# Partial Fractions - Linear Factors

**Partial fraction decomposition** is a technique used when one writes a rational function as the sum of simpler fractions. In certain cases, a rational function can be expressed as the sum of fractions whose denominators are **linear polynomials**. For example,

\[\frac{2}{x^2-1} = \frac{1}{x-1} - \frac{1}{x+1}.\]

#### Contents

## Identifying when linear factors exist

Note that repeated factors are **not** handled in this page, but have a separate link Partial Fractions - Repeated Factors.

A rational function has linear factors only when the denominator can be factorized as a product of linear polynomials. For the purposes of this page, the factors are not repeated, i.e. none of the factors has a power greater than one. In other words, this page will discuss rational functions in which the denominator can be factorized into \[(x-x_1)(x-x_2)\cdots(x-x_n),\] where \(x_1, x_2,\ldots, x_n\) are all constants.

For example,

- \(\dfrac{5}{x^2+x+1}\) has no linear factors since \(x^2+x+1\) cannot be factorized into linear polynomials;
- \(\dfrac{5}{(x+1)(x^2+x+1)}\) has a linear factor, along with a quadratic factor that cannot be further factorized into linear factors, which is outside the scope of this page;
- \(\dfrac{1}{x^2+x-2} = \dfrac{1}{(x-1)(x+2)}\) has only linear factors, and is the type of rational function which will be handled in this page;
- \(\dfrac{3}{(x-1)(x+2)^2}\) has only linear factors, but one of the factors is repeated \(((x+2)^2\) is a repeated factor since it has a power of \(2)\), so it is outside the scope of this page.

## Determining the Partial Fractions

There are three methods:

- Comparing coefficients of the left and right side
- Setting an identical equation, assigning any suitable valid value to the variable, and solving the resulting simultaneous equations
- Heaviside's cover-up method.

All three methods are used in the example below.

## Heaviside's Cover-up Method

We will first see an example before discussing the method: \[\dfrac{x−7}{(x−1)(x + 2)}=\dfrac A {x−1} + \dfrac B {x+2}\] To get \(A\), cover up \((x - 1)\) on the left side and substitute \((x - 1) = 0,\) or \(x=1,\) in the remaining \(\dfrac{x−7}{x + 2}: \) \[\dfrac{1−7}{1+2} = - 2 = A. \] Similarly, for \(B\), substitute \(x = -2 \ \) in \( \ \dfrac{x−7}{x−1} \) to get \[\dfrac{-2−7}{-2-1} = 3 = B. \]

So we finally have \( \dfrac{x−7}{(x−1)(x + 2)} = \dfrac{-2}{x−1} + \dfrac{3}{x+2}. \)

In other words, to find \(A\), we are multiplying both sides by the denominator of \(A\) and get \[\left. \dfrac{x−7}{\color{red}{(x−1)} (x + 2)} \times \color{red}{(x−1)} = \dfrac{A}{\color{red}{(x−1)}}\times \color{red}{(x−1)} + \dfrac{B}{x+2} \times \color{red}{(x−1)}\right|_{x=1}.\] By substituting \( (x - 1) = 0, \) that is, \( x = 1, \) we now get the value of \(A\).

## Additional Examples and Problem Solving

Express \(\dfrac{2}{1-x^2}\) as partial fractions.

Solution 1:

The denominator \((1-x^2)\) can be factorized as \((1-x)(1+x)\). Therefore, let \(A\) and \(B\) be constants such that: \[ \frac{2}{1-x^2} = \frac{A}{1-x} + \frac{B}{1+x}. \] Multiply both sides by \((1-x^2)\) to get \[\begin{align} 2 &= A (1+x) + B (1-x) \\ &= (A+B) + (A-B) x. \end{align}\] By comparing like terms, we have the following set of equations: \[\begin{cases}A+B=2\\A-B=0.\end{cases}\] We solve it and get \(A=1\) and \(B=1\). Therefore, \[\frac{2}{1-x^2} = \frac{1}{1-x} + \frac{1}{1+x}.\ _\square\]

Solution 2:

We may find \(A\) and \(B\) by inserting two differentvalidvalues to \(x\) and solving the simultaneous equations in \(A\) and \(B\).Substitute \( x=2 \implies 2 = 3A - B, \) and \( x = -2 \implies 2 = - A +3B. \)

So \( 4A=4B \implies A=B \) and \( A = 1,\) as above.

Note the word "valid." You cannot take \( x=1 \) or \( x=-1 \) here.Often \( x=0, x=1, \) and \( x=-1 \) are good choices.

Always try to take values for \( x \) that simplify calculations.

Say we have \(\dfrac 2 {x-3} \). Then \( x=1 \) is a good choice if it suits the other expression, or \( x=5.\ _\square\)\[\]

Solution 3:

Heaviside's cover-up method: (cannot be used for quadratic factors)

We have \[\begin{align} \dfrac{2}{(1-x)(1+x)} &= \dfrac{A}{1-x} + \dfrac{B}{1+x} \\\\ \left. A=\dfrac{2}{1+1}\right|_{x= 1}\ &=\dfrac 2 2 \\&=1 \\\\ \left. B=\dfrac{2}{1-(-1)}\right|_{x=-1}\ &=\dfrac 2 2 \\&=1, \end{align}\] which is the same as above. \(_\square\)

Write \(\dfrac{2x+3}{(x+2)(x+5)}\) in the form of partial fractions.

First let's write the given rational fraction as follows: \[\dfrac{2x+3}{(x+2)(x+5)} = \dfrac{A}{x+2} + \dfrac{B}{x+5}.\] By taking the LCM, the right side can also be written as \[\dfrac{2x+3}{(x+2)(x+5)} = \dfrac{A(x+5)+B(x+2)}{(x+2)(x+5)}\] Now cancelling the denominators of LHS and RHS, we get \[A(x+5) + B(x+2) = 2x + 3. \] To find out the value of the term containing \(B\), we must eliminate the term containing \(A\) value.

Substitute \(x=-5\) for the whole term \(A(x+5)\) to be equal to \(0\), then \[B(-5+2) = 2(-5) + 3 \implies B = \frac{7}{3}. \] Now, to find the value of \(A\), we must eliminate \(B\). This means \(x=-2,\) which gives \[A(-2+5) = 2(-2) + 3 \implies A = -\frac{1}{3}. \]Therefore, the required partial fractions are

\[\dfrac{2x+3}{(x+2)(x+5)} = -\dfrac{1}{3(x+2)} + \dfrac{7}{3(x+5)}.\ _\square\]

**Cite as:**Partial Fractions - Linear Factors.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/partial-fractions-linear-factors/