Partial Fractions - Decomposition Method, Examples

TL;DR

Partial fractions break a single complicated fraction like ( \frac{4x + 12}{x^2 + 4x} ) into a sum of simpler ones, ( \frac{3}{x} + \frac{1}{x + 4} ), the exact reverse of adding fractions over a common denominator. This article covers the cases (distinct linear, repeated linear, irreducible quadratic), the decomposition steps, and the mistakes to avoid.

What Are Partial Fractions?

Partial fractions are the simpler fractions you get when you split a single rational expression — a fraction whose numerator and denominator are both polynomials — into a sum of fractions with smaller denominators. The process of doing this is called partial fraction decomposition.

Formally, a rational expression ( \frac{P(x)}{Q(x)} ) is rewritten as a sum of fractions, each having one factor of ( Q(x) ) as its denominator. The method depends entirely on how ( Q(x) ) factors, which is why factorization is the first step every time.

One condition matters before you start: the expression must be proper, meaning the degree of the numerator is less than the degree of the denominator. If it is improper (numerator degree ≥ denominator degree), do polynomial long division first to split off the whole-number part, then decompose what remains.

What Are the Cases for Partial Fraction Decomposition?

The form of the decomposition is set by the type of factor in the denominator. There are three you need at this level, plus the repeated-quadratic extension.

Denominator factor	Decomposition form
Distinct linear, ( (ax+b) )	( \frac{A}{ax + b} )
Repeated linear, ( (ax+b)^n )	( \frac{A_1}{ax + b} + \frac{A_2}{(ax + b)^2} + \cdots + \frac{A_n}{(ax + b)^n} )
Irreducible quadratic, ( (ax^2 + bx + c) )	( \frac{Ax + B}{ax^2 + bx + c} )

Two rules drive the whole table. A linear factor gets a constant numerator ( A ). A quadratic factor (one that does not factor further over the reals) gets a linear numerator ( Ax+B ). And a repeated factor raised to the power ( n ) needs one term for every power from 1 up to ( n ), not just the highest.

How Do You Decompose Into Partial Fractions?

The procedure is the same every time, once you know which case you are in:

Factor the denominator completely and confirm the expression is proper (do long division first if it is not).
Write the decomposition with unknown constants (A, B, …) using the case table above.
Clear the denominators by multiplying both sides by the original denominator.
Solve for the constants — either by substituting smart values of x (each chosen to zero out a factor) or by comparing coefficients of like powers.
Write the final sum with the constants filled in.

Step 4 has two routes, and choosing well saves time. Substituting strategic x-values (the cover-up idea) is fastest when the denominator has distinct linear factors. Comparing coefficients is the reliable fallback for repeated or quadratic factors. You can mix them.

Examples of Partial Fractions

Example 1

Decompose ( \frac{4x + 12}{x^2 + 4x} ) into partial fractions.

Factor the denominator: ( x^2 + 4x = x(x + 4) ), two distinct linear factors. Write the form:

[ \frac{4x + 12}{x(x + 4)} = \frac{A}{x} + \frac{B}{x + 4} ]\n Clear denominators by multiplying through by ( x(x+4) ):

[ 4x + 12 = A(x + 4) + Bx ]\n Substitute ( x=0 ): ( 12 = A(4) \Rightarrow A=3 ). Substitute ( x=-4 ): ( -4 = -4B \Rightarrow B=1 ).

[ \frac{4x + 12}{x^2 + 4x} = \frac{3}{x} + \frac{1}{x + 4} ]

Final answer: ( \frac{3}{x} + \frac{1}{x + 4} ).

Example 2

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

The decomposition uses unknown constants, not copies of the numerator:

[ \frac{5x - 4}{(x - 1)(x + 2)} = \frac{A}{x - 1} + \frac{B}{x + 2} ]\n Clear denominators:

[ 5x - 4 = A(x + 2) + B(x - 1) ]\n Substitute ( x=1 ): ( 1=3A \Rightarrow A=\frac{1}{3} ). Substitute ( x=-2 ): ( -14 = -3B \Rightarrow B=\frac{14}{3} ).

[ \frac{5x - 4}{(x - 1)(x + 2)} = \frac{1/3}{x - 1} + \frac{14/3}{x + 2} ]

Final answer: ( \frac{1}{3(x - 1)} + \frac{14}{3(x + 2)} ).

Example 3

Decompose ( \frac{3x + 1}{(x - 2)^2} ), a repeated linear factor.

[ \frac{3x + 1}{(x - 2)^2} = \frac{A}{x - 2} + \frac{B}{(x - 2)^2} ]\n Substituting ( x=2 ): ( 7 = B ); Comparing x-coefficients: ( 3 = A ).

Final answer: ( \frac{3}{x - 2} + \frac{7}{(x - 2)^2} ).

Example 4

Decompose ( \frac{2x^2 + 3}{x(x^2 + 1)} ), with an irreducible quadratic factor.

[ \frac{2x^2 + 3}{x(x^2 + 1)} = \frac{A}{x} + \frac{Bx + C}{x^2 + 1} ]\n Substituting ( x=0 ): ( 3 = A ); Comparing x^2-coefficients: ( 2 = A + B \Rightarrow B=-1 ); Comparing x-coefficients: ( 0 = C ).

Final answer: ( \frac{3}{x} - \frac{x}{x^2 + 1} ).

Example 5

Decompose ( \frac{x^2 + 1}{x^2 - 1} ).

The numerator and denominator have the same degree, so divide first: [ \frac{x^2 + 1}{x^2 - 1} = 1 + \frac{2}{x^2 - 1} ]
Decompose the proper remainder: [ \frac{2}{(x - 1)(x + 1)} = \frac{A}{x - 1} + \frac{B}{x + 1} ] Substituting values: [ 2 = 2A \Rightarrow A=1 ] [ 2 = -2B \Rightarrow B=-1 ].

Final answer: ( 1 + \frac{1}{x - 1} - \frac{1}{x + 1} ).

Example 6

Decompose ( \frac{x + 7}{x^2 - x - 6} ).

Factor the denominator: ( x^2 - x - 6 = (x - 3)(x + 2) ). [ \frac{x + 7}{(x - 3)(x + 2)} = \frac{A}{x - 3} + \frac{B}{x + 2} ] Clear denominators: [ x + 7 = A(x + 2) + B(x - 3) ] Substituting values: [ 10 = 5A \Rightarrow A=2 ] [ 5 = -5B \Rightarrow B=-1 ].

Final answer: ( \frac{2}{x - 3} - \frac{1}{x + 2} ).

Why Partial Fractions Matter

Partial fractions exist for one practical reason: a sum of simple fractions is enormously easier to work with than one complicated fraction.

Where the method pays off:

Integration in calculus. Many operations that are impossible become routine on decomposed pieces.
Inverse Laplace and Z-transforms. Engineers decompose a system's transfer function into partial fractions.
Series expansions. Splitting a fraction makes its power-series expansion tractable.

Where Decomposition Goes Off the Rails

Mistake 1: Wrong numerator form over a quadratic factor

Correct way: A quadratic factor needs a linear numerator: ( \frac{Ax + B}{x^2 + 1} ).

Mistake 2: Missing terms for a repeated factor

Correct way: A factor of power n needs every power from 1 to n: ( \frac{A}{x-2} + \frac{B}{(x-2)^2} ).

Mistake 3: Forgetting to divide first when the fraction is improper

Correct way: Decomposition only works on proper fractions. Do long division first to peel off the polynomial part.

Conclusion

Partial fractions split one rational expression into a sum of simpler fractions.
Solve for constants by substituting strategic x-values or comparing coefficients.
The most common mistakes are the wrong numerator form, missing terms for a repeated factor, and skipping the long-division.