# 11.4 Partial fractions  (Page 2/7)

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Given a rational expression with distinct linear factors in the denominator, decompose it.

1. Use a variable for the original numerators, usually $\text{\hspace{0.17em}}A,B\text{,\hspace{0.17em}}$ or $\text{\hspace{0.17em}}C,\text{\hspace{0.17em}}$ depending on the number of factors, placing each variable over a single factor. For the purpose of this definition, we use $\text{\hspace{0.17em}}{A}_{n}\text{\hspace{0.17em}}$ for each numerator
$\frac{P\left(x\right)}{Q\left(x\right)}=\frac{{A}_{1}}{\left({a}_{1}x+{b}_{1}\right)}+\frac{{A}_{2}}{\left({a}_{2}x+{b}_{2}\right)}+\cdots \text{+}\frac{{A}_{n}}{\left({a}_{n}x+{b}_{n}\right)}$
2. Multiply both sides of the equation by the common denominator to eliminate fractions.
3. Expand the right side of the equation and collect like terms.
4. Set coefficients of like terms from the left side of the equation equal to those on the right side to create a system of equations to solve for the numerators.

## Decomposing a rational function with distinct linear factors

Decompose the given rational expression    with distinct linear factors.

$\frac{3x}{\left(x+2\right)\left(x-1\right)}$

We will separate the denominator factors and give each numerator a symbolic label, like $\text{\hspace{0.17em}}A,B\text{\hspace{0.17em},}$ or $\text{\hspace{0.17em}}C.$

$\frac{3x}{\left(x+2\right)\left(x-1\right)}=\frac{A}{\left(x+2\right)}+\frac{B}{\left(x-1\right)}$

Multiply both sides of the equation by the common denominator to eliminate the fractions:

$\left(x+2\right)\left(x-1\right)\left[\frac{3x}{\left(x+2\right)\left(x-1\right)}\right]=\overline{)\left(x+2\right)}\left(x-1\right)\left[\frac{A}{\overline{)\left(x+2\right)}}\right]+\left(x+2\right)\overline{)\left(x-1\right)}\left[\frac{B}{\overline{)\left(x-1\right)}}\right]$

The resulting equation is

$3x=A\left(x-1\right)+B\left(x+2\right)$

Expand the right side of the equation and collect like terms.

$\begin{array}{l}3x=Ax-A+Bx+2B\\ 3x=\left(A+B\right)x-A+2B\end{array}$

Set up a system of equations associating corresponding coefficients.

$\begin{array}{l}3=\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}A+B\\ 0=-A+2B\end{array}$

Add the two equations and solve for $\text{\hspace{0.17em}}B.$

$\begin{array}{l}\underset{¯}{\begin{array}{l}3=\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}A+B\\ 0=-A+2B\end{array}}\\ 3\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}0\text{\hspace{0.17em}}\text{\hspace{0.17em}}+3B\\ 1=B\end{array}$

Substitute $\text{\hspace{0.17em}}B=1\text{\hspace{0.17em}}$ into one of the original equations in the system.

$\begin{array}{l}3=A+1\\ 2=A\end{array}$

Thus, the partial fraction decomposition is

$\frac{3x}{\left(x+2\right)\left(x-1\right)}=\frac{2}{\left(x+2\right)}+\frac{1}{\left(x-1\right)}$

Another method to use to solve for $\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$ or $\text{\hspace{0.17em}}B\text{\hspace{0.17em}}$ is by considering the equation that resulted from eliminating the fractions and substituting a value for $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ that will make either the A - or B -term equal 0. If we let $\text{\hspace{0.17em}}x=1,\text{\hspace{0.17em}}$ the
$A-$ term becomes 0 and we can simply solve for $\text{\hspace{0.17em}}B.$

Next, either substitute $\text{\hspace{0.17em}}B=1\text{\hspace{0.17em}}$ into the equation and solve for $\text{\hspace{0.17em}}A,\text{\hspace{0.17em}}$ or make the B -term 0 by substituting $\text{\hspace{0.17em}}x=-2\text{\hspace{0.17em}}$ into the equation.

We obtain the same values for $\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}B\text{\hspace{0.17em}}$ using either method, so the decompositions are the same using either method.

$\frac{3x}{\left(x+2\right)\left(x-1\right)}=\frac{2}{\left(x+2\right)}+\frac{1}{\left(x-1\right)}$

Although this method is not seen very often in textbooks, we present it here as an alternative that may make some partial fraction decompositions easier. It is known as the Heaviside method , named after Charles Heaviside, a pioneer in the study of electronics.

Find the partial fraction decomposition of the following expression.

$\frac{x}{\left(x-3\right)\left(x-2\right)}$

$\frac{3}{x-3}-\frac{2}{x-2}$

## Decomposing $\text{\hspace{0.17em}}\frac{P\left(x\right)}{Q\left(x\right)}\text{\hspace{0.17em}}$ Where Q(x) Has repeated linear factors

Some fractions we may come across are special cases that we can decompose into partial fractions with repeated linear factors. We must remember that we account for repeated factors by writing each factor in increasing powers.

## Partial fraction decomposition of $\text{\hspace{0.17em}}\frac{P\left(x\right)}{Q\left(x\right)}:Q\left(x\right)\text{\hspace{0.17em}}$ Has repeated linear factors

The partial fraction decomposition of $\text{\hspace{0.17em}}\frac{P\left(x\right)}{Q\left(x\right)},\text{\hspace{0.17em}}$ when $\text{\hspace{0.17em}}Q\left(x\right)\text{\hspace{0.17em}}$ has a repeated linear factor occurring $\text{\hspace{0.17em}}n\text{\hspace{0.17em}}$ times and the degree of $\text{\hspace{0.17em}}P\left(x\right)\text{\hspace{0.17em}}$ is less than the degree of $\text{\hspace{0.17em}}Q\left(x\right),\text{\hspace{0.17em}}$ is

$\frac{P\left(x\right)}{Q\left(x\right)}=\frac{{A}_{1}}{\left(ax+b\right)}+\frac{{A}_{2}}{{\left(ax+b\right)}^{2}}+\frac{{A}_{3}}{{\left(ax+b\right)}^{3}}+\cdot \cdot \cdot +\frac{{A}_{n}}{{\left(ax+b\right)}^{n}}$

Write the denominator powers in increasing order.

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x exposent4+4x exposent3+8x exposent2+4x+1=0
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