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