Partial Fractions Done Right

Introduction

The purpose of these notes is to present the best methods for calculating partial fractions expansions by hand. The intended audience is high school and college students.

The brute-force approach

A sure-fire solution to any partial fractions expansion problem is the brute-force approach where one writes out the intended expansion in terms of undetermined coefficients, then clears it of fractions and equates like powers of the independent variable. Even for small problems this leads to large amounts of computations. For example, the expansion \[ \frac{1}{(x^2+1)(x^2+x+1)^2} = \frac{Ax+B}{x^2+1} + \frac{Cx+D}{x^2+x+1} + \frac{Ex+F}{(x^2+x+1)^2} \] results in the identity \[ 1 = (Ax+B)(x^2+x+1)^2 + (Cx+D)(x^2+1)(x^2+x+1)+(Ex+F)(x^2+1), \] and calls for solving the linear system: \[ \begin{pmatrix} 1 & 0 & 1 & 0 & 0 & 0 \\ 2 & 1 & 1 & 1 & 0 & 0 \\ 3 & 2 & 2 & 1 & 1 & 0 \\ 2 & 3 & 1 & 2 & 0 & 1 \\ 1 & 2 & 1 & 1 & 1 & 0 \\ 0 & 1 & 0 & 1 & 0 & 1 \end{pmatrix} \begin{pmatrix} A \\ B \\ C \\ D \\ E \\ F \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 1 \end{pmatrix}. \] Probably this is the worst way to solve this problem. The partial partial fractions expansion method that I advocate in these notes, in effect decouples the unknowns, allowing the computation of the coefficients $A$ through $F$ more or less independently of each other and very quickly.

Partial partial fractions

By partial partial fractions expansions (no, the doubled ‘partial’ is not a typo) I mean expansions of the form:

The case of a simple real root:

\( \ds \frac{p(x)}{(x-a)\phi(x)} = \frac{A}{x-a} + \frac{q(x)}{\phi(x)} \)

The case of a simple complex root ($a^2 - 4b < 0$):

\( \ds\frac{p(x)}{(x^2+ax+b)\phi(x)} =\frac{Ax + B}{x^2+ax+b} + \frac{q(x)}{\phi(x)} \)

The case of repeated real roots $(r>1)$:

\( \ds\frac{p(x)}{(x-a)^r\phi(x)} = \frac{A_r}{(x-a)^r} + \frac{A_{r-1}}{(x-a)^{r-1}} + \cdots + \frac{A_1}{x-a} + \frac{q(x)}{\phi(x)} \)

The case of repeated complex roots ($r>1$ and $a^2 - 4b < 0$):

\( \ds\frac{p(x)}{(x^2+ax+b)^r\phi(x)} = \frac{A_r x + B_r}{(x^2+ax+b)^r} + \frac{A_{r-1}x + B_{r-1}}{(x^2+ax+b)^{r-1}} + \cdots + \frac{A_1x + B_1}{x^2+ax+b} + \frac{q(x)}{\phi(x)} \)

I call these forms partial partial fractions because they correspond to performing the partial fractions expansion only partially. Each focuses on those components of the (full) partial fractions expansion that arise from only one of the denominator's roots; the rest of the terms are lumped in the ‘remainder’ term $q(x)/\phi(x)$.

The bulk of these notes is devoted to describing efficient methods for constructing partial partial expansions.

From partial partial fractions to (full) partial fractions

To compute the traditional (full) partial fractions expansion of a given fraction, we compute the partial partial fractions corresponding to each of the denominator's roots. The sum of the resulting fractions is the desired partial fractions expansion.

For example, let us look at the problem of computing partial fractions expansion of the fraction \[ \frac{4x^2+2x+1}{(x-1)(x-2)^2(x^2+x+1)}. \]

In these notes we will learn that the partial partial fractions expansions corresponding to each of the denominator's roots is easy to find: \begin{align*} \frac{4x^2+2x+1}{\bbox[background-color: yellow, 1pt]{(x-1)} \; (x-2)^2 \; (x^2+x+1)} &\hskip0.7em=\hskip0.7em \color{blue}{\frac{7}{3(x-1)}} + \cdots, \\ ~ \\ \frac{4x^2+2x+1}{(x-1) \; \bbox[background-color: yellow, 1pt]{(x-2)^2} \; (x^2+x+1)} &\hskip0.7em=\hskip0.7em \color{blue}{\frac{3}{(x-2)^2}} - \color{blue}{\frac{18}{7(x-2)}} + \cdots, \\ ~ \\ \frac{4x^2+2x+1}{(x-1) \; (x-2)^2 \; \bbox[background-color: yellow, 1pt]{(x^2+x+1)}} &\hskip0.7em=\hskip0.7em \color{blue}{\frac{5x+1}{21(x^2+x+1)}} + \cdots. \end{align*} The ‘$\cdots$’ stand for the remainder terms that are shown as $q(x)/\phi(x)$ in the general forms near the top of this page. They are not relevant to our goal.

The (full) partial fractions expansion is the sum of the fractions computed in partial partial fractions expansions: \[ \bbox[border: 1px solid blue, #eeeeff, 20px]{ \frac{4x^2+2x+1}{(x-1)(x-2)^2(x^2+x+1)} = \frac{7}{3(x-1)} + \frac{3}{(x-2)^2} - \frac{18}{7(x-2)} + \frac{5x+1}{21(x^2+x+1)}. } \]

Note that the terms in the expansion are obtained independently of each other. There is no system of equations to solve!

This web page was created in June 2008