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:

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.

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.

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:

Step 4:  Clear Fractions and Solve for the Unknown Constants

Clear fractions:

(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 ($\,\dagger\,$):

Let $x = 3\,$ in ($\,\dagger\,$):

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 $\,x = 0\,$: