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Mathematics · Ch 2 — Basic Algebra

Partial Fractions

2.7.2

Partial Fractions

A rational expression f(x)g(x)\dfrac{f(x)}{g(x)} is a proper fraction if deg⁡f<deg⁡g\deg f<\deg g; every proper fraction whose denominator factors into linear and irreducible-quadratic pieces can be written uniquely as a sum of simpler pieces -- its partial fraction decomposition:

  1. for each linear factor (x−a)(x-a) appearing to power kk in g(x)g(x): A1x−a+A2(x−a)2+⋯+Ak(x−a)k\dfrac{A_1}{x-a}+\dfrac{A_2}{(x-a)^2}+\cdots+\dfrac{A_k}{(x-a)^k};
  2. for each irreducible quadratic factor (x2+ax+b)(x^2+ax+b) (no real zeros) appearing to power kk: B1x+C1x2+ax+b+B2x+C2(x2+ax+b)2+⋯+Bkx+Ck(x2+ax+b)k\dfrac{B_1x+C_1}{x^2+ax+b}+\dfrac{B_2x+C_2}{(x^2+ax+b)^2}+\cdots+\dfrac{B_kx+C_k}{(x^2+ax+b)^k}. Finding the constants. After multiplying through by g(x)g(x), the unknown constants can be found either by substituting the roots of each linear factor directly (the cover-up shortcut -- e.g. for x(x+3)(x−4)=Ax+3+Bx−4\dfrac x{(x+3)(x-4)}=\dfrac A{x+3}+\dfrac B{x-4}, setting x=4x=4 isolates BB and x=−3x=-3 isolates AA instantly), or -- needed whenever an irreducible quadratic factor or a repeated factor is present -- by expanding and comparing coefficients of matching powers of xx on both sides, often combined with a convenient extra substitution such as x=0x=0. …