Partial Fraction Decomposition

In algebra, partial fraction decomposition (sometimes called partial fraction expansion) is a procedure used to reduce the degree of either the numerator or the denominator of a rational function. It involves splitting one ratio up into multiple simpler ratios. 

Here's an example of one ratio being split into a sum of three simpler ratios:  

$\displaystyle \frac{8x^2 + 3x - 21}{x^3 -7x -6} = \frac{1}{x+2} + \frac{3}{x-3} + \frac{4}{x+1}$

In mathematical terms, partial fraction expansion is used to change a rational function in the form $\frac{f(x)}{g(x)}$, where $f$ and $g$ are polynomials, into a function of the form $\sum_{j}\frac{f_{j}(x)}{g_{j}(x)}$. The denominators of the terms of this summation, $g_{j}(x)$, are polynomials that are factors of $g(x)$, and in general are of lower degree. 

The main motivation to decompose a rational function into a sum of simpler fractions is to make it easier to perform linear operations on the sum. Reducing complex mathematical problems via partial fraction decomposition allows us to focus on computing each single element of the decomposition rather than the more complex rational function.

Steps to Decomposing a Rational Function

Say we have a rational function $R(x) = \frac{f(x)}{g(x)}$, where the degree of the numerator is less than the degree of the denominator. Assume $R(x)$ has a denominator that factors into other expressions, as $g(x)=P(x)\cdot Q(x)$, and that there are no repeated roots.

The first step to decomposing the function $R(x)$ is to factor its denominator:

$\displaystyle R(x) = \frac{f(x)}{(x - a_1)(x - a_2)\cdots (x - a_p)}$

where $a_1, ... , a_p$ are the roots of $g(x)$.

We can then write $R(x)$ as the sum of partial fractions:

$R(x) = \frac{c_1}{(x - a_1)}+ \frac{c_2}{(x - a_2)}+ \cdots + \frac{c_p}{(x - a_p)}$

where $c_1, ... , c_p$ are constants.

To complete the process, we must determine the values of these $c_i$ coefficients.  To find a coefficient, multiply the denominator associated with it by the rational function $R(x)$:

$c_i = (x - a_i)R(x)$ 

This will yield an expression with an $x$-value. Substitute the associated root $a_i$ in for $x$, and solve for the constant. The following problems provide an examples of this technique.

Example 1

Apply decomposition to the rational function $f(x)=\frac{1}{x^{2}+2x-3}$

Factoring the denominator, we have:

$x^{2}+2x-3=(x+3)(x-1)$

So we have the partial fraction decomposition:

$f(x)=\frac{1}{x^{2}+2x-3}=\frac{c_1}{x+3}+\frac{c_2}{x-1}$

Now let's solve for the constant $c_1$:

$c_1 = \frac{1}{x^{2}+2x-3} (x+3) = \frac{x+3}{(x+3)(x-1)} = \frac{1}{x-1}$

Substituting $x=-3$ into this equation gives $c_1 = -\frac{1}{4}$.

Use the same process to solve for $c_2$:

$c_2 = \frac{1}{x^{2}+2x-3} (x-1) = \frac{x-1}{(x+3)(x-1)} = \frac{1}{x+3}$

Substituting $x=1$ gives $c_2 = \frac{1}{4}$. Substituting these coefficients into the decomposed function, we have:

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

We have rewritten the initial rational function in terms of partial fractions. This is the most simplified form possible, so we are finished.

Example 2

Apply decomposition to the rational function $g(x) = \frac{8x^2 + 3x - 21}{x^3 - 7x - 6}$

Factoring the denominator, we have:

$x^3 - 7x - 6=(x+2)(x-3)(x+1)$

So we have the partial fraction decomposition:

$g(x)=\frac{8x^2 + 3x - 21}{x^3 - 7x - 6}=\frac{c_1}{(x+2)} + \frac{c_2}{(x-3)}+ \frac{c_3}{(x+1)}$

We will now solve for each constant $c_i$:

$c_1 = \frac{8x^2 + 3x - 21}{x^3 - 7x - 6} (x+2) = \frac{8x^2 + 3x - 21}{(x-3)(x+1)} $

Substituting $x=-2$, we have: 

$\begin {aligned} c_1 &= \frac{8(-2)^2 + 3(-2) - 21}{(-2-3)(-2+1)} \\&= \frac {32-27}{(-5)(-1)} \\&=1 \end {aligned}$

$c_2 = \frac{8x^2 + 3x - 21}{x^3 - 7x - 6} (x-3) = \frac{8x^2 + 3x - 21}{(x+2)(x+1)} $

Substituting $x=3$, we have: 

$\begin {aligned} c_2 &= \frac{8(3)^2 + 3(3) - 21}{(3+2)(3+1)} \\&= \frac {72-12}{15} \\&= 4 \end {aligned}$

$c_3 = \frac{8x^2 + 3x - 21}{x^3 - 7x - 6} (x+1) = \frac{8x^2 + 3x - 21}{(x+2)(x-3)} $

Substituting $x=-1$, we have: 

$\begin {aligned} c_3&=\frac{8(-1)^2 + 3(-1) - 21}{(-1+2)(-1-3)} \\ &= \frac {8-24}{-4} \\ &= 4 \end {aligned}$

We have solved for each constant and have our partial fraction expansion:  

$g(x)=\frac{8x^2 + 3x - 21}{x^3 - 7x - 6}=\frac{1}{(x+2)} + \frac{4}{(x-3)}+ \frac{4}{(x+1)}$

Additional Considerations 

There are some important cases to note, for which partial fraction decomposition becomes more complicated. Decomposition in each of the below cases involves steps in addition to those described above.

• If there are repeated roots in the denominator of a rational function (for example, consider $G(x) = \frac{x+2}{(x-1)^2(x+3)}$, for which $x=1$ is a repeated root), additional steps must be taken to decompose the function.
• For a rational function $R(x) = \frac{f(x)}{g(x)}$, if the degree of $f(x)$ is greater than or equal to the degree of $g(x)$, the function cannot be decomposed in a straightforward way. It is necessary to perform the Euclidean division of $f$ by $g$ using polynomial long division, giving $f(x) = E(X)g(x) + h(x)$. Dividing through by $g(x)$ gives $\frac{f(x)}{g(x)}=E(x)+\frac{h(x)}{g(x)}$, which you can then perform the decomposition on $\frac{h(x)}{g(x)}$.