Why Check by Recombining?
A proposed partial-fraction decomposition may look plausible while containing a wrong coefficient, sign, or factor. Recombining the fractions gives a direct check: put them over a common denominator, simplify the numerator, and compare the result with the numerator of the original rational expression.
This check is useful after finding coefficients, but it is not another method for finding them. Its purpose is to test whether the proposed expression is actually equal to the original one. In particular, a check should account for every factor and exponent in the denominator; an answer that has the right general shape can still be wrong.
The Recombination Check
To recombine fractions, use a common denominator and multiply each fraction’s numerator by the factors missing from its denominator. For repeated factors, keep track of the exponent: a term with denominator \((x-a)\) needs an additional factor of \((x-a)\) when the common denominator contains \((x-a)^2\).
After finding the common denominator, expand and collect like terms in the combined numerator. Then compare this numerator with the original numerator. You can compare expanded expressions directly, or subtract one from the other and simplify. The check succeeds only if the difference is the zero polynomial.
Record the denominators and their factors, including repeated powers.
For factored denominators, include each factor to the greatest exponent appearing in any term.
For each term, multiply its numerator by the factors needed to reach the common denominator.
Collect like terms and check whether the result equals the original numerator exactly.
Why Numerator Comparison Works
The check rests on a basic fact about polynomial identities. If two rational expressions have the same denominator after being written over a common denominator, then they are equal wherever that denominator is nonzero exactly when their numerators are equal there. Moreover, if two polynomials agree for every \(x\) in an interval, they are the same polynomial; a nonzero polynomial can have only finitely many roots.
To justify the result, first suppose \(P=Q\) as polynomials. At any \(x\) with \(D(x)\ne 0\), division by \(D(x)\) gives \(P(x)/D(x)=Q(x)/D(x)\). Conversely, suppose the rational expressions agree wherever \(D(x)\ne 0\). Multiplying that equality by \(D(x)\) gives \(P(x)=Q(x)\) for every such \(x\). There are infinitely many such values of \(x\), since a nonzero polynomial denominator has only finitely many zeros. Therefore \(P-Q\) has infinitely many roots, so it must be the zero polynomial. Thus \(P=Q\).
This explains why checking just a few selected values of \(x\) is not enough: two different numerator polynomials can happen to agree at a few inputs. The full algebraic comparison establishes the identity.
Worked Examples
Worked Example: Distinct Linear Factors
Check whether the proposed decomposition of \(\frac{3x+3}{(x-1)(x+2)}\) is correct:
The common denominator is \((x-1)(x+2)\). The first fraction needs a factor of \(x+2\), and the second needs a factor of \(x-1\). Recombine:
Expand and collect terms in the numerator:
This is exactly the original numerator. Therefore the proposed decomposition is correct for \(x\ne 1\) and \(x\ne -2\). The denominator has not changed, and the recombined numerator matches exactly, so this is a complete check.
Worked Example: A Repeated Linear Factor
Verify the proposed decomposition of \(\frac{5x+1}{(x-1)^2(x+2)}\):
Use the common denominator \((x-1)^2(x+2)\). The term with denominator \(x-1\) needs the factors \((x-1)(x+2)\). The term with denominator \((x-1)^2\) needs \(x+2\). The term with denominator \(x+2\) needs \((x-1)^2\). Thus the combined numerator is:
Expand each part, keeping the minus sign on the entire final square:
The recombined numerator equals the original numerator \(5x+1\), so the decomposition is correct for \(x\ne 1\) and \(x\ne -2\). The repeated factor matters: the common denominator contains \((x-1)^2\), not just \(x-1\).
Worked Example: A Linear Factor and an Irreducible Quadratic
Check whether this proposed decomposition is correct:
The common denominator is \((x^2+1)(x-2)\). The first fraction needs the factor \(x-2\), while the second needs \(x^2+1\). Therefore:
The resulting numerator is exactly the original numerator. The decomposition is correct wherever the common denominator is nonzero. Since \(x^2+1\) is positive for every real \(x\), the only real value excluded here is \(x=2\).
Common Pitfall: Losing a Factor from a Repeated Denominator
When checking a term with denominator \(x+2\) against the common denominator \((x-1)^2(x+2)\), the missing factor is \((x-1)^2\), not merely \(x-1\). Forgetting the square changes the numerator and can make a correct decomposition appear incorrect.
For example, in the repeated-factor example, an incorrect recombination would be:
The last term was multiplied by only \(x-1\), but it needs to be multiplied by \((x-1)^2\). The incorrect numerator simplifies to:
This does not match \(5x+1\), but that mismatch does not expose an error in the proposed decomposition. It exposes an error in the check. The correct contribution of the last fraction is \(-(x-1)^2\), giving the numerator \((x-1)(x+2)+2(x+2)-(x-1)^2=5x+1\). A reliable check writes out each missing factor before expanding.
What a Successful Check Establishes
Once the recombined numerator matches, the proposed expression is verified as a rational-function identity on its domain. In an integration problem, that confirms the algebraic rewriting step before any antiderivatives are taken. If the numerators do not match, do not proceed as if the decomposition were correct: locate the discrepancy in the proposed terms or in the recombination algebra.
Keep the domain in view. The original expression is undefined where its denominator is zero. A proposed expression may also have denominators that vanish at additional values. The identity is asserted only where both sides are defined; canceling a factor in a simplified expression does not restore a value excluded from the original expression.
For an expression that has a polynomial part as well as proper fractions, include that polynomial part when recombining. The same principle still applies: put all terms over a common denominator and compare the resulting numerator with the original one. The check is about equality of the whole expression, not just the fractional terms.
Check Your Understanding
For each question, focus on how the common denominator determines the recombined numerator.
- What common denominator should be used to recombine \(\frac{A}{x-3}+\frac{B}{(x-3)^2}\)?
- In \(\frac{1}{x+1}+\frac{2}{(x-1)^2}\), what factor multiplies the numerator \(1\) when using the common denominator \((x+1)(x-1)^2\)?
- Recombine \(\frac{2}{x-1}-\frac{1}{x+1}\). What numerator do you obtain over the common denominator?
- Why is matching the numerator at only three selected values of \(x\) not, in general, a complete verification?
- If a proposed decomposition recombines to a numerator that differs from the original by a nonzero constant, is the decomposition correct? Explain.
