Partial Fractions

Partial fractions are a powerful mathematical technique that involves breaking down more complex fractions into simpler, component fractions. This simplification is vital in varied mathematical disciplines, including Calculus, Algebra, and Differential Equations. They infuse ease and precision in integral calculations, Laplace transformations, and solving linear ordinary differential equations. Engineers, scientists, and data analyst often utilize partial fractions in countless real-world applications for simplifying complex problems.

When a single fraction with multiple factors in the denominator is separated into more than one fraction, this is known as partial fractions. All the usual fractional arithmetic apply to algebraic fractions. This includes addition, subtraction, multiplication and division but partial fractions relies heavily on the addition of fractions. Recall that when adding fractions, we look for a common denominator. Need a quick reminder on how to do arithmetic with algebraic fractions?

Partial Fractions with Linear Factors

Suppose that we want to separate the fraction $\frac{2x+19}{(x-4)(2x+1)}$ into two fractions – one for each of the factors in the denominator. We use partial fractions and assume that the fractions can be written as $\frac{A}{x-4}+\frac{B}{2x+1}$ . The idea is to find the values of the constants $A$ and $B$ . We start by adding the fractions by giving them a common denominator:

$\frac{A}{x-4}+\frac{B}{2x+1}\equiv\frac{A(2x+1)}{(x-4)(2x+1)}+\frac{B(x-4)}{(x-4)(2x+1)}\equiv\frac{A(2x+1)+B(x-4)}{(x-4)(2x+1)}$

We can now find the values of $A$ and $B$ by comparing this numerator with that in the original fraction, i.e. $A(2x+1)+B(x-4)=2x+19$ . In other words, $(2A+B)x+A-4B=2x+19$ . The $x$ coefficient and the constant must be the same in each expression and so we have the simultaneous equations $2A+B=2$ and $A-4B=19$ . Solving these gives $A=3$ and $B=-4$ . See more on solving simultaneous equations. Hence, the original fraction can be parted into two fractions as follows:

$\frac{2x+19}{(x-4)(2x+1)}\equiv\frac{3}{x-4}-\frac{4}{2x+1}$

It’s always a good idea to check the partial fractions by adding them to see if you get back the original fraction. Notice that in this example the denominator was already factorised. The question may expect you to factorise first and it could have more factors – see Factorising Example.

Repeated Factors

Consider the fraction $\frac{x^2-3x+1}{(x+2)^2(x-5)}$ . This fraction has a repeated factor in the denominator ( $x+2$ appears twice). In this case, we must look for partial fractions of the form $\frac{A}{x+2}+\frac{B}{(x+2)^2}+\frac{C}{x-5}$ . This is because $\frac{A}{x+2}+\frac{B}{x+2}$ is simply just $\frac{A+B}{x+2}$ and doesn’t allow the factor to repeat. It follows that

$\frac{A}{x+2}+\frac{B}{(x+2)^2}+\frac{C}{x-5}\equiv\frac{A(x+2)(x-5)}{(x+2)^2(x-5)}+\frac{B(x-5)}{(x+2)^2(x-5)}+\frac{C(x+2)^2}{(x+2)^2(x-5)}\equiv\frac{A(x+2)(x-5)+B(x-5)+C(x+2)^2}{(x+2)^2(x-5)}$

Comparing the numerator with that of the original fraction gives $A(x+2)(x-5)+B(x-5)+C(x+2)^2=x^2-3x+1$ . Expanding and collecting like terms gives $\left(A+C\right)x^2+\left(-3A+B+4C\right)x+\left(-10A-5B+4C\right)=x^2-3x+1$ . Hence, $A+C=1$ , $-3A+B+4C=-3$ and $-10A-5B+4C=1$ . Using the simultaneous equations solver gives $A=\frac{38}{49}$ , $B=-\frac{11}{7}$ and $C=\frac{11}{49}$ . The partial fractions are given by

$\frac{x^2-3x+1}{(x+2)^2(x-5)}\equiv\frac{\frac{38}{49}}{x+2}-\frac{\frac{11}{7}}{(x+2)^2}+\frac{\frac{11}{49}}{x-5}\equiv\frac{38}{49(x+2)}-\frac{11}{7(x+2)^2}+\frac{11}{49(x-5)}$

See the Repeated Factor Example for another illustration of this.

Improper Partial Fractions

In the same way that a normal fraction can be improper, an algebraic fraction can also be improper. An algebraic fraction is improper if the numerator has an order that is equal or higher than that of the denominator. What is the order (or degree) of a polynomial? In the above notes, both original fractions were proper – the numerators were lower in order. If the order is the same and there are two linear factors in the denominator then we must try the form $A+\frac{B}{\text{factor 1}}+\frac{C}{\text{factor 2}}$ . Or if the order is one higher on the top, for example, the numerator is cubic and the denominator is quadratic, then we must try $Ax+B+\frac{C}{\text{factor 1}}+\frac{D}{\text{factor 2}}$ . We make similar adjustments to the above when there is a repeated factor. See Example 3 for an illustration of an improper fraction with a repeated factor being parted.

Examples

Factorising Example

Separate $\frac{5-3x}{x^2+x-6}$ into partial fractions.

Solution:

We must factorise the denominator to identify the linear factors first:

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

It follows that we must find $A$ and $B$ such that

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

It follows that $5-3x=A(x-2)+B(x+3)=(A+B)x-2A+3B$ That is, $A+B=-3$ and $-2A+3B=5$ . Solving gives $A=-\frac{14}{5}$ and $=-\frac{1}{5}$ . Hence,

$\frac{5-3x}{x^2+x-6}=-\frac{\frac{14}{5}}{x+3}-\frac{\frac{1}{5}}{x-2}=-\frac{14}{5(x+3)}-\frac{1}{5(x-2)}$

Repeated Factor Example

Separate $\frac{5x-2}{x^3-2x^2+x}$ into partial fractions.

Solution:

We begin by factorising the denominator. Factoring out the $x$ first, then factorising the quadratic in the brackets gives:

$x^3-2x^2+x=x\left(x^2-2x+1\right)=x\left(x-1\right)^2$ .

Since $x-1$ is a repeated factor, we find $A$ , $B$ and $C$ such that:

$\frac{5x-2}{x^3-2x^2+x}=\frac{A}{x}+\frac{B}{x-1}+\frac{C}{(x-1)^2}$

We write this as a single fraction and compare the numerator with that of the original fraction:

$\begin{array}{lll}\frac{A}{x}+\frac{B}{x-1}+\frac{C}{(x-1)^2}&=&\frac{A(x-1)^2}{x(x-1)^2}+\frac{Bx(x-1)}{x(x-1)^2}+\frac{Cx}{x(x-1)^2}\\&=&\frac{A(x-1)^2+Bx(x-1)+Cx}{x(x-1)^2}\end{array}$

It follows that $A(x-1)^2+Bx(x-1)+Cx=5x-2$ Equating the coefficients gives $A+B=0$ , $-2A-B+C=5$ and $A=-2$ . Hence, $A=-2$ , $B=2$ and $C=3$ giving

$\frac{5x-2}{x^3-2x^2+x}=-\frac{2}{x}+\frac{2}{x-1}+\frac{3}{(x-1)^2}$

Improper Fraction Example

Express $\frac{3x^2+8x-5}{x^2-6x+9}$ in the form $a+\frac{b}{x-3}+\frac{c}{(x-3)^2}$ where $a$ , $b$ and $c$ are constants to be found.

Solution:

The given form is used because, even though there is a $3x^2$ on the top and an $x^2$ on the bottom, the order is the same and so the fraction is improper. Adding the fractions by finding a common denominator gives:

$\begin{array}{lll}a+\frac{b}{x-3}+\frac{c}{(x-3)^2}&=&\frac{a(x-3)^2}{(x-3)^2}+\frac{b(x-3)}{(x-3)^2}+\frac{c}{(x-3)^2}\\&=&\frac{a(x-3)^2+b(x-3)+c}{(x-3)^2}\end{array}$