Partial Fractions Decompositions

Tutorial on decomposing complicated fractions into simpler manageable fractions. One of its important applications is in Integration Using Partial Fractions in calculus.

Rules of Decomposition

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

1 - Factor completely polynomial $Q(x)$ into factors of the form

$(ax + b)^m$ and $(ax^2 + b x + c)^n$


Example

Let $f(x) = \dfrac {(2x - 1)}{(x^3+ 2 x^2 + 4 x)}$

The denominator is factored as follows

$(x^3 + 2 x^2 + 4 x) = x(x^2 + 2 x + 4)$

The quadratic term $x^2 + 2 x + 4$ is irreducible over the real.

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

$\dfrac{C1}{ax + b}+\dfrac{C2}{(ax + b)^2}+...+\dfrac{Cm}{(ax + b)^m}$


Example

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

$\dfrac{2}{(x-2)^3}=\dfrac{C1}{x-2}+\dfrac{C2}{(x-2)^2}+\dfrac{C3}{(x-2)^3}$

3 - For each factor of the form $(a x^2 + b x + c)^n$, the decomposition includes the following sum of fractions

$\dfrac{A1 x + B1}{a x^2 + b x + c} + \dfrac{A2 x + B2}{(a x^2 + b x + c)^2} + ... + \dfrac{An x + Bn}{a x^2 + b x + c)^n}$


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, 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

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



Exercises: Decompose the following fractions into partial fractions.

1. $\dfrac{-x+10}{x^2+x-2}$

2. $\dfrac{2 x - 3}{x-3}$

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

Answers to Above Exercises

1. $\dfrac{3}{x-1}-\dfrac{4}{x+2}$

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

3. $-\dfrac{2}{x+4}+\dfrac{2 x-1}{x^2+5x+10}$


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Updated: 2 April 2013

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