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# Partial Fraction Expansions

## Decompose rational expressions into the sum of rational expressions with unlike denominators.

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Partial Fractions

When given a rational expression like $\frac{4x-9}{x^2-3x}$  it is very helpful in calculus to be able to write it as the sum of two simpler fractions like $\frac{3}{x}+\frac{1}{x-3}$ .  The challenging part is trying to get from the initial rational expression to the simpler fractions

You may know how to add fractions and go from two or more separate fractions to a single fraction, but how do you go the other way around?

#### Guidance

Partial fraction decomposition is a procedure that reverses adding fractions with unlike denominators.  The most challenging part is coming up with the denominators of each individual partial fraction.  See if you can spot the pattern.

$\frac{6x-1}{x^2 (x-1)(x^2+2)}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x-1}+\frac{Dx+E}{x^2+2}$

In this example each individual factor must be represented.  Linear factors that are raised to a power greater than one must have each successive power included as a separate denominator.  Quadratic terms that do not factor to be linear terms are included with a numerator that is a linear function of $x$ .

Example A

Use partial fractions to decompose the following rational expression.

$\frac{7x^2+x+6}{x^3+3x}$

Solution:  First factor the denominator and identify the denominators of the partial fractions.

$\frac{7x^2+x+6}{x(x^2+3)}=\frac{A}{x}+\frac{Bx+C}{x^2+3}$

When the fractions are eliminated by multiplying through by the LCD the equation becomes:

$7x^2+x+6 &= A(x^2+3)+x(Bx+C)\\7x^2+x+6 &= Ax^2+3A+Bx^2+Cx$

Notice the squared term, linear term and constant term form a system of three equations with three variables.

$A+B &= 7\\C &= 1\\3A &= 6$

In this case it is easy to see that $A=2, B=5, C=1$ . Often, the resulting system of equations is more complex and would benefit from your knowledge of solving systems using matrices.

$\frac{7x^2+x+6}{x(x^2+3)}=\frac{2}{x}+\frac{5x+1}{x^2+3}$

Example B

Decompose the following rational expression.

$\frac{5x^4-3x^3-x^2+4x-1}{(x-1)^3x^2}$

Solution:  First identify the denominators of the partial fractions.

$\frac{5x^4-3x^3-x^2+4x-1}{(x-1)^3x^2}=\frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{C}{(x-1)^3}+\frac{D}{x}+\frac{E}{x^2}$

When the entire fraction is multiplied through by $(x-1)^3x^2$  the equation results to:

$&5x^4-3x^3-x^2+4x-1\\&=A(x-1)^2x^2+B(x-1)x^2+Cx^2+D(x-1)^3x+E(x-1)^3$

Multiplication of each term can be done separately to be extra careful.

$& Ax^4-2Ax^3+Ax^2\\& Bx^3-Bx^2\\& Cx^2\\& Dx^4-3Dx^3+3Dx^2-Dx\\& Ex^3-3Ex^2+3Ex-E$

Group terms with the same power of  $x$ and set equal to the corresponding term.

$5x^4 &= Ax^4+Dx^4\\-3x^3 &= -2Ax^3+Bx^3-3D^3+Ex^3\\-x^2 &= Ax^2-Bx^2+Cx^2+3Dx^2-3Ex^2\\4x &= -Dx+3Ex\\-1 &= -E$

From these 5 equations, every  $x$ can be divided out.  Assume that $x \neq 0$  because if it were, then the original expression would be undefined.

$5 &= A+D\\-3 &= -2A+B-3D+E\\-1 &= A-B+C+3D-3E\\4 &= -D+3E\\-1 &= E$

This is a system of equations of five variables and 5 equations.  Some of the equations can be solved using logic and substitution like $E=-1, \ D=-7, \ A=12$ .  You can use any method involving determinants or matrices.  In this case it is easiest to substitute known values into equations with one unknown value to get more known values and repeat.

$B &= 1\\C &= 6\\\frac{5x^4-3x^3-x^2+4x-1}{(x-1)^3x^2} &= \frac{12}{x-1}+\frac{1}{(x-1)^2}+\frac{6}{(x-1)^3}+\frac{-7}{x}+\frac{-1}{x^2}$

Example C

Use matrices to complete the partial fraction decomposition of the following rational expression.

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

Solution:

$\frac{2x+4}{(x-1)(x+3)} &= \frac{A}{x+1}+\frac{B}{x+3}\\2x+4 &= Ax+3A+Bx+B\\$

$2 &= A+B\\4 &= 3A+B\\$

$\begin{bmatrix}\begin{matrix}1 & 1\\3 & 1\end{matrix} \left| \begin{matrix} 2\\4\end{matrix}\right.\end{bmatrix} \begin{matrix}& \rightarrow &\\\rightarrow & -3 \cdot I & \rightarrow\end{matrix} & \begin{bmatrix}\begin{matrix}1 & 1\\0 & -2\end{matrix} \left| \begin{matrix} 2\\-2\end{matrix}\right.\end{bmatrix} \begin{matrix}& \rightarrow &\\\rightarrow & \div -2 & \rightarrow\end{matrix} \begin{bmatrix}\begin{matrix}1 & 1\\0 & 1\end{matrix} \left| \begin{matrix} 2\\1\end{matrix}\right.\end{bmatrix} \begin{matrix}\rightarrow & -II & \rightarrow\\& \rightarrow &\end{matrix} \begin{bmatrix}\begin{matrix}1 & 0\\0 & 1\end{matrix} \left| \begin{matrix} 1\\1\end{matrix}\right.\end{bmatrix}\\$

$A=1, \ & B=1\\$

$\frac{2x+4}{(x-1)(x+3)} &= \frac{1}{x+1}+\frac{1}{x+3}$

Concept Problem Revisited

To decompose the rational expression into the sum of two simpler fractions you need to use partial fraction decomposition.

$\frac{4x-9}{x^2-3x} &= \frac{A}{x}+\frac{B}{x-3}\\4x-9 &= A(x-3)+Bx\\4x-9 &= Ax-3A+Bx$

Notice that the constant term -9 must be equal to the constant term $-3A$  and that the terms with $x$  must be equal as well.

$-9 &= -3A\\4 &= A+B$

Solving this system yields:

$A=3, \quad B=1$

Therefore,

$\frac{4x-9}{x^2-3x}=\frac{3}{x}+\frac{1}{x-3}$

#### Vocabulary

Partial fraction decomposition is a procedure that undoes the operation of adding fractions with unlike denominators.  It separates a rational expression into the sum of rational expressions with unlike denominators.

#### Guided Practice

$\frac{5x-2}{(2x-1)(3x+4)}$

2. Confirm Example C by adding the partial fractions.

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

3. Confirm Guided Practice #1 by adding the partial fractions.

$\frac{5x-2}{(2x-1)(3x+4)}=\frac{\frac{1}{11}}{2x-1}+\frac{-\frac{26}{11}}{3x+4}$

1. $\frac{5x-2}{(2x-1)(3x+4)} &= \frac{A}{2x-1}+\frac{B}{3x+4}\\5x-2 &= A(3x+4)+B(2x-1)\\5x-2 &= 3Ax+4A+2Bx-B\\5 &= 3A+2B\\-2 &= 4A-B\\\begin{bmatrix}\begin{matrix}3 & 2\\4 & -1\end{matrix} \left| \begin{matrix}5\\-2\end{matrix}\right.\end{bmatrix} \begin{matrix}\rightarrow & \cdot 4 & \rightarrow\\\rightarrow & \cdot 3 & \rightarrow\end{matrix} & \begin{bmatrix}\begin{matrix}12 & 8\\12 & -3\end{matrix} \left| \begin{matrix}20\\-6\end{matrix}\right.\end{bmatrix} \begin{matrix}& \rightarrow &\\\rightarrow & -I & \rightarrow\end{matrix} \begin{bmatrix}\begin{matrix}12 & 8\\0 & -11\end{matrix} \left| \begin{matrix}20\\-26\end{matrix}\right.\end{bmatrix} \begin{matrix}\rightarrow & \cdot 11 & \rightarrow\\\rightarrow & \cdot 8 & \rightarrow\end{matrix} \begin{bmatrix}\begin{matrix}132 & 88\\0 & -88\end{matrix} \left| \begin{matrix}220\\-208\end{matrix}\right.\end{bmatrix} \\& \begin{matrix}\rightarrow & +II & \rightarrow\\& \rightarrow &\end{matrix} \begin{bmatrix}\begin{matrix}132 & 0\\0 & -88\end{matrix} \left| \begin{matrix}12\\-208\end{matrix}\right.\end{bmatrix} \begin{matrix}\rightarrow & \div 132 & \rightarrow\\\rightarrow & \div -88 & \rightarrow\end{matrix} \begin{bmatrix}\begin{matrix}1 & 0\\0 & 1\end{matrix} \left| \begin{matrix}\frac{1}{11}\\\frac{26}{11}\end{matrix}\right.\end{bmatrix}\\A = \frac{1}{11}, \ & B =-\frac{26}{11}\\\frac{5x-2}{(2x-1)(3x+4)} &= \frac{\frac{1}{11}}{2x-1}+\frac{\frac{26}{11}}{3x+4}$

2.  $\frac{1}{x+1}+\frac{1}{x+3}=\frac{x+3}{(x+1)(x+3)}+\frac{x+1}{(x+1)(x+3)}=\frac{2x+4}{(x+1)(x+3)}$

3. $\frac{5x-2}{(2x-1)(3x+4)} &= \frac{\frac{1}{11}}{2x-1}+\frac{\frac{26}{11}}{3x+4}\\5x-2 &= \frac{1}{11}(3x+4)+\frac{26}{11}(2x-1)\\55x-22 &= 3x+4+26(2x-1)\\55x-22 &= 3x+4+52x-26\\55x-22 &= 55x-22$

#### Practice

Decompose the following rational expressions.  Practice using matrices with at least one of the problems.

1.  $\frac{3x-4}{(x-1)(x+4)}$

2.  $\frac{2x+1}{x^2(x-3)}$

3.  $\frac{x+1}{x(x-5)}$

4. $\frac{x^2+3x+1}{x(x-3)(x+6)}$

5.  $\frac{3x^2+2x-1}{x^2(x+2)}$

6.  $\frac{x^2+1}{x(x-1)(x+1)}$

7.  $\frac{4x^2-9}{x^2(x-4)}$

8.  $\frac{2x-4}{(x+7)(x-3)}$

9. $\frac{3x-4}{x^2(x^2+1)}$

10. $\frac{2x+5}{(x-3)(x^2+4)}$

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

### Vocabulary Language: English

partial fraction decomposition

partial fraction decomposition

Partial fraction decomposition is a procedure that undoes the operation of adding fractions with unlike denominators. It separates a rational expression into the sum of rational expressions with unlike denominators.
Rational Expression

Rational Expression

A rational expression is a fraction with polynomials in the numerator and the denominator.