Partial Fraction Expansion (PFE) renames a fraction of polynomials (i.e., a rational function), using smaller, simpler ‘pieces’. This is the third of three sections covering PFE:
The steps indicated in the examples below follow the summary of the previous section.
Find the partial fraction expansion of $\displaystyle \frac{2x^4 + 4x^2 - 5x + 2}{x^3 - 1} \,$.
The equation that the unknowns ($\,A\,$, $\,B\,$ and $\,C\,$) must satisfy is repeated here for your convenience: $$ \cssId{s67}{4x^2 - 3x + 2 = A(x^2 + x + 1) + (Bx + C)(x-1)} \qquad \cssId{s68}{(\ddagger)} $$ Here's the same equation with the right-hand side in standard form, so it's easier to compare the coefficients of each term type. In practice, however, try to ‘see’ all this from ($\,\ddagger\,$), without multiplying it out. $$ \cssId{s71}{4x^2 - 3x + 2 = x^2(A + B) + x(A + C - B) + (A - C)} $$ Recall that it has already been determined that $\,A = 1\,$. | |||
(good) Equate $\,x^2\,$ coefficients in ($\,\ddagger\,$) to solve for $\,B\,$. $$ \begin{gather} \cssId{s75}{4 = A + B}\cr \cssId{s76}{4 = 1 + B}\cr \cssId{s77}{\color{green}{B = 3}} \end{gather} $$ Equate constant terms in ($\,\ddagger\,$) to solve for $\,C\,$: $$ \begin{gather} \cssId{s79}{2 = A - C}\cr \cssId{s80}{2 = 1 - C}\cr \cssId{s81}{\color{red}{C = -1}} \end{gather} $$ |
(okay) Equate constant terms in ($\,\ddagger\,$) to solve for $\,C\,$: $$ \begin{gather} \cssId{s84}{2 = A - C}\cr \cssId{s85}{2 = 1 - C}\cr \cssId{s86}{\color{red}{C = -1}} \end{gather} $$ Equate $\,x\,$ coefficients in ($\,\ddagger\,$) to solve for $\,B\,$. $$ \begin{gather} \cssId{s88}{-3 = A + C - B}\cr \cssId{s89}{-3 = 1 - 1 - B}\cr \cssId{s90}{-3 = -B}\cr \cssId{s91}{\color{green}{B = 3}} \end{gather} $$ |
(good) Let $\,x = 0\,$ in ($\,\ddagger\,$): $$\begin{gather} \cssId{s94}{2 = A - C}\cr \cssId{s95}{2 = 1 - C}\cr \cssId{s96}{\color{red}{C = -1}} \end{gather} $$ Equate $\,x^2\,$ coefficients in ($\,\ddagger\,$) to solve for $\,B\,$. $$ \begin{gather} \cssId{s98}{4 = A + B}\cr \cssId{s99}{4 = 1 + B}\cr \cssId{s100}{\color{green}{B = 3}} \end{gather} $$ |
(horrible) Equate $\,x\,$ coefficients in ($\,\ddagger\,$): $$ \begin{gather} \cssId{s103}{-3 = A + C - B}\cr \cssId{s104}{-3 = 1 + C - B}\cr \cssId{s105}{-4 = C - B} \end{gather} $$ Let $\,x = 2\,$ (arbitrarily chosen) in ($\,\ddagger\,$): $$ \begin{gather} \cssId{s107}{4\cdot 2^2 - 3\cdot 2 + 2 = A(2^2 + 2 + 1) + (B\cdot 2 + C)(2-1)}\cr \cssId{s108}{12 = 7A + 2B + C}\cr \cssId{s109}{12 = 7 + 2B + C}\cr \cssId{s110}{5 = 2B + C} \end{gather} $$ Solve the system of equations: $$\begin{gather} \cssId{s112}{-4 = C - B}\cr \cssId{s113}{5 = 2B + C} \end{gather} $$ Using substitution, the first equation gives $\,C = -4 + B\,$. Substitution into the second equation gives $\,5 = 2B - 4 + B\,$. Solving for $\,B\,$ gives $\,9 = 3B\,$, so $\,\color{green}{B = 3}\,$. Then, the first equation gives $\,-4 = C - 3\,$, so $\,\color{red}{C = -1}\,$. |
Find the partial fraction expansion of $\displaystyle\,\frac{3x^2 - 5}{x^4 + 6x^2 + 9}\,$.
(The following solution is much more compact than the prior example.)
On this exercise, you will not key in your answer. However, you can check to see if your answer is correct. |
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