(2 = 2A + B + C + D\) - \(x\): \(0 = A + 2B + C\) - Constant: \(2 = B + C + D\) 37. Solve system: From first: \(C = 2 - A\) Substitute into third: $$ 0 = A + 2B + (2 - A) = 2B + 2 \Rightarrow 2B = -2 \Rightarrow B = -1 $$ Substitute \(B\) and \(C\) into second: $$ 2 = 2A -1 + (2 - A) + D = (2A -1 + 2 - A) + D = (A + 1) + D \Rightarrow D = 2 - A - 1 = 1 - A $$ Substitute \(B, C, D\) into constant term: $$ 2 = -1 + (2 - A) + (1 - A) = 2 - 2A $$ $$ 2A = 2 - 2 = 0 \Rightarrow A = 0 $$ Then: $$ C = 2 - 0 = 2 $$ $$ D = 1 - 0 = 1 $$ 38. Final decomposition: $$ \frac{2x^3 + 2x^2 + 2}{(x^2 + 1)(x+1)^2} = \frac{0 \cdot x - 1}{x^2 + 1} + \frac{2}{x+1} + \frac{1}{(x+1)^2} = \frac{-1}{x^2 + 1} + \frac{2}{x+1} + \frac{1}{(x+1)^2} $$ --- **Problem 11 (f):** \( \frac{2x^2 - 9x - 35}{(x+1)(x-2)(x+3)} \) 39. Partial fractions form: $$ \frac{2x^2 - 9x - 35}{(x+1)(x-2)(x+3)} = \frac{A}{x+1} + \frac{B}{x-2} + \frac{C}{x+3} $$ 40. Multiply both sides: $$ 2x^2 - 9x - 35 = A(x-2)(x+3) + B(x+1)(x+3) + C(x+1)(x-2) $$ 41. Substitute convenient values: - \(x = -1\): Left \(2(1) + 9 - 35 = 2 - 9 - 35 = -42\), Right \(A(-3)(2) = -6A\) so \(-42 = -6A \Rightarrow A = 7\). - \(x = 2\): Left \(2(4) - 18 - 35 = 8 - 18 - 35 = -45\), Right \(B(3)(5) = 15B\) so \(-45 = 15B \Rightarrow B = -3\). - \(x = -3\): Left \(2(9) + 27 - 35 = 18 + 27 - 35 = 10\), Right \(C(-2)(-5) = 10C\) so \(10 = 10C \Rightarrow C = 1\). 42. Final decomposition: $$ \frac{2x^2 - 9x - 35}{(x+1)(x-2)(x+3)} = \frac{7}{x+1} - \frac{3}{x-2} + \frac{1}{x+3} $$ --- **Problem 12:** \( \frac{4x - 1}{x^2 (x^2 - 4)} \) 43. Factor denominator: \(x^2 (x-2)(x+2)\). 44. Partial fractions form: $$ \frac{4x - 1}{x^2 (x-2)(x+2)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x-2} + \frac{D}{x+2} $$ 45. Multiply both sides by denominator: $$ 4x - 1 = A x (x-2)(x+2) + B (x-2)(x+2) + C x^2 (x+2) + D x^2 (x-2) $$ 46. Expand \((x-2)(x+2) = x^2 - 4\). 47. So: $$ 4x - 1 = A x (x^2 - 4) + B (x^2 - 4) + C x^2 (x+2) + D x^2 (x-2) $$ 48. Expand terms: - \(A x^3 - 4 A x\) - \(B x^2 - 4 B\) - \(C x^3 + 2 C x^2\) - \(D x^3 - 2 D x^2\) 49. Sum: $$ (A + C + D) x^3 + (B + 2 C - 2 D) x^2 + (-4 A) x + (-4 B) $$ 50. Equate coefficients with left side \(4x - 1 = 0 x^3 + 0 x^2 + 4 x - 1\): - \(x^3\): \(0 = A + C + D\) - \(x^2\): \(0 = B + 2 C - 2 D\) - \(x\): \(4 = -4 A\) \(\Rightarrow A = -1\) - Constant: \(-1 = -4 B\) \(\Rightarrow B = \frac{1}{4}\) 51. Substitute \(A = -1\) into first: $$ 0 = -1 + C + D \Rightarrow C + D = 1 $$ 52. Substitute \(B = \frac{1}{4}\) into second: $$ 0 = \frac{1}{4} + 2 C - 2 D \Rightarrow 2 C - 2 D = -\frac{1}{4} \Rightarrow C - D = -\frac{1}{8} $$ 53. Solve system: $$ C + D = 1 $$ $$ C - D = -\frac{1}{8} $$ Add: $$ 2 C = 1 - \frac{1}{8} = \frac{7}{8} \Rightarrow C = \frac{7}{16} $$ Then: $$ D = 1 - C = 1 - \frac{7}{16} = \frac{9}{16} $$ 54. Final decomposition: $$ \frac{4x - 1}{x^2 (x^2 - 4)} = \frac{-1}{x} + \frac{1/4}{x^2} + \frac{7/16}{x-2} + \frac{9/16}{x+2} $$ --- **Problem 13:** \( \frac{x^2 + 1}{x^2 - 3x + 2} \) 55. Factor denominator: \( (x-1)(x-2) \). 56. Partial fractions form: $$ \frac{x^2 + 1}{(x-1)(x-2)} = \frac{A}{x-1} + \frac{B}{x-2} $$ 57. Multiply both sides: $$ x^2 + 1 = A(x-2) + B(x-1) = (A + B) x - 2 A - B $$ 58. Equate coefficients: - \(x^2\) term: Left has 1, right has 0, so polynomial division needed. 59. Since numerator degree (2) \(\geq\) denominator degree (2), divide: Divide \(x^2 + 1\) by \(x^2 - 3x + 2\): - Leading term: 1 - Multiply divisor by 1: \(x^2 - 3x + 2\) - Subtract: \((x^2 + 1) - (x^2 - 3x + 2) = 3x - 1\) 60. So: $$ \frac{x^2 + 1}{x^2 - 3x + 2} = 1 + \frac{3x - 1}{(x-1)(x-2)} $$ 61. Decompose remainder: $$ \frac{3x - 1}{(x-1)(x-2)} = \frac{A}{x-1} + \frac{B}{x-2} $$ 62. Multiply both sides: $$ 3x - 1 = A(x-2) + B(x-1) = (A + B) x - 2 A - B $$ 63. Equate coefficients: - \(x\): \(3 = A + B\) - Constant: \(-1 = -2 A - B\) 64. Solve: From first: \(B = 3 - A\) Substitute into second: $$ -1 = -2 A - (3 - A) = -2 A - 3 + A = -A - 3 $$ $$ -A = 2 \Rightarrow A = -2 $$ Then: $$ B = 3 - (-2) = 5 $$ 65. Final decomposition: $$ \frac{x^2 + 1}{x^2 - 3x + 2} = 1 - \frac{2}{x-1} + \frac{5}{x-2} $$ --- **Problem 14:** \( \frac{x^3 - 2x^2 - 4x - 4}{x^2 + x - 2} \) 66. Factor denominator: \( (x+2)(x-1) \). 67. Degree numerator (3) \(>\) denominator (2), perform division: Divide \(x^3 - 2x^2 - 4x - 4\) by \(x^2 + x - 2\): - Leading term: \(x^3 / x^2 = x\) - Multiply divisor by \(x\): \(x^3 + x^2 - 2x\) - Subtract: \((x^3 - 2x^2 - 4x - 4) - (x^3 + x^2 - 2x) = -3x^2 - 2x - 4\) Next term: - Leading term: \(-3x^2 / x^2 = -3\) - Multiply divisor by \(-3\): \(-3x^2 - 3x + 6\) - Subtract: \((-3x^2 - 2x - 4) - (-3x^2 - 3x + 6) = x - 10\) 68. So: $$ \frac{x^3 - 2x^2 - 4x - 4}{x^2 + x - 2} = x - 3 + \frac{x - 10}{x^2 + x - 2} $$ 69. Decompose remainder: $$ \frac{x - 10}{(x+2)(x-1)} = \frac{A}{x+2} + \frac{B}{x-1} $$ 70. Multiply both sides: $$ x - 10 = A(x-1) + B(x+2) = (A + B) x + (-A + 2 B) $$ 71. Equate coefficients: - \(x\): \(1 = A + B\) - Constant: \(-10 = -A + 2 B\) 72. Solve: From first: \(B = 1 - A\) Substitute into second: $$ -10 = -A + 2(1 - A) = -A + 2 - 2 A = 2 - 3 A $$ $$ -12 = -3 A \Rightarrow A = 4 $$ Then: $$ B = 1 - 4 = -3 $$ 73. Final decomposition: $$ \frac{x^3 - 2x^2 - 4x - 4}{x^2 + x - 2} = x - 3 + \frac{4}{x+2} - \frac{3}{x-1} $$ --- **Problem 15 (a):** \( \frac{12}{x^2 - 9} \) 74. Factor denominator: \( (x-3)(x+3) \). 75. Partial fractions: $$ \frac{12}{(x-3)(x+3)} = \frac{A}{x-3} + \frac{B}{x+3} $$ 76. Multiply both sides: $$ 12 = A(x+3) + B(x-3) = (A + B) x + (3 A - 3 B) $$ 77. Equate coefficients: - \(x\): \(0 = A + B\) - Constant: \(12 = 3 A - 3 B\) 78. From first: \(B = -A\) Substitute into second: $$ 12 = 3 A - 3 (-A) = 3 A + 3 A = 6 A \Rightarrow A = 2 $$ Then: $$ B = -2 $$ 79. Final decomposition: $$ \frac{12}{x^2 - 9} = \frac{2}{x-3} - \frac{2}{x+3} $$ --- **Problem 15 (b):** \( \frac{4(4 - x)}{x^2 - 2x - 3} \) 80. Factor denominator: \( (x-3)(x+1) \). 81. Rewrite numerator: $$ 4(4 - x) = 16 - 4x $$ 82. Partial fractions: $$ \frac{16 - 4x}{(x-3)(x+1)} = \frac{A}{x-3} + \frac{B}{x+1} $$ 83. Multiply both sides: $$ 16 - 4x = A(x+1) + B(x-3) = (A + B) x + (A - 3 B) $$ 84. Equate coefficients: - \(x\): \(-4 = A + B\) - Constant: \(16 = A - 3 B\) 85. Solve: From first: \(B = -4 - A\) Substitute into second: $$ 16 = A - 3(-4 - A) = A + 12 + 3 A = 4 A + 12 $$ $$ 4 A = 4 \Rightarrow A = 1 $$ Then: $$ B = -4 - 1 = -5 $$ 86. Final decomposition: $$ \frac{4(4 - x)}{x^2 - 2x - 3} = \frac{1}{x-3} - \frac{5}{x+1} $$ --- **Problem 15 (c):** \( \frac{x^2 - 3x + 6}{x(x-1)(x-2)} \) 87. Partial fractions: $$ \frac{x^2 - 3x + 6}{x(x-1)(x-2)} = \frac{A}{x} + \frac{B}{x-1} + \frac{C}{x-2} $$ 88. Multiply both sides: $$ x^2 - 3x + 6 = A(x-1)(x-2) + B x (x-2) + C x (x-1) $$ 89. Expand: - \(A(x^2 - 3x + 2) = A x^2 - 3 A x + 2 A\) - \(B(x^2 - 2x) = B x^2 - 2 B x\) - \(C(x^2 - x) = C x^2 - C x\) 90. Sum: $$ (A + B + C) x^2 + (-3 A - 2 B - C) x + 2 A $$ 91. Equate coefficients: - \(x^2\): \(1 = A + B + C\) - \(x\): \(-3 = -3 A - 2 B - C\) - Constant: \(6 = 2 A\) \(\Rightarrow A = 3\) 92. Substitute \(A=3\) into first two: - \(1 = 3 + B + C \Rightarrow B + C = -2\) - \(-3 = -9 - 2 B - C \Rightarrow -3 + 9 = -2 B - C \Rightarrow 6 = -2 B - C\) 93. From first: \(C = -2 - B\) Substitute into second: $$ 6 = -2 B - (-2 - B) = -2 B + 2 + B = -B + 2 $$ $$ -B = 4 \Rightarrow B = -4 $$ Then: $$ C = -2 - (-4) = 2 $$ 94. Final decomposition: $$ \frac{x^2 - 3x + 6}{x(x-1)(x-2)} = \frac{3}{x} - \frac{4}{x-1} + \frac{2}{x-2} $$