# Partial Fraction Decomposition

## Solved Problems

### Example 7.

Determine a partial fraction decomposition for the function $\frac{{30x}}{{\left( {x + 1} \right)\left( {x - 2} \right)\left( {x + 3} \right)}}.$

Solution.

The partial fraction decomposition has the form

$\frac{{30x}}{{\left( {x + 1} \right)\left( {x - 2} \right)\left( {x + 3} \right)}} = \frac{A}{{x + 1}} + \frac{B}{{x - 2}} + \frac{C}{{x + 3}}.$

We combine the partial fractions on the right side into a single fraction and then equate the numerators on both sides. This yields:

$30x = A\left( {x - 2} \right)\left( {x + 3} \right) + B\left( {x + 1} \right)\left( {x + 3} \right) + C\left( {x + 1} \right)\left( {x - 2} \right),$
$30x = A\left( {{x^2} + x - 6} \right) + B\left( {{x^2} + 4x + 3} \right) + C\left( {{x^2} - x - 2} \right),$
$30x = \left( {A + B + C} \right){x^2} + \left( {A + 4B - C} \right)x + \left( { - 6A + 3B - 2C} \right).$

Hence we obtain the following system of equations:

$\left\{ \begin{array}{l} A + B + C = 0\\ A + 4B - C = 30\\ - 6A + 3B - 2C = 0 \end{array} \right..$

Solving it, we find the unknown coefficients:

$A = 5,\;B = 4,\;C = - 9.$

The partial fraction decomposition is given by

$\frac{{30x}}{{\left( {x + 1} \right)\left( {x - 2} \right)\left( {x + 3} \right)}} = \frac{5}{{x + 1}} + \frac{4}{{x - 2}} - \frac{9}{{x + 3}}.$

### Example 8.

Determine a partial fraction decomposition for the function $\frac{{{x^2} + 2}}{{{x^3} - 3{x^2} + 2x}}.$

Solution.

First we factor the cubic function in the denominator:

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

As the denominator contains only linear factors, the decomposition has the form

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

Combining the partial fractions on the right side into a single fraction and equating the numerators on both sides, we have:

${x^2} + 2 = A\left( {x - 1} \right)\left( {x - 2} \right) + Bx\left( {x - 2} \right) + Cx\left( {x - 1} \right),$
${x^2} + 2 = A\left( {{x^2} - 3x + 2} \right) + B\left( {{x^2} - 2x} \right) + C\left( {{x^2} - x} \right),$
${x^2} + 2 = \left( {A + B + C} \right){x^2} + \left( { - 3A - 2B - C} \right)x + 2A.$

The system of equations for the unknown coefficients is written as

$\left\{ \begin{array}{l} A + B + C = 1\\ - 3A - 2B - C = 0\\ 2A = 2 \end{array} \right..$

It has the following solution:

$\left\{ \begin{array}{l} A = 1\\ B = - 3\\ C = 3 \end{array} \right.,$

so the decomposition of the original rational function is expressed by the equation:

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

### Example 9.

Decompose $\frac{6}{{x\left( {{x^2} + x + 3} \right)}}$ using the partial fractions.

Solution.

Given the irreducible quadratic factor $${{x^2} + x + 3}$$ in the denominator, we write the partial fraction expansion in the form

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

The equation for the unknown coefficients is given by

$6 = A\left( {{x^2} + x + 3} \right) + B{x^2} + Cx,$

or

$6 = \left( {A + B} \right){x^2} + \left( {A + C} \right)x + 3A.$

We obtain the system of equations:

$\left\{ \begin{array}{l} A + B = 0\\ A + C = 0\\ 3A = 6 \end{array} \right..$

This yields:

$A = 2,\;B = - 2,\;C = - 2.$

Hence, the partial fraction decomposition is written as

$\frac{6}{{x\left( {{x^2} + x + 3} \right)}} = \frac{2}{x} - \frac{{2x + 2}}{{{x^2} + x + 3}}.$

### Example 10.

Expand the rational function ${\frac{{16}}{{\left( {{x^2} + x + 2} \right){{\left( {x - 1} \right)}^2}}}}$ into partial fractions.

Solution.

Note that the denominator $$Q\left( x \right)$$ contains the irreducible quadratic factor $${{x^2} + x + 2}$$ and the linear factor $${{{\left( {x - 1} \right)}^2}}$$ with multiplicity $$2.$$ Hence, the partial fraction decomposition is written as

$\frac{{16}}{{\left( {{x^2} + x + 2} \right){{\left( {x - 1} \right)}^2}}} = \frac{{Ax + B}}{{{x^2} + x + 2}} + \frac{C}{{x - 1}} + \frac{D}{{{{\left( {x - 1} \right)}^2}}}.$

To determine the unknown coefficients $$A, B, C, D$$ we clear fractions multiplying both sides by the denominator. This gives

$16 = \left( {Ax + B} \right){\left( {x - 1} \right)^2} + C\left( {x - 1} \right)\left( {{x^2} + x + 2} \right) + D\left( {{x^2} + x + 2} \right),$
$16 = \left( {Ax + B} \right)\left( {{x^2} - 2x + 1} \right) + \left( {Cx - C} \right)\left( {{x^2} + x + 2} \right) + D{x^2} + Dx + 2D,$
$\color{blue}{16} = \color{magenta}{A{x^3}} + \color{darkgreen}{B{x^2}} - \color{darkgreen}{2A{x^2}} - \color{red}{2Bx} + \color{red}{Ax} + \color{blue}B + \color{magenta}{C{x^3}} - \cancel{\color{darkgreen}{C{x^2}}} + \cancel{\color{darkgreen}{C{x^2}}} - \color{red}{Cx} + \color{red}{2Cx} - \color{blue}{2C} + \color{darkgreen}{D{x^2}} + \color{red}{Dx} + \color{blue}{2D},$
$\color{blue}{16} = \left( {\color{magenta}A + \color{magenta}C} \right)\color{magenta}{x^3} + \left( { - \color{darkgreen}{2A} + \color{darkgreen}B + \color{darkgreen}D} \right)\color{darkgreen}{x^2} + \left( {\color{red}A - \color{red}{2B} + \color{red}C + \color{red}D} \right)\color{red}x + \color{blue}B - \color{blue}{2C} + \color{blue}{2D}.$

By equating coefficients of similar powers on both sides, we have

\left\{ \begin{align} \color{magenta}A + \color{magenta}C & = \color{magenta}0\\ - \color{darkgreen}{2A} + \color{darkgreen}B + \color{darkgreen}D & = \color{darkgreen}0\\ \color{red}A - \color{red}{2B} + \color{red}{C} + \color{red}D & = \color{red}0\\ \color{blue}B - \color{blue}{2C} + \color{blue}{2D} & = \color{blue}{16} \end{align} \right..

We solve the resulting system and find

$A = 3,\;\;B = 2,\;\;C = - 3,\;\;D = 4.$

Hence, the partial fraction decomposition of the rational function is given by

$\frac{{16}}{{\left( {{x^2} + x + 2} \right){{\left( {x - 1} \right)}^2}}} = \frac{{3x + 2}}{{{x^2} + x + 2}} - \frac{3}{{x - 1}} + \frac{4}{{{{\left( {x - 1} \right)}^2}}}.$

### Example 11.

Decompose $\frac{{{x^2} - 6x}}{{\left( {x - 1} \right)\left( {{x^2} + 2x + 2} \right)}}$ using the partial fractions.

Solution.

The denominator has the irreducible quadratic factor $${{x^2} + 2x + 2}.$$ Hence, the partial fraction decomposition has the form

$\frac{{{x^2} - 6x}}{{\left( {x - 1} \right)\left( {{x^2} + 2x + 2} \right)}} = \frac{A}{{x - 1}} + \frac{{Bx + C}}{{{x^2} + 2x + 2}}.$

By equating the numerators on both sides of the equation, we have

${x^2} - 6x = A\left( {{x^2} + 2x + 2} \right) + \left( {Bx + C} \right)\left( {x - 1} \right),$
${x^2} - 6x = A{x^2} + 2Ax + 2A + B{x^2} + Cx - Bx - C,$
${x^2} - 6x = \left( {A + B} \right){x^2} + \left( {2A + C - B} \right)x + \left( {2A - C} \right).$

We get the following system of equations:

$\left\{ \begin{array}{l} A + B = 1\\ 2A + C - B = - 6\\ 2A - C = 0 \end{array} \right..$

It has the solution

$A = -1,\;B = 2,\;C = - 2.$

Hence, the partial fraction decomposition is given by

$\frac{{{x^2} - 6x}}{{\left( {x - 1} \right)\left( {{x^2} + 2x + 2} \right)}} = \frac{{2x - 2}}{{{x^2} + 2x + 2}} - \frac{1}{{x - 1}}.$