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 $\dfrac{2}{(x2)^3}$ is decomposed as
$\dfrac{2}{(x2)^3}=\dfrac{C1}{x2}+\dfrac{C2}{(x2)^2}+\dfrac{C3}{(x2)^3}$
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
$\dfrac{2 x + 5}{x^2x2}$
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
$\dfrac{2 x + 5}{x^2x2}=\dfrac{A}{x2}+\dfrac{B}{x+1}$
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
$\dfrac{2 x + 5}{x^2x2}=\dfrac{3}{x2}\dfrac{1}{x+1}$
As an exercise, group terms on the right to obtain the left side
Example 2: Decompose into partial fractions
$\dfrac{12 x}{x^2+2x+1}$
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
$\dfrac{12 x}{x^2+2x+1}=\dfrac{A}{x+1}+\dfrac{B}{(x+1)^2}$
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
$\dfrac{12 x}{x^2+2x+1}=\dfrac{2}{x+1}+\dfrac{3}{(x+1)^2}$
Example 3: Decompose into partial fractions
$\dfrac{4x^2x+8}{(x2)(x^2+2x+3)}$
Solution to Example 3:
Use the rule above to decomposed the given fraction as follows
$\dfrac{4x^2x+8}{(x2)(x^2+2x+3)}=\dfrac{A}{x2}+\dfrac{B x+C}{x^2+2x+3}$
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
$\dfrac{4x^2x+8}{(x2)(x^2+2x+3)}=\dfrac{2}{x2}+\dfrac{2 x1}{x^2+2x+3}$
Exercises: Decompose the following fractions into partial fractions.
1. $\dfrac{x+10}{x^2+x2}$
2. $\dfrac{2 x  3}{x3}$
3. $\dfrac{3 x  24}{(x+4)(x^2+5x+10)}$
Answers to Above Exercises
1. $\dfrac{3}{x1}\dfrac{4}{x+2}$
2. $\dfrac{2}{x3}+\dfrac{3}{(x3)^2}$
3. $\dfrac{2}{x+4}+\dfrac{2 x1}{x^2+5x+10}$
