# 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](/content/math/algebra/factorization-of-algebraic-expressions/index.html) 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](/content/math/algebra/long-division-of-polynomials/index.html) 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:

1. **Factor the denominator** completely and confirm the expression is proper (do long division first if it is not).
2. **Write the decomposition** with unknown constants (A, B, …) using the case table above.
3. **Clear the denominators** by multiplying both sides by the original denominator.
4. **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.
5. **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.