Why Denominator Structure Matters
Partial fraction decomposition rewrites a rational function as a sum of simpler rational functions. The denominator determines the kinds of terms that can appear in that sum. A real linear factor leads to a constant numerator; an irreducible quadratic factor leads to a numerator of degree at most one. When a factor is repeated, the decomposition includes terms for its successive powers.
These forms are not guesses about which expression will be easiest to integrate. They follow from polynomial division and the structure of the denominator. First, however, the rational function must be proper: its numerator must have degree less than its denominator.
Proper and Improper Rational Functions
The degree of a nonzero polynomial is its highest exponent with a nonzero coefficient. For example, \(3x^2-1\) has degree \(2\), while \(x^3+2x\) has degree \(3\). Properness compares degrees; it does not mean that the fraction’s numerical value is less than one.
If numerator and denominator have a common polynomial factor, cancel it first when describing the rational function’s reduced form. The denominator whose factors determine the decomposition is the denominator after any such cancellation.
Worked Example: Classifying Proper and Improper Functions
Classify \(f(x)=(2x+1)/(x^2-4)\). The numerator has degree \(1\), and the denominator has degree \(2\). Since \(1<2\), the function is proper.
Now consider \(g(x)=(x^2+1)/(x-1)\). Its numerator has degree \(2\), while its denominator has degree \(1\). Since \(2\geq1\), it is improper.
Check: The comparison uses only the polynomial degrees. In particular, it does not depend on choosing a value of \(x\) and comparing the values of the two functions.
Polynomial Division Comes First
Polynomial long division writes any quotient of polynomials in the form \(S(x)+R(x)/Q(x)\), where \(S\) is a polynomial and either \(R=0\) or \(\deg R<\deg Q\). Thus an improper rational function has a polynomial part as well as a proper rational part. Partial fraction terms are used for the proper part; the polynomial part remains separate.
Worked Example: Separating a Polynomial Part
Divide \(x^2+1\) by \(x-1\). The first quotient term is \(x\), since \(x^2/x=x\). Subtracting \(x(x-1)=x^2-x\) leaves \(x+1\). The next quotient term is \(1\); subtracting \(1(x-1)=x-1\) leaves remainder \(2\). Therefore:
Divide both sides by \(x-1\), for \(x\ne1\):
Check: Combining the right-hand side gives \(\big((x+1)(x-1)+2\big)/(x-1)=(x^2-1+2)/(x-1)=(x^2+1)/(x-1)\), as required. The remainder term is proper because its numerator has degree \(0\), less than the denominator’s degree \(1\).
How Real Factors Set the Partial Fraction Forms
Suppose a proper rational function has been written with a denominator factored over the real numbers. Its factors are real linear polynomials or irreducible real quadratic polynomials. An irreducible quadratic here means a quadratic that has no real linear factors. The factorization is taken with each distinct factor listed once and its multiplicity recorded.
For instance, if a linear factor is \(x-a\), the numerator over each of its powers is a constant \(A\). If an irreducible quadratic is \(x^2+bx+c\), the numerator over each of its powers has the form \(Bx+C\). The coefficients are determined by the rational function; the factor type determines the shape of the numerator.
For repeated factors, the exponent \(j\) runs from \(1\) through the factor’s multiplicity. For example, a factor \((x-a)^3\) calls for terms over \(x-a\), \((x-a)^2\), and \((x-a)^3\). An irreducible quadratic squared calls for one term over the quadratic and another over its square. The coefficient in a particular term may turn out to be zero, but the full form includes all the indicated powers.
Why the Numerator Degrees and Term Types Are Correct
The degree restriction on each numerator comes from polynomial division: dividing a polynomial by a factor \(p\) leaves a remainder of degree less than \(\deg p\). The theorem’s form follows by applying that principle to each factor power. The following argument also establishes that the listed forms are sufficient for every proper rational function with the stated denominator.
Write \(d=\deg D=\sum_{i=1}^{r}m_i\deg p_i\). For each power \(p_i^j\), a numerator of degree less than \(\deg p_i\) has \(\deg p_i\) coefficients. The total number of coefficients is therefore \(d\).
Multiplying a term \(R_{ij}/p_i^j\) by \(D\) gives \(R_{ij}(D/p_i^j)\). Its degree is at most \((\deg p_i-1)+(d-j\deg p_i)\), which is at most \(d-1\). Thus every proposed sum has a numerator of degree less than \(d\).
Suppose a proposed sum is zero and some numerator for a fixed factor \(p_i\) is nonzero. Choose the largest index \(j\) for which \(R_{ij}\ne0\), and put \(H_i=\prod_{k\ne i}p_k^{m_k}\). Multiply the identity by \(p_i^jH_i\). Reducing the resulting polynomial identity modulo \(p_i\) removes all terms except \(R_{ij}H_i\): lower powers for this factor contain \(p_i\), higher-power coefficients are zero, and terms for other factors contain \(p_i^j\). Hence \(p_i\) divides \(R_{ij}H_i\). Because \(p_i\) is distinct from every factor in \(H_i\), it is relatively prime to \(H_i\); therefore \(p_i\) divides \(R_{ij}\). This contradicts \(\deg R_{ij}<\deg p_i\). So every numerator must be zero.
The coefficient choices form a \(d\)-dimensional real vector space. The numerators they produce lie in the \(d\)-dimensional space of polynomials of degree less than \(d\). The zero-sum argument proves that the linear map between these spaces is one-to-one. A one-to-one linear map between finite-dimensional spaces of equal dimension is onto, so every numerator of degree less than \(d\) is produced. This proves the stated form and its uniqueness.
This justification explains both parts of the rule. The degree bound on a numerator is tied to the factor over which it appears, and the repeated powers are included as distinct terms in the full form. It does not provide a procedure for calculating the coefficients; the examples below focus on identifying and checking the forms.
Worked Decompositions and Checks
Worked Example: Two Distinct Linear Factors
Consider \(1/((x-1)(x+2))\). The denominator has two distinct linear factors, each appearing once, so each partial fraction term has a constant numerator. A decomposition is:
To check it, combine the right-hand terms over their common denominator:
The check confirms the identity for all \(x\) where the original denominator is nonzero, namely \(x\ne1,-2\).
Worked Example: A Repeated Linear Factor
Consider \((2x+1)/(x-1)^2\). The numerator degree is \(1\), less than the denominator degree \(2\), so the function is proper. The factor \(x-1\) is repeated twice. The form therefore has a constant over \(x-1\) and a constant over \((x-1)^2\). One decomposition is:
Combine the terms to check:
The recovered numerator is \(2x+1\), and the denominator is the original \((x-1)^2\).
Worked Example: A Repeated Irreducible Quadratic
Consider \((x^2+2x)/(x^2+1)^2\). The numerator has degree \(2\), while the denominator has degree \(4\), so the function is proper. The denominator contains a repeated irreducible quadratic. Each term therefore has a linear-or-constant numerator, and the terms use the first and second powers of the quadratic. For this example, a decomposition is:
Check by putting the right-hand side over the common denominator:
The numerator over the squared quadratic, \(2x-1\), has degree \(1\), as required. The numerator over the first power is the constant \(1\), which also meets the degree bound.
Worked Example: A Linear Factor and an Irreducible Quadratic
Suppose the denominator is already written as \((x-1)(x^2+1)\). The linear factor requires a constant numerator, and the quadratic requires a numerator of degree at most \(1\). The following identity illustrates both forms:
Verify by combining the right-hand side. The numerator from the first term is \(2(x^2+1)=2x^2+2\). The numerator from the second is \((x+3)(x-1)=x^2+2x-3\). Adding gives:
Thus the right-hand side has precisely the original numerator and denominator. The original function is proper because its numerator has degree \(2\) and its denominator has degree \(3\).
A Common Mistake
A frequent error is to put a linear numerator over a linear factor. For example, one might propose \( (Ax+B)/(x-1)\) as the term associated with a single linear factor. That is not the required partial fraction form: since the linear factor has degree \(1\), its numerator must have degree less than \(1\), hence be constant. The correct term is \(A/(x-1)\).
For a quadratic factor, the reverse mistake is common: using only a constant numerator. A constant is allowed, but it is not the full general form. Since a quadratic has degree \(2\), its numerator may have degree \(0\) or \(1\), so the general form is \((Bx+C)/(x^2+bx+c)\). Restricting it to a constant could leave some proper rational functions unrepresented.
Why These Forms Matter
The forms identify what a decomposition must contain before any coefficients are calculated. This prevents missing terms, especially when factors are repeated, and prevents numerators that are too large in degree. It also makes clear why properness matters: if the original numerator is at least as large in degree as the denominator, polynomial division must first separate the polynomial part.
When the factorization is known, a useful checklist is:
- Confirm the rational function is proper, or divide out its polynomial part first.
- Use one constant-numerator term for every power of each linear factor.
- Use one degree-at-most-one numerator term for every power of each irreducible quadratic factor.
- Check a proposed decomposition by recombining it over the common denominator.
Check Your Understanding
Use the factorization and degree rules from this tutorial to answer the questions.
- Is \((x^2+1)/(x^3-2)\) proper or improper? Give the two degrees that determine your answer.
- What numerator form belongs over a single factor \(x+4\)? What numerator form belongs over a single irreducible quadratic?
- What terms are required by the repeated factor \((x-2)^3\)?
- What terms are required by the repeated factor \((x^2+4)^2\), assuming \(x^2+4\) is irreducible over the reals?
- Why must an improper rational function be divided into a polynomial part and a proper rational part before applying the partial fraction forms?
