__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

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