﻿ Partial Fraction Expansion: Linear Factors

# Partial Fraction Expansion: Linear Factors

• PRACTICE (online exercises and printable worksheets)

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 $\,\displaystyle \frac{N(x)}{D(x)} = Q(x) + \frac{R(x)}{D(x)}\,$.
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, $\,\displaystyle\frac{R(x)}{D(x)}\,$.
• 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: $$\frac{N(x)}{D(x)} = \text{a sum of terms that are determined by the types of factors in \,D(x)\,}$$ 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:

$\displaystyle\frac{\color{red}{A_1}}{ax+b} + \frac{\color{red}{A_2}}{(ax+b)^2} + \cdots + \frac{\color{red}{A_n}}{(ax+b)^n}$

$\,ax^2 + bx + c\,$, where $\,a\ne 0\,$ and $\,b^2 - 4ac < 0\,$
a single term and $\,2\,$ unknown constants:

$\displaystyle\frac{\color{red}{A}x + \color{red}{B}}{ax^2+bx + c}$

$\,(ax^2 + bx + c)^n\,$, where $\,a\ne 0\,$, $\,b^2 - 4ac < 0\,$, and $\,n = 2,3,4,\ldots\,$
$\,n\,$ terms and $\,2n\,$ unknown constants:

$\displaystyle\frac{\color{red}{A_1}x + \color{red}{B_1}}{ax^2 +bx + c} + \frac{\color{red}{A_2}x + \color{red}{B_2}}{(ax^2+bx + c)^2} + \cdots + \frac{\color{red}{A_n}x + \color{red}{B_n}}{(ax^2 +bx + c)^n}$
• 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:

## PFE Example: Repeated Linear Factors

Find the partial fraction expansion of $\displaystyle\,\frac{3x^2 + 15x + 8}{(x+1)^2(x-3)}\,$.

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.
• 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: $$\frac{3x^2 + 15x + 8}{(x+1)^2(x-3)} = \frac{\color{red}{A}}{x-3} + \frac{\color{red}{B}}{x+1} + \frac{\color{red}{C}}{(x+1)^2}$$
• STEP 4:
Clear fractions and solve for the unknown constants.
Clear fractions: \begin{align} \frac{3x^2 + 15x + 8}{(x+1)^2(x-3)}\cdot(x+1)^2(x-3) \ \ &=\ \ \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)\cr\cr 3x^2 + 15x + 8 \ \ &=\ \ \color{red}{A}(x+1)^2 + \color{red}{B}(x+1)(x-3) + \color{red}{C}(x-3)\qquad(\dagger)\cr\cr \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:
$x = -1\,$: $$\begin{gather} 3(-1)^2 + 15(-1) + 8 = \color{red}{C}(-1-3)\cr -4 = -4\color{red}{C}\cr \color{red}{C} = 1 \end{gather}$$ $x = 3\,$: $$\begin{gather} 3 \cdot 3^2 + 15\cdot 3 + 8 = \color{red}{A}(3+1)^2\cr 27 + 45 + 8 = 16\color{red}{A}\cr 80 = 16\color{red}{A}\cr \color{red}{A} = 5 \end{gather}$$ 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.
$$\begin{gather} 3 = \color{red}{A} + \color{red}{B}\cr 3 = 5 + \color{red}{B}\cr \color{red}{B} = -2 \end{gather}$$
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!
$$\begin{gather} 8 = \color{red}{A} - 3\color{red}{B} - 3\color{red}{C}\cr 8 = 5 - 3\color{red}{B} - 3(1)\cr 6 = -3\color{red}{B}\cr \color{red}{B} = -2 \end{gather}$$
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 $\,\color{red}{C}\,$.
$$\begin{gather} 15 = 2\color{red}{A} -2\color{red}{B} + \color{red}{C}\cr 15 = 2\cdot 5 -2\color{red}{B} + 1\cr 4 = -2\color{red}{B}\cr \color{red}{B} = -2 \end{gather}$$
4. [slightly harder than (1)]
In ($\dagger$), let $\,x\,$ be any number other than $\,-1\,$ or $\,3\,$.
Here, we choose $\,x = 0\,$: $$\begin{gather} 3\cdot 0^2 + 15\cdot 0 + 8 = \color{red}{A}(0+1)^2 + \color{red}{B}(0+1)(0-3) + \color{red}{C}(0 - 3)\cr 8 = \color{red}{A}(1) + \color{red}{B}(-3) + \color{red}{C}(-3)\cr 8 = (5)(1) + \color{red}{B}(-3) + (1)(-3)\cr 6 = -3\color{red}{B}\cr \color{red}{B} = -2 \end{gather}$$
• STEP 5:
$$\frac{3x^2 + 15x + 8}{(x+1)^2(x-3)} = \frac{5}{x-3} - \frac{2}{x+1} + \frac{1}{(x+1)^2}$$ For a spot-check, let $\,x = 0\,$: $$\begin{gather} -\frac{8}{3} \overset{?}{=} -\frac{5}{3} - 2 + 1\cr -\frac 83 = -\frac 83 \qquad \text{Check!} \end{gather}$$
Master the ideas from this section