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
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
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
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
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