Why Verify a Decomposition?
A proposed partial fraction decomposition is a claim that a rational function can be written as a sum of simpler rational functions. A direct way to check that claim is to reverse the process: place the proposed terms over a common denominator and compare the resulting numerator with the original one.
This check is useful because it tests the complete proposed equality, not just a few selected values of \(x\). It also works whether the denominator contains distinct linear factors, repeated factors, irreducible quadratic factors, or a mixture of these. The method here is verification: the terms are given, and we test whether they recombine to the stated rational function.
The Recombination Rule
Suppose the original rational function is \(P(x)/D(x)\), and a proposed decomposition has terms \(p_i(x)/d_i(x)\). Choose a common denominator \(M(x)\) for the original function and all proposed terms. For each term, multiply its numerator by the factor needed to turn its denominator into \(M(x)\).
Adding the rewritten terms gives one fraction with denominator \(M(x)\). If the original function also has denominator \(M(x)\), compare the two numerators. When the original denominator is already the least common denominator of the proposed terms, use it directly. If a proposed term has a denominator that does not divide the original denominator, a common denominator may need to be larger.
The order of work matters: first identify the common denominator; then multiply each numerator by its own missing factor; then add and simplify the numerators. The missing factor depends on that term’s denominator, not on the other numerators.
Why the Numerator Identity Is Enough
Here is the justification behind the test. Suppose the original function and the proposed sum have been written over the same denominator \(D(x)\). Let \(P(x)\) be the original numerator and let \(Q(x)\) be the numerator obtained by recombining the proposed terms. At every \(x\) with \(D(x)\ne 0\), their difference is
If \(P(x)=Q(x)\) as polynomials, the numerator of this difference is zero. Therefore the rational functions have the same value at every point in their shared domain. Conversely, if the rational functions agree at every point of their shared domain, then \(P(x)-Q(x)=0\) at infinitely many values of \(x\). A nonzero polynomial has only finitely many roots, so \(P-Q\) must be the zero polynomial. Thus equality of the rational functions is equivalent to the numerator identity, when they are put over the same denominator.
The domain condition is essential. A rational expression is undefined where its denominator is zero. A verified decomposition does not make those excluded inputs valid, even if some algebraic simplification appears to cancel a factor.
Worked Examples
Worked Example: Verifying Two Distinct Linear Factors
Check the proposed decomposition
The common denominator is \((x-1)(x+2)\). The first term needs the factor \(x+2\), and the second needs the factor \(x-1\). Recombine:
The recombined numerator is \(5x+1\), exactly the original numerator. Thus the proposed decomposition is verified for \(x\ne 1\) and \(x\ne -2\). Substituting the recombined expression back into the original claim gives the original fraction, so the check is complete.
Worked Example: Verifying a Term with a Repeated Factor
Check the proposed decomposition
Use \((x-1)^2(x+2)\) as the common denominator. The term with denominator \(x-1\) needs the factor \((x-1)(x+2)\). The term with denominator \((x-1)^2\) needs \(x+2\). The term with denominator \(x+2\) needs \((x-1)^2\). Therefore,
Expand each numerator contribution:
Add the contributions by powers of \(x\): the \(x^2\) coefficient is \(2+3=5\), the \(x\) coefficient is \(2-1-6=-5\), and the constant is \(-4-2+3=-3\). Hence
This is the original numerator, so the decomposition is verified for \(x\ne 1\) and \(x\ne -2\). Recombining all three terms reproduces the stated rational function with its original denominator.
Worked Example: Verifying a Linear and an Irreducible Quadratic Factor
Check the proposed decomposition
The common denominator is \((x^2+1)(x-2)\). The first term is missing the factor \(x-2\); the second is missing \(x^2+1\). Recombine and simplify:
The numerator identity holds: \((x+1)(x-2)+2(x^2+1)=3x^2-x\). Since \(x^2+1\) is nonzero for real \(x\), the only excluded real input is \(x=2\). The recombined expression is the original function throughout its real domain, verifying the claim.
A Common Pitfall: Forgetting the Missing Factor
When putting a term over a common denominator, multiplying only its denominator is not enough. Its numerator must be multiplied by the same missing factor. For example, in the first worked example the term \(3/(x+2)\) is missing the factor \(x-1\). A mistaken calculation might write
This does not correctly rewrite the second term: its numerator \(3\) must also be multiplied by \(x-1\). The corrected calculation is
A useful check is to divide each new numerator contribution by the common denominator mentally: \(2(x+2)/[(x-1)(x+2)]=2/(x-1)\), and \(3(x-1)/[(x-1)(x+2)]=3/(x+2)\). Each term must return to its original form.
A Reliable Verification Sequence
Use this sequence whenever a proposed decomposition needs checking. It keeps the focus on whether the entire equality holds and makes the algebra easy to audit.
Keep its numerator and denominator visible so the final comparison is unambiguous.
Usually this is the factored denominator of the original function when it is also a common denominator for the proposed terms.
For each term, multiply its numerator by exactly the factor missing from its denominator.
Add the resulting numerator contributions and collect like powers of \(x\).
Confirm that the numerator matches the original polynomial, and exclude every real input that makes a denominator zero.
Why This Check Matters
Coefficient work can involve several equations and many algebraic steps. Recombination provides an independent final check: it tests whether the proposed terms, taken together, reproduce the numerator and denominator of the starting function. It is also a useful way to present a result clearly, because each term’s contribution to the combined numerator is visible.
Do not treat agreement at a few sample values as the verification. Two different rational functions can happen to have the same value at particular inputs. The numerator identity establishes equality across the full shared domain. Likewise, a numerator mismatch means the proposed equality has not passed the check; verifying the decomposition does not require explaining how the mismatch arose.
Check Your Understanding
For each question, show the recombination steps and compare the resulting numerator with the original one.
- Verify or reject \(\frac{4x+1}{(x-1)(x+3)}=\frac{1}{x-1}+\frac{3}{x+3}\).
- What factor is needed to rewrite \(\frac{5}{(x+2)^2}\) over the denominator \((x+2)^2(x-1)\)? What must happen to its numerator?
- Recombine \(\frac{2}{x-1}+\frac{1}{(x-1)^2}\) over the common denominator \((x-1)^2\), and state the resulting numerator.
- Why does an identical numerator establish equality only where the common denominator is nonzero?
- In \(\frac{1}{x+1}+\frac{2}{x-2}\), identify the missing factor for each term when the common denominator is \((x+1)(x-2)\).
