Partial Fraction Decomposition — Mathematics Knowledge System

Overview

Partial fraction decomposition reverses the process of combining fractions over a common denominator: it breaks a single complicated rational expression into a sum of simpler fractions, each with a lower-degree denominator. This is the key algebraic step that makes rational functions tractable for integration in calculus and for inverse Laplace/Z-transforms in engineering.

Intuition

If you know 1/(x-1) + 2/(x+1) combine into a single fraction (3x+1)/(x²-1), partial fractions runs that process backward: given the combined fraction, recover the simpler pieces it came from. The technique works by matching coefficients — since a rational expression's simplified form is unique, forcing the decomposition to reproduce the original numerator pins down every unknown constant.

Formal Definition

Definition

For a proper rational function p(x)/q(x) (deg p < deg q) with q(x) factored into linear and irreducible quadratic factors, the decomposition assigns one fraction per factor (or repeated factor):

\frac{p(x)}{(x-a)(x-b)} = \frac{A}{x-a} + \frac{B}{x-b}

Distinct linear factors

\frac{p(x)}{(x-a)^2} = \frac{A}{x-a} + \frac{B}{(x-a)^2}

Repeated linear factor

\frac{p(x)}{x^2+bx+c} = \frac{Ax+B}{x^2+bx+c}

Irreducible quadratic factor

Properties

Properness requirement

\text{If } \deg p \geq \deg q, \text{ first divide via polynomial long division before decomposing}

Uniqueness

\text{The decomposition of a given proper rational function is unique}

Applications

Inverse Laplace transforms, essential for solving linear differential equations in control systems, require decomposing a transfer function into partial fractions with known inverse-transform forms.

Worked Examples

Set up the decomposition with unknowns A, B and clear denominators.
$\frac{3x+1}{(x-1)(x+1)} = \frac{A}{x-1}+\frac{B}{x+1} \implies 3x+1 = A(x+1)+B(x-1)$
Substitute x=1 to isolate A (this kills the B term).
$3(1)+1 = A(2) \implies A = 2$
Substitute x=-1 to isolate B (this kills the A term).
$3(-1)+1 = B(-2) \implies -2 = -2B \implies B=1$

Answer: (3x+1)/((x-1)(x+1)) = 2/(x-1) + 1/(x+1)

Practice Problems

Difficulty 6/10

Decompose (5x-4)/((x-2)(x+3)) into partial fractions.

Difficulty 7/10

Decompose (2x+3)/(x²(x+1)) into partial fractions.

Common Mistakes

Common Mistake

Applying partial fraction decomposition directly to an improper fraction (numerator degree ≥ denominator degree).

You must first perform polynomial long division to write the expression as a polynomial plus a proper remainder fraction; only the proper fractional part gets decomposed into partial fractions.

Common Mistake

Using only one unknown constant for a repeated linear factor like (x-a)², e.g. writing A/(x-a)² alone.

A repeated factor (x-a)² requires TWO terms: A/(x-a) + B/(x-a)². Omitting the lower-power term A/(x-a) makes the decomposition unable to match the original numerator in general.

Summary

Partial fraction decomposition splits a rational function into a sum of simpler fractions, one per factor of the denominator.
Distinct linear factors get constant numerators; repeated linear factors need one term per power; irreducible quadratics need a linear numerator (Ax+B).
The expression must be proper (numerator degree < denominator degree) before decomposing — divide first if not.
Substituting the roots of each linear factor is the fastest way to solve for unknown constants (the 'cover-up' method).

References

BookSullivan, M. Algebra and Trigonometry, 10th ed. Ch. 11 — Systems of Equations and Inequalities.
WebsiteWikipedia — Partial fraction decomposition