Partial fraction decomposition: repeated linear factors for the denominator

Basic functions: Rational functions

Partial fraction decomposition: repeated linear factors for the denominator

Let us start with three simple examples that we can generalise.

$\begin{aligned}\frac{3x+2}{x^2+2x+1} &= \frac{3x+2}{(x+1)^2}\\[0.25cm] &= \frac{3(x+1)-1}{(x+1)^2}\\[0.25cm]&= \frac{3}{x+1}-\frac{1}{(x+1)^2}\end{aligned}$

$\begin{aligned}\frac{4x^2+3x+2}{(x+1)^3} &= \frac{4(x^2+2x+1)-5x-2}{(x+1)^3} &\blue{\text{note that }(x+1)^2=x^2+2x+1}\\[0.25cm]&=\frac{4(x+1)^2-5(x+1)+3}{(x+1)^3}&\blue{\text{create } x+1\text{ terms}}\\[0.25cm]&=\frac{3}{(x+1)^3}-\frac{5}{(x+1)^2}+\frac{4}{x+1}&\blue{\text{decompose the fraction}}\end{aligned}$ $\phantom{abc}$

Rewriting the numerator as a sum of powers of $x+1$ is a bit of a puzzle; those who don't like this much can use the direct method below.

Assume that $\frac{4x^2+3x+2}{(x+1)^3}=\frac{a}{x+1}+\frac{b}{(x+1)^2}+\frac{c}{(x+1)^3}$ for certain numbers $a$ , $b$ and $c$ , the values of which must be determined. For the right-hand side we have $\begin{aligned}\frac{a}{x+1}+\frac{b}{(x+1)^2}+\frac{c}{(x+1)^3} &= \frac{a(x+1)^2}{(x+1)^3}+\frac{b(x+1)}{(x+1)^3}+\frac{c}{(x+1)^3}\\[0.25cm] &\phantom{abcdefg}\blue{\text{reduction to a common denominator}} \\[0.25cm]&=\frac{a(x+1)^2+b(x+1)+c}{(x+1)^3} \\[0.25cm] &\phantom{abcdefg}\blue{\text{addition of numerators}} \\[0.25cm]&=\frac{a(x^2+2x+1)+b(x+1)+c}{(x+1)^3}\\[0.25cm] &\phantom{abcdefg}\blue{\text{expansion of power}}\\[0.25cm] &=\frac{ax^2+(b+2a)x+(a+b+c)}{(x+1)^3}\\[0.25cm]&\phantom{abcdefg}\blue{\text{collection of terms}}\end{aligned}$ Then we have $ax^2+(b+2a)x+(a+b+c)=4x^2+3x+2$ for all $x$ .
But polynomials are equal only when their coefficients are equal.
This is here only possible if $a=4$ , $b+2a=3$ and $a+b+c=2$ .
It follows that $a=4$ , $b=-5$ and $c=3$ .
In this way we have found the sum that we started with at the beginning of this example.

Assume that $\frac{x^2+2x+3}{x(x-1)^2}=\frac{a}{x}+\frac{b}{x-1}+\frac{c}{(x-1)^2}$ for certain numbers $a$ , $b$ and $c$ , the values of which must be determined. For the right-hand side we have $\begin{aligned}\frac{a}{x}+\frac{b}{x-1}+\frac{c}{(x-1)^2} &= \frac{a(x-1)^2}{x(x-1)^2} +\frac{bx(x-1)}{x(x-1)^2}+\frac{cx}{x(x-1)^2}\\[0.25cm] &\phantom{abcdefg}\blue{\text{reduction to a common denominator}} \\[0.25cm]&=\frac{a(x-1)^2+bx(x-1)+cx}{x(x-1)^2} \\[0.25cm] &\phantom{abcdefg}\blue{\text{addition of numerators}} \\[0.25cm]&=\frac{a(x^2-2x+1)+b(x^2-x)+cx}{x(x-1)^2}\\[0.25cm] &\phantom{abcdefg}\blue{\text{expansion of terms at }a\text{ and }b}\\[0.25cm] &=\frac{(a+b)x^2+(c-b-2a)x+c}{x(x-1)^2}\\[0.25cm]&\phantom{abcdefg}\blue{\text{collection of terms}}\end{aligned}$ Then we have $(a+b)x^2+(c-b-2a)x+a=x^2+2x+3$ for all $x$ .
But polynomials are equal only when their coefficients are equal.
This is here only possible if $a+b=1$ , $c-b-2a=2$ and $a=3$ . It follows that $a=3$ , $b=-2$ and $c=6$ .
In this way we have determined the partial fraction decomposition: $\frac{x^2+2x+3}{x(x-1)^2}=\frac{3}{x}-\frac{2}{x-1}+\frac{6}{(x-1)^2}$

$\phantom{abc}$

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 $q(x)$ still only consists of linear factors, but with repetitions. Every distinct factor may therefore be raised to a certain power with positive natural exponent in the factorisation of the denominator.

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 $n$ disticnt linear factors, say, $q(x)=(x-\alpha_1)^{m_1}(x-\alpha_2)^{m_1}\cdots(x-\alpha_n)^{m_n}$ with $a_i\neq\alpha_j$ for $i\neq j, 1\le i,j,\le n$ and $m_i\in\{1,2,3,\ldots\}$ for $1\le i\le n$ . Then, each power of a factor, say $(x-\alpha)^m$ , in the partial fraction decomposition of $\frac{p(x)}{q(x)}$ corresponds with the following sum of fractions: $\frac{a_1}{x-\alpha}+\frac{a_2}{(x-\alpha)^2}+\cdots +\frac{a_m}{(x-\alpha)^m}$ All coefficients can be computed by adding up all fractions in the decomposition and equating the coefficients of like powers of $x$ in the numerator with those in $p(x)$ , followed by solving the obtained system of equations.