Why Uniqueness Matters
A partial-fraction decomposition is useful only if its terms and coefficients are determined reliably. Recombining a proposed decomposition checks that it gives the original rational function, but a further question remains: could a different set of coefficients give the same function using the same factors?
For a proper rational function, the answer is no when the denominator factorization is fixed and each numerator has the required degree. This uniqueness means that once a valid complete decomposition has been found, its coefficients are not arbitrary choices. It also makes coefficient-solving methods meaningful: any correct method must produce the same coefficients.
The Form of a Complete Decomposition
Work over the real numbers. Suppose a proper rational function has denominator
where the \(p_i\) are distinct monic irreducible real polynomials and each \(m_i\) is a positive integer. Over the real numbers, each such factor is either linear, \(x-a\), or an irreducible quadratic. A factor is irreducible if it cannot be factored into lower-degree polynomials with real coefficients.
The complete partial-fraction form includes one term for every power of every factor. For a linear factor, the numerator is a constant. For an irreducible quadratic, the numerator is linear. In general, the numerator over \(p_i^k\) is a polynomial of degree less than \(\deg p_i\).
Here the rational function is proper, so \(\deg P<\deg D\), and the displayed identity is understood wherever the denominators are nonzero. A zero numerator is allowed. For example, if \(p_i(x)=x-a\), then \(R_{i,k}\) is a constant. If \(p_i\) is quadratic, then \(R_{i,k}(x)=A_{i,k}x+B_{i,k}\), with constants \(A_{i,k}\) and \(B_{i,k}\).
The degree restriction is part of the form, not an optional convenience. It prevents a numerator from containing a whole copy of its denominator factor, which could otherwise be moved between terms.
Uniqueness Theorem
The theorem is about uniqueness for a fixed factorization. It does not say that every proposed list of coefficients is correct, nor does it provide a method for finding those coefficients. Checking a decomposition by recombining terms establishes that it is valid; the theorem then guarantees that no other valid complete decomposition for the same fixed factors can differ.
Why the Decomposition Is Unique
Suppose there are two complete decompositions of the same rational function. Subtract them. For each \(i\) and \(k\), let \(S_{i,k}(x)\) be the difference between the two numerators over \(p_i^k\). Then \(\deg S_{i,k}<\deg p_i\), and the difference of the decompositions is zero:
Multiplying by \(D(x)\) gives a polynomial identity. To show that all the differences \(S_{i,k}\) vanish, focus first on one factor \(p_i\) and its highest power \(p_i^{m_i}\).
Multiply the zero identity by \(D(x)\). Each resulting term is a polynomial, so the result is a polynomial identity that can be considered modulo \(p_i\).
Every term from a different factor \(p_j\) contains \(p_i^{m_i}\), so it is divisible by \(p_i\). Terms from \(p_i^k\) with \(k<m_i\) contain \(p_i^{m_i-k}\), so they too are divisible by \(p_i\). The only term that can remain is \(S_{i,m_i}(x)D(x)/p_i(x)^{m_i}\).
The factor \(D/p_i^{m_i}\) has no factor \(p_i\), so it is relatively prime to \(p_i\). The reduced identity therefore implies that \(p_i\) divides \(S_{i,m_i}\).
Since \(\deg S_{i,m_i}<\deg p_i\), divisibility by \(p_i\) is possible only if \(S_{i,m_i}=0\). Repeat the argument for \(p_i^{m_i-1}\), then \(p_i^{m_i-2}\), down to \(p_i\). This proves that every \(S_{i,k}\) is zero.
For clarity, the divisibility step uses a basic property of polynomials: if \(p_i\) divides a product and is relatively prime to one factor of that product, then \(p_i\) divides the other factor. Here, reducing the polynomial identity modulo \(p_i\) gives
Because the second factor is relatively prime to \(p_i\), it follows that \(p_i\) divides \(S_{i,m_i}\). But a nonzero polynomial divisible by \(p_i\) has degree at least \(\deg p_i\), contradicting the strict degree bound on \(S_{i,m_i}\). Thus \(S_{i,m_i}=0\). Once that highest-power difference is zero, the same reasoning applies to the next-highest power. Repeating it proves that every difference is zero, which is the claimed uniqueness.
Worked Examples
Worked Example: Uniqueness with Two Linear Factors
Consider the proposed decomposition
Multiplying by \((x-1)(x+2)\) gives the polynomial identity
Equating the coefficients of \(x\) and the constant terms gives \(A+B=3\) and \(2A-B=5\). Adding these equations gives \(3A=8\), so \(A=\frac{8}{3}\). Then \(B=3-\frac{8}{3}=\frac{1}{3}\).
Check by recombining the terms:
Thus the decomposition is valid. Since the factors \(x-1\) and \(x+2\) are distinct and the numerators are constants, the uniqueness theorem says that no other constants \(A\) and \(B\) can give a valid decomposition with these factors.
Worked Example: A Linear Factor and an Irreducible Quadratic
Find the coefficients in this complete form:
The quadratic \(x^2+1\) is irreducible over the real numbers because it has no real roots. Multiplying by the denominator gives
Expand the right-hand side:
Equating coefficients gives \(A+B=1\), \(C-B=4\), and \(A-C=-1\). From the last equation, \(C=A+1\). The middle equation then gives \(B=C-4=A-3\). Substituting into the first equation gives \(A+(A-3)=1\), so \(2A=4\) and \(A=2\). Therefore \(B=-1\) and \(C=3\).
Check by recombining:
The recombination verifies the proposed coefficients. The degree restrictions are satisfied: the numerator \(A\) over a linear factor is constant, and \(Bx+C\) over the quadratic has degree at most one. The theorem ensures these coefficients are the only ones for this fixed factorization.
Worked Example: Uniqueness with a Repeated Linear Factor
Consider the complete form for the denominator \((x-1)^2(x+1)\):
Multiplying by the denominator gives
Expanding each term yields
Equating coefficients gives \(A+C=0\), \(B-2C=0\), and \(-A+B+C=1\). The first equation gives \(C=-A\), and the second gives \(B=2C=-2A\). The constant equation becomes \(-A-2A-A=1\), so \(-4A=1\) and \(A=-\frac14\). Hence \(C=\frac14\) and \(B=\frac12\).
Check the result by substitution into the recombined numerator:
Indeed, the \(x^2\) terms cancel, the \(x\) terms \(-\frac12x+\frac12x\) cancel, and the constants \(\frac14+\frac12+\frac14\) sum to \(1\). The repeated factor requires a term for each power, and the theorem guarantees that these three coefficients are unique.
A Common Pitfall: Dropping the Degree Restriction
The numerator degree bounds are essential. If they are ignored, even a fixed list of denominator powers can allow different-looking expressions for the same function. For example, consider the fixed denominator \((x^2+1)^2\). The rational function \(1/(x^2+1)\) can be written in either of these ways:
These are not two different admissible complete decompositions. For a quadratic factor, the numerator over each power must have degree less than \(2\). The numerator \(x^2+1\) in the second expression has degree \(2\), so that expression violates the required form. Calling it a valid alternative would incorrectly suggest that uniqueness fails.
Another mistake is to treat a repeated factor as though its powers were interchangeable or to omit a power from the complete form. The uniqueness result applies to the specified complete form: every power is included, and every numerator satisfies its degree bound. Those conditions are what make the coefficient list well-defined.
Why the Result Is Useful
Uniqueness gives a firm basis for checking and comparing decompositions. If two valid calculations use the same fixed factorization, their coefficients must agree, even if the calculations took different routes. If their results disagree, at least one calculation contains an error or one expression does not meet the required form.
The theorem also clarifies what “the” partial-fraction decomposition means. It is not merely one convenient representation among many admissible ones: once the factorization and the complete form are fixed, there is at most one. The theorem proves at most one; whenever a valid decomposition has been found, it is therefore the unique one in that form.
Check Your Understanding
Use the stated factorization and degree restrictions when answering these questions.
- For a linear factor \(x-a\), what degrees are allowed for the numerator over \((x-a)^k\)?
- For an irreducible quadratic \(q(x)\), what is the largest possible degree of the numerator over \(q(x)^k\)?
- In the uniqueness proof, why does reducing modulo \(p_i\) eliminate terms belonging to other factors \(p_j\)?
- Why does \(p_i\mid S_{i,m_i}\), together with \(\deg S_{i,m_i}<\deg p_i\), imply \(S_{i,m_i}=0\)?
- Why is \((x^2+1)/(x^2+1)^2\) not an admissible numerator term in the complete form for a repeated irreducible quadratic?
