Partial Fraction Expansion: Linear Factors
Partial Fraction Expansion (PFE) renames a fraction of polynomials using smaller, simpler ‘pieces’.
The preceding section introduces PFE, reviews all needed concepts, and presents a simple example (distinct linear factors).
This current section builds on that one, giving:
- A summary of the complete ‘how-to’ of PFE
- An example that shows how to handle repeated linear factors
Summary: The Complete ‘How-To’ of Partial Fraction Expansion
Notation
Let $\displaystyle\,R(x) := \frac{N(x)}{D(x)}\,$ be a rational function. More precisely:
- $N(x)\,$ (the Numerator) and $\,D(x)\,$ (the Denominator) are both polynomials with real number coefficients
- $D(x)\,$ is nonzero
Step 1: Check That the Degree of $\,N(x)\,$ is Strictly Less Than the Degree of $\,D(x)$
If not, then do a long division first, writing:
$$\cssId{s15}{\frac{N(x)}{D(x)} = Q(x) + \frac{R(x)}{D(x)}}$$One of two cases will occur:
- $R(x) = 0\,,$ in which case you don't need PFE
- $R(x) \ne 0\,,$ and the degree of $\,R(x)\,$ is strictly less than the degree of $\,D(x)\,.$ In this case, apply PFE to the new fraction, $\,\frac{R(x)}{D(x)}\,.$
Step 2: Completely Factor $\,D(x)\,$ Into Linear Factors and Irreducible Quadratics
Since $\,D(x)\,$ has real number coefficients, this can theoretically be done—though it may not be easy!
Step 3: Different Types of Factors in $\,D(x)\,$ Give Rise to Different Term(s) in the Partial Fraction Expansion
That is, you will write: $\frac{N(x)}{D(x)} = $ a sum of terms that are determined by the types of factors in $\,D(x)\,.$
The table below summarizes the term(s) required for each type of factor in $\,D(x)\,.$ Constants (like $\,\color{red}{A}\,$) that are colored in red are unknown constants, and will be solved for in Step 4.
| Type of Factor(s) in $\,D(x)$ |
| Corresponding Term(s) in the PFE |
|
Distinct Linear Factor: $ax + b$ where $\,a\ne 0$ |
|
A Single Term
and a
Single
Unknown Constant:
$\displaystyle\frac{\color{red}{A}}{ax+b}$ |
|
Repeated Linear Factor: $(ax + b)^n$ where $\,a\ne 0\,$ and $\,n = 2,3,4,\ldots$ |
|
$\,n\,$ Terms
and
$n\,$ Unknown Constants:
$\displaystyle\frac{\color{red}{A_1}}{ax+b} + \frac{\color{red}{A_2}}{(ax+b)^2} + \cdots + \frac{\color{red}{A_n}}{(ax+b)^n}$ |
|
Distinct Irreducible Quadratic Factor: $ax^2 + bx + c$ where $\,a\ne 0\,$ and $\,b^2 - 4ac \lt 0$ |
|
A Single Term and $\,2\,$ Unknown Constants: $\displaystyle\frac{\color{red}{A}x + \color{red}{B}}{ax^2+bx + c}$ |
|
Repeated Irreducible Quadratic Factor: $(ax^2 + bx + c)^n$ where $\,a\ne 0\,,$ $\,b^2 - 4ac \lt 0\,,$ and $\,n = 2,3,4,\ldots$ |
|
$\,n\,$ Terms and
$\,2n\,$ Unknown Constants:
$\displaystyle
\begin{align}
&\frac{\color{red}{A_1}x + \color{red}{B_1}}{ax^2 +bx + c}
+ \frac{\color{red}{A_2}x + \color{red}{B_2}}{(ax^2+bx + c)^2}\cr\cr
&\qquad + \cdots + \frac{\color{red}{A_n}x + \color{red}{B_n}}{(ax^2 +bx + c)^n}
\end{align}$
|
Step 4: Clear Fractions and Solve for the Unknown Constants
One or more of the following methods will be used:
-
When an equation is true for all values of $\,x\,,$ then it must be true for any particular value of $\,x\,.$
Choose value(s) that make all the unknown constants disappear, except one.
This method is easiest, when it is available.
-
When two polynomials are equal for all values of $\,x\,,$ then they must be equal term-by-term. That is: constant terms must be equal, $\,x\,$ terms must be equal, $\,x^2\,$ terms must be equal (and so on).
Equate coefficients of like terms, which usually leads to a system of equations to be solved for the unknowns.
Step 5: Report the Answer. Spot-Check to Gain Confidence in Your Result.
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PFE Example: Repeated Linear Factors
Note: Here, the denominator (a cubic polynomial) is already completely factored—rejoice! The denominator has one distinct linear factor, $\,x-3\,,$ and a repeated linear factor, $\,(x+1)^2\,.$
Step 1: Check That the Degree of the Numerator is Strictly Less Than the Degree of the Denominator
The degree of the numerator is $\,2\,.$ The degree of the denominator is $\,3\,.$ Check!
Step 2: Completely Factor the Denominator into Linear Factors and Irreducible Quadratics
Already done! Hooray!
Step 3: Different Types of Factors in the Denominator Give Rise to Different Term(s) in the Partial Fraction Expansion
The distinct linear factor $\,x-3\,$ gives rise to a single term in the PFE: $\displaystyle\,\frac{\color{red}{A}}{x-3}$
The repeated linear factor $\,(x+1)^2\,$ gives rise to two terms in the PFE: $\displaystyle\,\frac{\color{red}{B}}{x+1} + \frac{\color{red}{C}}{(x+1)^2}$
Of course, other names can be used for the unknown constants—but make sure they're all different! Thus, we have:
$$ \begin{align} &\cssId{s82}{\frac{3x^2 + 15x + 8}{(x+1)^2(x-3)}}\cr\cr &\qquad \cssId{s83}{= \frac{\color{red}{A}}{x-3} + \frac{\color{red}{B}}{x+1} + \frac{\color{red}{C}}{(x+1)^2}} \end{align} $$Step 4: Clear Fractions and Solve for the Unknown Constants
Clear fractions:
$$\begin{align} &\cssId{s87}{\frac{3x^2 + 15x + 8}{(x+1)^2(x-3)}\cdot(x+1)^2(x-3)}\cr &\quad \cssId{s88}{= \left(\frac{\color{red}{A}}{x-3} + \frac{\color{red}{B}}{x+1} + \frac{\color{red}{C}}{(x+1)^2}\right)\cdot(x+1)^2(x-3)} \end{align} $$$$ \begin{align} &\cssId{s89}{3x^2 + 15x + 8}\cr &\quad \cssId{s90}{= \color{red}{A}(x+1)^2 + \color{red}{B}(x+1)(x-3) + \color{red}{C}(x-3)} \quad \cssId{s91}{(\dagger)} \end{align} $$
(My husband Ray says that you can identify the scariest equations in math textbooks by the dagger ($\,\dagger\,$) beside them ☺)
Solve for the unknowns:
Let $x = -1\,$ in ($\,\dagger\,$):
$$ \begin{gather} \cssId{s95}{3(-1)^2 + 15(-1) + 8 = \color{red}{C}(-1-3)}\cr\cr \cssId{s96}{-4 = -4\color{red}{C}}\cr\cr \cssId{s97}{\color{red}{C} = 1} \end{gather} $$Let $x = 3\,$ in ($\,\dagger\,$):
$$\begin{gather} \cssId{s99}{3 \cdot 3^2 + 15\cdot 3 + 8 = \color{red}{A}(3+1)^2}\cr\cr \cssId{s100}{27 + 45 + 8 = 16\color{red}{A}}\cr\cr \cssId{s101}{80 = 16\color{red}{A}}\cr\cr \cssId{s102}{\color{red}{A} = 5} \end{gather} $$Only $\,\color{red}{B}\,$ remains to be found. Four different methods for finding $\,\color{red}{B}\,$ are shown below. In this example, the first method is easiest, and is preferred.
Of course, the same value of $\,\color{red}{B}\,$ is found by correct application of any of these methods. Understand them all, since different methods may be easier in different problems.
1. [Easy; the Preferred Method]
Equate $\,x^2\,$ Coefficients in
($\dagger$)
That is, the coefficient of $\,x^2\,$ on the left side of ($\dagger$) must equal the (combined) coefficient of $\,x^2\,$ on the right.
Notes:
- In the expression $\,3x^2 + 15x + 8\,,$ the coefficient of the $\,x^2\,$ term is $\,3\,.$
- In the expression $\,\color{red}{A}(x+1)^2\,,$ the coefficient of the $\,x^2\,$ term is $\,\color{red}{A}\,.$ This can be seen without multiplying anything out!
- In the expression $\,\color{red}{B}(x+1)(x-3)\,,$ the coefficient of the $\,x^2\,$ term is $\,\color{red}{B}\,.$ This can be seen without multiplying anything out!
- The expression $\,\color{red}{C}(x-3)\,$ does not have an $\,x^2\,$ term.
2. [Slightly Harder than (1)]
Equate Constant Terms in
($\dagger$)
That is, the constant term on the left side of ($\dagger$) must equal the constant term on the right.
Notes:
- In the expression $\,3x^2 + 15x + 8\,,$ the constant term is $\,8\,.$
- In the expression $\,\color{red}{A}(x+1)^2\,,$ the constant term is $\,\color{red}{A}\,.$ This can be seen without multiplying anything out!
- In the expression $\,\color{red}{B}(x+1)(x-3)\,,$ the constant term is $\,-3\color{red}{B}\,.$ This can be seen without multiplying anything out!
- In the expression $\,\color{red}{C}(x-3)\,,$ the constant term is $\,-3\color{red}{C}\,.$ This can be seen without multiplying anything out!
3. [Slightly Harder than (1)]
Equate $\,x\,$ Coefficients in
($\dagger$)
Notes:
- In the expression $\,3x^2 + 15x + 8\,,$ the coefficient of the $\,x\,$ term is $\,15\,.$
- In the expression $\,\color{red}{A}(x+1)^2\,,$ the coefficient of the $\,x\,$ term is $\,2\color{red}{A}\,,$ since $\,(x+1)^2 = x^2 + 2x + 1\,.$
- In the expression $\,\color{red}{B}(x+1)(x-3)\,,$ the coefficient of the $\,x\,$ term is $\,-2\color{red}{B}\,,$ since $\,(x+1)(x-3) = x^2 - 2x - 3\,.$
- In the expression $\,\color{red}{C}(x-3)\,,$ the coefficient of the $\,x\,$ term is $\,\color{red}{C}\,.$
4. [Slightly Harder than (1)]
In
($\dagger$),
Let $\,x\,$ Be Any Number Other Than
$\,-1\,$ or $\,3$
Here, we choose $\,x = 0\,$:
Step 5: Report the Answer. Spot-Check to Gain Confidence in Your Result.
For a spot-check, let $\,x = 0\,$:
$$ \begin{gather} \cssId{s161}{-\frac{8}{3} \overset{?}{=} -\frac{5}{3} - 2 + 1}\cr\cr \cssId{s162}{-\frac 83 = -\frac 83} \quad \cssId{s163}{\text{Check!}} \end{gather} $$
Concept Practice
- Choose a specific problem type, or click ‘New problem’ for a random question.
- Think about your answer.
- Click ‘Check your answer’ to check!