Notes & theory for this topic

Partial Fractions

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1. Decompose into partial fractions (distinct linear factors).
  1. \(y = \frac{7x + 6}{6x^{2} + 15x + 9}\)
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    Factorise the denominator: \(6x^{2} + 15x + 9 = (3x + 3)(2x + 3)\). Write \(y = \frac{A}{3x + 3} + \frac{B}{2x + 3}\), so \(7x + 6 = A(2x + 3) + B(3x + 3)\). Matching coefficients gives \(A = -1,\; B = 3\).
    \(y = -\frac{1}{3x + 3} + \frac{3}{2x + 3}\)
  2. \(y = \frac{-3x - 4}{3x^{2} - 7x + 2}\)
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    Factorise the denominator: \(3x^{2} - 7x + 2 = (x - 2)(3x - 1)\). Write \(y = \frac{A}{x - 2} + \frac{B}{3x - 1}\), so \(-3x - 4 = A(3x - 1) + B(x - 2)\). Matching coefficients gives \(A = -2,\; B = 3\).
    \(y = -\frac{2}{x - 2} + \frac{3}{3x - 1}\)
2. Decompose into partial fractions (a repeated factor).
  1. \(y = \frac{-2x}{4x^{2} + 4x + 1}\)
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    The denominator is \((2x + 1)^2\). Write \(y = \frac{A}{2x + 1} + \frac{B}{(2x + 1)^2}\), so \(-2x = A(2x + 1) + B\). Matching coefficients gives \(A = -1,\; B = 1\).
    \(y = -\frac{1}{2x + 1} + \frac{1}{(2x + 1)^{2}}\)
  2. \(y = \frac{-x - 2}{x^{2} - 2x + 1}\)
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    The denominator is \((x - 1)^2\). Write \(y = \frac{A}{x - 1} + \frac{B}{(x - 1)^2}\), so \(-x - 2 = A(x - 1) + B\). Matching coefficients gives \(A = -1,\; B = -3\).
    \(y = -\frac{1}{x - 1} - \frac{3}{(x - 1)^{2}}\)
3. Decompose into partial fractions (an irreducible quadratic factor).
  1. \(y = \frac{3x^{2} + 3x - 15}{(x^{2} - 2x + 2)(2x - 3)}\)
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    The quadratic has no real roots, so write \(y = \frac{Ax+B}{x^{2} - 2x + 2} + \frac{C}{2x - 3}\). Then \(3x^{2} + 3x - 15 = (Ax+B)(2x - 3) + C(x^{2} - 2x + 2)\). Matching coefficients gives \(A = 3,\; B = 3,\; C = -3\).
    \(y = \frac{3x + 3}{x^{2} - 2x + 2} - \frac{3}{2x - 3}\)
  2. \(y = \frac{-12x^{2} - 6x - 13}{(x^{2} + 2x + 4)(2x + 1)}\)
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    The quadratic has no real roots, so write \(y = \frac{Ax+B}{x^{2} + 2x + 4} + \frac{C}{2x + 1}\). Then \(-12x^{2} - 6x - 13 = (Ax+B)(2x + 1) + C(x^{2} + 2x + 4)\). Matching coefficients gives \(A = -4,\; B = 3,\; C = -4\).
    \(y = \frac{-4x + 3}{x^{2} + 2x + 4} - \frac{4}{2x + 1}\)
4. Decompose into partial fractions (improper fraction — divide out first).
  1. \(y = \frac{6x^{2} + 25x + 30}{2x^{2} + 9x + 9}\)
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    The fraction is improper (top and bottom are both degree 2). Dividing out gives \(y = 3 + \frac{-2x + 3}{2x^{2} + 9x + 9}\). Factorise \(2x^{2} + 9x + 9 = (2x + 3)(x + 3)\) and decompose the remainder: \(A = 4,\; B = -3\).
    \(y = 3 + \frac{4}{2x + 3} - \frac{3}{x + 3}\)
  2. \(y = \frac{12x^{2} + 14x + 3}{4x^{2} + 6x + 2}\)
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    The fraction is improper (top and bottom are both degree 2). Dividing out gives \(y = 3 + \frac{-4x - 3}{4x^{2} + 6x + 2}\). Factorise \(4x^{2} + 6x + 2 = (2x + 2)(2x + 1)\) and decompose the remainder: \(A = -1,\; B = -1\).
    \(y = 3 - \frac{1}{2x + 2} - \frac{1}{2x + 1}\)
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