Partial Fractions Decomposition

Rules, Step-by-Step Examples with Collapsible Solutions, Practice Questions, and Answers

The decomposition of fractions [1] into simpler manageable fractions is presented. One of its important applications is in rational functions integral computation in calculus. Examples and questions and their solutions are included.

An online partial fractions decomposition calculator may be used to check answers to the examples and questions.

Decomposition into Partial Fractions Rules

How to decompose a rational function \( \dfrac{P(x)}{Q(x)} \) into partial fractions?

1 - Factor completely polynomial \( Q(x) \) in the denominator of the above rational function into factors of the form:

\[ (ax + b)^m \quad \text{and} \quad (ax^2 + bx + c)^n \]

Example: Let \( f(x) = \dfrac{2x-1}{x^3 + 2x^2 + 4x} \). The denominator is factored as \( x^3 + 2x^2 + 4x = x(x^2 + 2x + 4) \). The quadratic term \( x^2 + 2x + 4 \) is irreducible over the reals.

2 - For each factor of the form \( (ax + b)^m \), the decomposition includes the following sum of fractions:

\[ \dfrac{C_1}{ax + b} + \dfrac{C_2}{(ax + b)^2} + \dots + \dfrac{C_m}{(ax + b)^m} \]

Example: The fraction \( \dfrac{2}{(x-2)^3} \) is decomposed as \( \dfrac{C_1}{x-2} + \dfrac{C_2}{(x-2)^2} + \dfrac{C_3}{(x-2)^3} \).

3 - For each factor of the form \( (ax^2 + bx + c)^n \), the decomposition includes the following sum of fractions:

\[ \dfrac{A_1 x + B_1}{ax^2 + bx + c} + \dfrac{A_2 x + B_2}{(ax^2 + bx + c)^2} + \dots + \dfrac{A_n x + B_n}{(ax^2 + bx + c)^n} \]

Examples with Detailed Solutions

Click on each example to view its detailed step-by-step solution.

Example 1: Decompose into partial fractions \( \displaystyle \dfrac{2x + 5}{x^2 - x - 2} \)

Solution to Example 1:

We start by factoring the denominator:

\[ x^2 - x - 2 = (x - 2)(x + 1) \]

Both factors are linear, with power \( 1 \) each, hence the given fraction is decomposed as follows:

\[ \dfrac{2x + 5}{x^2 - x - 2} = \dfrac{A}{x - 2} + \dfrac{B}{x + 1} \]

Multiply both sides by the least common denominator \( (x - 2)(x + 1) \) and simplify:

\[ 2x + 5 = A(x + 1) + B(x - 2) \]

Expand the right side and group like terms:

\[ 2x + 5 = x(A + B) + A - 2B \]

Equating coefficients of polynomials:

\[ 2 = A + B \quad \text{and} \quad 5 = A - 2B \]

Solving this system yields \( A = 3 \) and \( B = -1 \). Substituting back:

\[ \dfrac{2x + 5}{x^2 - x - 2} = \dfrac{3}{x - 2} - \dfrac{1}{x + 1} \]
Example 2: Decompose into partial fractions \( \displaystyle \dfrac{1 - 2x}{x^2 + 2x + 1} \)

Solution to Example 2:

Factor the denominator:

\[ x^2 + 2x + 1 = (x + 1)^2 \]

Using the rule for repeated linear factors:

\[ \dfrac{1 - 2x}{x^2 + 2x + 1} = \dfrac{A}{x + 1} + \dfrac{B}{(x + 1)^2} \]

Multiply both sides by \( (x + 1)^2 \):

\[ 1 - 2x = A(x + 1) + B \implies -2x + 1 = Ax + (A + B) \]

Equating coefficients:

\[ -2 = A \quad \text{and} \quad 1 = A + B \implies B = 3 \]

Substituting \( A \) and \( B \):

\[ \dfrac{1 - 2x}{x^2 + 2x + 1} = \dfrac{-2}{x + 1} + \dfrac{3}{(x + 1)^2} \]
Example 3: Decompose into partial fractions \( \displaystyle \dfrac{4x^2 - x + 8}{(x - 2)(x^2 + 2x + 3)} \)

Solution to Example 3:

Decompose using the linear and irreducible quadratic factor rules:

\[ \dfrac{4x^2 - x + 8}{(x - 2)(x^2 + 2x + 3)} = \dfrac{A}{x - 2} + \dfrac{Bx + C}{x^2 + 2x + 3} \]

Multiply both sides by \( (x - 2)(x^2 + 2x + 3) \):

\[ 4x^2 - x + 8 = A(x^2 + 2x + 3) + (Bx + C)(x - 2) \]

Let \( x = 2 \):

\[ 4(4) - 2 + 8 = A(4 + 4 + 3) \implies 22 = 11A \implies A = 2 \]

Let \( x = 0 \):

\[ 8 = A(3) + C(-2) \implies 8 = 2(3) - 2C \implies 2 = -2C \implies C = -1 \]

Let \( x = 1 \) (knowing \( A = 2, C = -1 \)):

\[ 4(1) - 1 + 8 = 2(1 + 2 + 3) + (B - 1)(1 - 2) \] \[ 11 = 12 - (B - 1) \implies B = 2 \]

The fraction decomposes as:

\[ \dfrac{4x^2 - x + 8}{(x - 2)(x^2 + 2x + 3)} = \dfrac{2}{x - 2} + \dfrac{2x - 1}{x^2 + 2x + 3} \]

Questions

Decompose the following fractions into partial fractions:

  1. \( \displaystyle \dfrac{-x + 10}{x^2 + x - 2} \)
  2. \( \displaystyle \dfrac{2x - 3}{(x - 3)^2} \)
  3. \( \displaystyle \dfrac{-3x - 24}{(x + 4)(x^2 + 5x + 10)} \)

Solutions to Above Questions

  1. Factor denominator: \( x^2 + x - 2 = (x - 1)(x + 2) \).
    Setup: \( \dfrac{-x + 10}{x^2 + x - 2} = \dfrac{A}{x - 1} + \dfrac{B}{x + 2} \).
    Answer: \( \dfrac{3}{x - 1} - \dfrac{4}{x + 2} \)

  2. Denominator is already factored.
    Setup: \( \dfrac{2x - 3}{(x - 3)^2} = \dfrac{A}{x - 3} + \dfrac{B}{(x - 3)^2} \).
    Answer: \( \dfrac{2}{x - 3} + \dfrac{3}{(x - 3)^2} \)

  3. The expression \( x^2 + 5x + 10 \) cannot be factored over the reals (\( \Delta = 5^2 - 4(1)(10) = -25 < 0 \)).
    Setup: \( \dfrac{-3x - 24}{(x + 4)(x^2 + 5x + 10)} = \dfrac{A}{x + 4} + \dfrac{Bx + C}{x^2 + 5x + 10} \).
    Answer: \( -\dfrac{2}{x + 4} + \dfrac{2x - 1}{x^2 + 5x + 10} \)

More References and Links

  1. University Calculus - Early Transcendentals - Joel Hass, Maurice D. Weir, George B. Thomas, Jr., Christopher Heil - ISBN-13: 978-0134995540
  2. Calculus - Gilbert Strang - MIT - ISBN-13: 978-0961408824
  3. Calculus - Early Transcendentals - James Stewart - ISBN-13: 978-0-495-01166-8
  4. Online Partial Fractions Decomposition Calculator