Partial Fractions – II

Partial Fractions – II

Rule – I: Let \(\frac{f(x)}{g(x)}\) be a proper fraction. To each non-repeated factor (ax +b) of g(x) there will be a partial fraction of the form \(\frac{A}{ax+b}\)where A is a zero-real number, to be determined.

Rule – II: Let \(\frac{f(x)}{g(x)}\) be a proper fraction. To each factor  (ax + b)ⁿ, a ≠ 0 where n is a positive integer, of g(x) there will be partial fractions of the form \(\frac{{{A}_{1}}}{ax+b}+\frac{{{A}_{2}}}{{{(ax+b)}^{2}}}+…..+\frac{{{A}_{n}}}{{{(ax+b)}^{n}}}\) where A₁, A₂ … A n are to be determined constants. Note that An ≠ 0 and Rule – I is a particular case of Rule – II foe n = 1.

Rule – III: Let \(\frac{f(x)}{g(x)}\) be a proper fraction. To each non-repeated quadratic factor (ax² + bx + c)ⁿ, a ≠ 0  of g(x) there will be a partial fraction of the form \(\frac{Ax+B}{a{{x}^{2}}+bx+c}\) where A, B are real numbers, to be determined.

Rule – IV: Let \(\frac{f(x)}{g(x)}\) be a proper fraction. is n(>1) ϵ N the largest exponent so (ax² + bx + c)ⁿ, a ≠ 0  is a factor of g(x), then corresponding to each such factor, there will be partial fractions of the form \(\frac{{{A}_{1}}x+{{B}_{1}}}{a{{x}^{2}}+bx+c}+\frac{{{A}_{2}}x+{{B}_{2}}}{{{(a{{x}^{2}}+bx+c)}^{2}}}+…+\frac{{{A}_{n}}x+{{B}_{n}}}{{{(a{{x}^{2}}+bx+c)}^{n}}}\). We note that one of An and Bn is different from zero.

Examples: \(\frac{5x-4}{{{x}^{2}}-x-2}\)

Step 1: Factor the bottom \(\frac{5x-4}{{{x}^{2}}-x-2}=\frac{5x-4}{(x-2)(x+1)}\)

Step 2: Write one partial fraction for each of those factors \(\frac{5x-4}{{{x}^{2}}-x-2}=\frac{A}{(x-2)}+\frac{B}{(x+1)}\)

Step 3: Multiply through by the bottom so we no longer have fractions 5x – 4 = A (x + 1) + B (x – 2)

Step 4: Now find the constants A and B

Substituting the roots, or “zeros”, of (x – 2) (x + 1) can help:

Root for (x + 1) is x = -1

5 (-1) – 4 = A (-1 + 1) + B (-1 – 2)

-9 = 0 + B (-3)

B = 3

Root for (x – 2) is x = 2

5(2) – 4 = A (2 + 1) + B (2 – 2)

6 = A (2 + 1) + B (0)

A = 2

\(\frac{5x-4}{(x-2)(x+1)}=\frac{2}{x-2}+\frac{3}{x+1}\).