Partial fraction decomposition: distinct linear factors for the denominator

Basic functions: Rational functions

Partial fraction decomposition: distinct linear factors for the denominator

Addition of two rational functions with distinct denominators is done by reducing the functions to a common denominator and the result is a fraction with a compound denominator.

Sometimes it is more convenient to go the opposite way and rewrite a rational function with a compound denominator as a sum of simpler fractions. This reverse process is called partial fraction decomposition. Let us first have a look at two simple examples, which we can generalise.

$\begin{aligned} \frac{1}{x}+\frac{1}{1-x} &= \frac{1-x}{x(1-x)}+\frac{x}{x(1-x)} & \blue{\text{reduction to a common denominator}}\\[0.25cm] &= \frac{1-x+x}{x(1-x)}& \blue{\text{addition of numerators}}\\[0.25cm] &= \frac{1}{x(1-x)}& \blue{\text{fraction with a compound denominator}}\end{aligned}$
But how do you get from $\frac{1}{x(1-x)}$ to the sum of $\frac{1}{x}+\frac{1}{1-x}$ ?
The basic idea is to write $\frac{1}{x(1-x)}$ as a sum of multiples of $\frac{1}{x}$ and $\frac{1}{1-x}$ .
In other words, we assume that $\frac{1}{x(1-x)}=\frac{a}{x}+\frac{b}{1-x}$ for certain, yet to be determined, constants $a$ and $b$ . For the right-hand side we have $\begin{aligned}\frac{a}{x}+\frac{b}{1-x} &= \frac{a(1-x)}{x(1-x)}+\frac{bx}{x(1-x)} & \blue{\text{reduction to a common denominator}} \\[0.25cm]&=\frac{a(1-x)+bx}{x(1-x)} & \blue{\text{addition of numerators}} \\[0.25cm] &=\frac{(b-a)x+a}{x(1-x)} & \blue{\text{collection of terms}}\end{aligned}$ Then we have $(b-a)x+a=1$ for all $x$ . This is only possible if $b-a=0$ and $a=1$ .
It follows that $a=1$ and $b=1$ (you also get these results by $x=0$ and $x=1$ in $(b-a)x+a=1$ ).
In this way we have found the sum that we started with at the beginning of this example.

Assume that $\frac{4x-3}{(x-2)(x-1)}=\frac{a}{x-2}+\frac{b}{x-1}$ for certain numbers $a$ and $b$ , the values of which must be determined. For the right-hand side we have $\begin{aligned}\frac{a}{x-2}+\frac{b}{x-1} &= \frac{a(x-1)}{(x-2)(x-1)}+\frac{b(x-2)}{(x-2)(x-1)} & \blue{\text{reduction to a common denominator}} \\[0.25cm]&=\frac{a(x-1)+b(x-2)}{(x-2)(x-1)} & \blue{\text{addition of numerators}} \\[0.25cm] &=\frac{(a+b)x-(a+2b)}{(x-2)(x-1)} & \blue{\text{collection of terms}}\end{aligned}$ Then we have $(a+b)x-(a+2b)=4x-3$ for all $x$ .
This is only possible if $a+b=4$ and $a+2b=3$ .
This is a system of two linear equations in two unknowns, which can easily be solved (do this!):
$a=5$ and $b=-1$ . So: $\frac{4x-3}{(x-2)(x-1)}=\frac{5}{x-2}-\frac{1}{x-1}$
Note that the scalar $5$ of $\frac{1}{x-2}$ is equal to the value at substitution of $x=2$ into $\frac{4x-3}{(x-1)}$ ; similarly, the scalar $-1$ of $\frac{1}{x-1}$ is equal to the value at substitution of $x=1$ into $\frac{4x-3}{(x-2)}$ .

$\phantom{abc}$

The above examples illustrate how partial fraction decomposition can be done for a rational function $\frac{p(x)}{q(x)}$ with the degree of $p(x)$ less than the degree of $q(x)$ and where the factorisation of the denominator $q(x)$ only consists of distinct linear factors (so without repetition).

We consider a rational function $\frac{p(x)}{q(x)}$ with the degree of $p(x)$ less than the degree of $q(x)$ and with $q(x)$ a polynomial that can be written as $n$ deistinct linear factors, say, $q(x)=(x-\alpha_1)(x-\alpha_2)\cdots(x-\alpha_n)$ where $a_i\neq\alpha_j$ for $i\neq j$ and $1\le i,j,\le n$ . Then the partial fraction decomposition of $\frac{p(x)}{q(x)}$ has the following form: $\frac{p(x)}{q(x)}=\frac{a_1}{x-\alpha_1}+\frac{a_2}{x-\alpha_2}+\cdots +\frac{a_n}{x-\alpha_n}$ The coefficients can be calculated by setting up and solving a system of $n$ linear equations with $n$ unknowns.

But there is also a direct method. If we multiply the above partial fraction decomposition on the left-hand and right-hand side by $x-\alpha_j$ for $1\le j\le n$ , we get $(x-\alpha_j)\cdot \frac{p(x)}{q(x)}=a_1\frac{x-\alpha_j}{x-\alpha_1}+\cdots +a_{j-1}\frac{x-\alpha_j}{x-\alpha_{j-1}}+a_j+a_{j+1}\frac{x-\alpha_j}{x-\alpha_{j+1}}+\cdots +a_n\frac{x-a_j}{x-\alpha_n}$ So: $\begin{aligned}a_j &=\lim_{x\to\alpha_j}(x-\alpha_j)\cdot\frac{p(x)}{q(x)}\\[0.25cm] &=\frac{p(\alpha_j)}{(\alpha_j-\alpha_1)\cdots(\alpha_j-\alpha_{j-1})\cdot (\alpha_{j}-\alpha_{j+1})\cdots (\alpha_{j}-\alpha_n)} \end{aligned}$ In practice, you calculate $a_j$ by cancelling the factor $x-\alpha_j$ in the denominator of $\frac{p(x)}{q(x)}$ followed by substituting $x=\alpha_j$ in the obtained rational function.