Partial Fractions Decompositions

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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
\[ (ax + b)^m \text{and} (a x^2 + b x + c )^n \]
Example
Let \( f(x) = \dfrac{2x-1}{x^3 + 2x^2 + 4x} \)
The denominator is factored as follows
\( x^3 + 2x^2 + 4x = x (x^2 + 2x + 4) \)
The quadratic term \( x^2 + 2 x + 4 \) is irreducible (cannot be factored) over the real.

2 - For each factor of the form \( (ax + b)^m \), the decomposition includes the following sum of fractions
\( \dfrac{C_1}{ax + b}+\dfrac{C_2}{(ax + b)^2}+...+\dfrac{C_m}{(ax + b)^m} \)


Example
The fraction \( \dfrac{2}{(x-2)^3} \) is decomposed as

\( \dfrac{2}{(x-2)^3}=\dfrac{C_1}{x-2}+\dfrac{C_2}{(x-2)^2}+\dfrac{C_3}{(x-2)^3} \)

3 - For each factor of the form \( (a x^2 + b x + c)^n \), the decomposition includes the following sum of fractions
\( \dfrac{A_1 x + B_1}{a x^2 + b x + c} + \dfrac{A_2 x + B_2}{(a x^2 + b x + c)^2} + ... + \dfrac{A_n x + B_n}{a x^2 + b x + c)^n} \)



Examples with Detailed Solutions

Example 1

Decompose into partial fractions

\( \dfrac{2 x + 5}{x^2-x-2} \)

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

\( \dfrac{2 x + 5}{x^2-x-2}=\dfrac{A}{x-2}+\dfrac{B}{x+1} \)

Multiply both side of the above equation by the least common denominator, \( (x - 2)(x + 1) \), and simplify to obtain an equation of the form
\( 2 x+5 = A(x + 1) + B(x - 2) \)
Expand the right side and group like terms
\( 2 x + 5 = x (A + B) + A - 2 B \)
For the right and left polynomials to be equal we need to have
\( 2 = A + B \) and \( 5 = A - 2 B \)
Solve the above system to obtain
\( A = 3 \) and \( B = -1 \)
Substitute \( A \) and \( B \) in the suggested decomposition above to obtain
\( \dfrac{2 x + 5}{x^2-x-2}=\dfrac{3}{x-2}-\dfrac{1}{x+1} \)

As an exercise, group terms on the right to obtain the left side




Example 2

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

Solution to Example 2:
We start by factoring the denominator
\( x^2 + 2 x + 1 = (x + 1)^2\)
Using the rule above, the given fraction is decomposed as follows

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

Multiply both side of the above equation by \( (x + 1)^2 \), and simplify to obtain an equation of the form
\( 1 - 2 x = A(x + 1) + B \)
Expand the right side and group like terms
\( -2x + 1 = A x + (A + B) \)
For the right and left polynomials to be equal we need to have
\( - 2 = A \) and \( 1 = A + B \)
Solve the above system to obtain
\( A = - 2 \) and \( B = 3 \)
Substitute \( A \) and \( B \) in the suggested decomposition above to obtain
\( \dfrac{1-2 x}{x^2+2x+1}=\dfrac{-2}{x+1}+\dfrac{3}{(x+1)^2} \)



Example 3

Decompose into partial fractions
\( \dfrac{4x^2-x+8}{(x-2)(x^2+2x+3)} \)

Solution to Example 3:
Use the rule above to decomposed the given fraction as follows

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

Multiply both side of the above equation by \( (x - 2)(x^2 + 2 x + 3) \), and simplify to obtain an equation of the form
\( 4 x^2 - x + 8 = A(x^2 + 2 x + 3) + (B x + C)(x - 2) \)
The above equality is true for all values of \( x \), let us use \( x = 2 \) to obtain an equation in \( A \)
\( 22 = 11 A \)
Solve for \( A \) to obtain
\( A = 2 \)
In order to find \( C \), we use \( x = 0 \) in the above equality
\( 8 = 6 - 2 C \)
Solve for \( C \) to obtain
\( C = -1 \)
To find \( B \), we now use \( x = 1 \) in the above equality
\( 11 = 12 + (B - 1)(1 - 2) \)
Solve for \( B \) to obtain
\( B = 2 \)
The given fraction can be decomposed as follows
\( \dfrac{4x^2-x+8}{(x-2)(x^2+2x+3)}=\dfrac{2}{x-2}+\dfrac{2 x-1}{x^2+2x+3} \)



Questions

Decompose the following fractions into partial fractions.
1. \( \dfrac{-x+10}{x^2+x-2} \)

2. \( \dfrac{2 x - 3}{(x-3)^2} \)

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

Solutions to Above Exercises

1.
Factor denomiantor: \( x^2+x-2 = (x - 1)(x + 2) \)
Hence according to
rules the decomposition is written as: \( \dfrac{-x+10}{x^2+x-2} = \dfrac{A}{x-1} + \dfrac{B}{x+2} \)
Use numerical values for \( x \) and a system of two equations with unknowns\( A \) and \( B \) then solve to to obtain
\( \dfrac{3}{x-1}-\dfrac{4}{x+2} \)

2.
The denominator is already in factored form.
Hence according to
rules the decomposition is written as: \( \dfrac{2 x - 3}{(x-3)^2} = \dfrac{A}{x-3} + \dfrac{B }{(x-3)^2} \)

Use numerical values for \( x \) and a system of two equations with unknowns\( A \) and \( B \) then solve to to obtain
\( \dfrac{2}{x-3}+\dfrac{3}{(x-3)^2} \)

3.
The expression \( x^2+5x+10 \) in the denominator cannot be factored over the real numbers because its discriminant \( \Delta = 5^2 - 4(1)(10) = -25 \) is negative and therefore the denominator is in factored form.
The
rules of the decomposition is used to write the decomposition as: \( \dfrac{-3 x - 24}{(x+4)(x^2+5x+10)} = \dfrac{A}{x+4} + \dfrac{B x + C }{x^2+5x+10} \)
Use numerical values for \( x \) and a system of three equations with unknowns\( A \), \( B \) and \( C \) then solve to to obtain
\( -\dfrac{2}{x+4}+\dfrac{2 x-1}{x^2+5x+10} \)



More References and Links

University Calculus - Early Transcendental - Joel Hass, Maurice D. Weir, George B. Thomas, Jr., Christopher Heil - ISBN-13 ? : ? 978-0134995540
Calculus - Gilbert Strang - MIT - ISBN-13 ? : ? 978-0961408824
Calculus - Early Transcendental - James Stewart - ISBN-13: 978-0-495-01166-8
online partial fractions decomposition calculator

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