Partial Fractions (Or, Partial Fraction Decomposition)

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There are some situations when it’s useful to write a rational function (a quotient of polynomials) as a sum of simpler such functions. As an example, we can see by cross multiplying that

$\begin{align*} \frac{1}{x-1}+\frac{1}{x+1} & = \frac{(x+1) + (x-1)}{(x-1)(x+1)}\\ & = \frac{2x}{x^2-1}. \end{align*}$

Partial fraction decomposition (often called just partial fractions) allows us to go backward, to figure out how to write $2n/(n^2-1)$ as a sum. It works like this: given a rational function whose denominator has higher degree than its numerator, we factor the denominator. Then we write the fraction as a sum of fractions whose denominators are the factors of the original denominator. The numerators of these fractions are yet to be determined, so we usually write them with variables $A$ , $B$ , and so on, whose values we have to figure out. Then we add together these fractions, getting an equation in the variables $A$ , $B$ , and so on by setting the numerator equal to the original numerator. Finally, we use this equation to solve for the $A$ , $B$ , and so on.

It sounds complicated, but it’s usually pretty simple. Let’s take a look at an example:

$\frac{1}{x^2-5x+6}$

We can factor $x^2-5x+6 = (x-2)(x-3)$ . So we want to find numbers $A$ and $B$ such that

$\frac{1}{x^2-5x+6} = \frac{A}{x-2} + \frac{B}{x-3}.$

To do this, cross multiply to get

$\frac{A}{x-2} + \frac{B}{x-3} = \frac{A(x-3) + B(x-2)}{(x-2)(x-3)}.$

Since we want this to be equal to the original fraction, and their denominators are equal, their numerators must be equal as well. Therefore we have the equation $A(x-3) + B(x-2) = 1$ . This holds for every value of $x$ . It’s convenient to put in the values $x=3$ and $x=2$ . When $x=3$ , we get $B=1$ . When $x=2$ , we get $A=-1$ . Putting these back in for $A$ and $B$ we get the desired decomposition

$\frac{1}{x^2-5x+6} = \frac{-1}{x-2} + \frac{1}{x-3}.$

Not too hard, huh?

What if there’s a repeated factor in the denominator? For example, how can we write

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

as a sum of simpler fractions? When there are repeated factors in the denominator, we have to repeat these factors in the sum as well: here we’ll take

$\frac{x+1}{(x-1)^2} = \frac{A}{x-1} + \frac{B}{(x-1)^2}$

as our decomposition. Cross multiplying we get

$\begin{align*} \frac{A}{x-1} + \frac{B}{(x-1)^2} &= \frac{A(x-1)^2 + B(x-1)}{(x-1)^3}\\ & = \frac{A(x-1) + B}{(x-1)^2}. \end{align*}$

The numerator must be equal to the original numerator, so we see that $A(x-1)+B = x+1$ . Putting in $x=1$ we get $B=2$ , and so $A=1$ . Therefore

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

Finally, we need to see what happens if there is an irreducible polynomial (one with no real roots, such as $x^2 + 1$ ) in the denominator. Here we have to take the numerator to be a degree 1 function instead of a constant. For example, let’s take a look at

$\frac{2x+2}{(x-1)(x^2+1)}.$

We want to write this as

$\frac{A}{x-1}+\frac{Bx+C}{x^2+1}.$

Note the linear term in the numerator of the second fraction. Cross multiplying we get

$\frac{A}{x-1}+\frac{Bx+C}{x^2+1} = \frac{A(x^2+1) + (Bx+C)(x-1)}{(x-1)(x^2+1)}.$

The numerator of this is the same as the numerator of the original fraction, so $(A+B)x^2 + (C-B)x + A-C = 2x+2$ . Solving for $A$ , $B$ , and $C$ , we get $A=2$ , $B=-2$ , and $C=0$ , so that