Example: Decomposing a Rational Expression with Distinct Linear Factors

Decompose the given rational expression with distinct linear factors.

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

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

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

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.

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

Add the two equations and solve for $B$.

\begin{array}\text{ }3=A+B \\ 0=-A+2B \hfill& \\ \text{_____________} \\ 3=0+3B \hfill& \\ 1=B \hfill& \end{array}

Substitute $B=1$ 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 $A$ or $B$ is by considering the equation that resulted from eliminating the fractions and substituting a value for $x$ that will make either the A- or B-term equal 0. If we let $x=1$, the $A-$ term becomes 0 and we can simply solve for $B$.

$\begin{array}{l}\text{ }3x=A\left(x - 1\right)+B\left(x+2\right)\hfill \\ 3\left(1\right)=A\left[\left(1\right)-1\right]+B\left[\left(1\right)+2\right]\hfill \\ \text{ }3=0+3B\hfill \\ \text{ }1=B\hfill \end{array}$

Next, either substitute $B=1$ into the equation and solve for $A$, or make the B-term 0 by substituting $x=-2$ into the equation.

$\begin{array}{l}\text{ }3x=A\left(x - 1\right)+B\left(x+2\right)\hfill \\ \text{ }3\left(-2\right)=A\left[\left(-2\right)-1\right]+B\left[\left(-2\right)+2\right]\hfill \\ \text{ }-6=-3A+0\hfill \\ \text{ }\frac{-6}{-3}=A\hfill \\ \text{ 2}=A\hfill \end{array}$

We obtain the same values for $A$ and $B$ 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.

Example: Decomposing a Rational Function with a Repeated Irreducible Quadratic Factor in the Denominator

Decompose the given expression that has a repeated irreducible factor in the denominator.

$\frac{{x}^{4}+{x}^{3}+{x}^{2}-x+1}{x{\left({x}^{2}+1\right)}^{2}}$

Answer: The factors of the denominator are $x,\left({x}^{2}+1\right)$, and ${\left({x}^{2}+1\right)}^{2}$. Recall that, when a factor in the denominator is a quadratic that includes at least two terms, the numerator must be of the linear form $Ax+B$. So, let’s begin the decomposition.

$\frac{{x}^{4}+{x}^{3}+{x}^{2}-x+1}{x{\left({x}^{2}+1\right)}^{2}}=\frac{A}{x}+\frac{Bx+C}{\left({x}^{2}+1\right)}+\frac{Dx+E}{{\left({x}^{2}+1\right)}^{2}}$

We eliminate the denominators by multiplying each term by $x{\left({x}^{2}+1\right)}^{2}$. Thus,

${x}^{4}+{x}^{3}+{x}^{2}-x+1=A{\left({x}^{2}+1\right)}^{2}+\left(Bx+C\right)\left(x\right)\left({x}^{2}+1\right)+\left(Dx+E\right)\left(x\right)$

Expand the right side.

$\begin{array}{l} \text{ }{x}^{4}+{x}^{3}+{x}^{2}-x+1=A\left({x}^{4}+2{x}^{2}+1\right)+B{x}^{4}+B{x}^{2}+C{x}^{3}+Cx+D{x}^{2}+Ex\hfill \\ \text{ }=A{x}^{4}+2A{x}^{2}+A+B{x}^{4}+B{x}^{2}+C{x}^{3}+Cx+D{x}^{2}+Ex\hfill \end{array}$

Now we will collect like terms.

${x}^{4}+{x}^{3}+{x}^{2}-x+1=\left(A+B\right){x}^{4}+\left(C\right){x}^{3}+\left(2A+B+D\right){x}^{2}+\left(C+E\right)x+A$

Set up the system of equations matching corresponding coefficients on each side of the equal sign.

$\begin{array}{l}\text{ }A+B=1\hfill \\ \text{ }C=1\hfill \\ 2A+B+D=1\hfill \\ \text{ }C+E=-1\hfill \\ \text{ }A=1\hfill \end{array}$

We can use substitution from this point. Substitute $A=1$ into the first equation.

$\begin{array}{l}1+B=1\hfill \\ \text{ }B=0\hfill \end{array}$

Substitute $A=1$ and $B=0$ into the third equation.

$\begin{array}{l}2\left(1\right)+0+D=1\hfill \\ \text{ }D=-1\hfill \end{array}$

Substitute $C=1$ into the fourth equation.

$\begin{array}{r}\hfill 1+E=-1\\ \hfill \text{ }E=-2\end{array}$

Now we have solved for all of the unknowns on the right side of the equal sign. We have $A=1$, $B=0$, $C=1$, $D=-1$, and $E=-2$. We can write the decomposition as follows:

$\frac{{x}^{4}+{x}^{3}+{x}^{2}-x+1}{x{\left({x}^{2}+1\right)}^{2}}=\frac{1}{x}+\frac{1}{\left({x}^{2}+1\right)}-\frac{x+2}{{\left({x}^{2}+1\right)}^{2}}$