Why Decomposition Errors Are Easy to Miss
A partial-fraction decomposition can look plausible while still being incomplete or numerically wrong. A missing term may not be obvious from the appearance of the fractions, and a single sign error can change several coefficients. The most reliable diagnosis is to compare the proposed form with the denominator’s factors, then clear denominators and check the resulting polynomial identity.
This tutorial concentrates on three common problems: leaving out a power of a repeated factor, using a constant-only numerator over an irreducible quadratic, and making arithmetic errors while solving for coefficients. Recombining the fractions at the end is essential: it confirms that the proposed terms really produce the original numerator.
First Check the Template
Before checking numerical coefficients, inspect the proposed kinds of terms. For a repeated linear factor \((x-a)^m\), the template includes one term for each power from \(1\) through \(m\):
Leaving out one of these terms is not justified just because the other terms look similar. A coefficient in the correct decomposition might happen to be zero, but that is something to discover by solving; it is not a reason to omit the term in advance.
For an irreducible quadratic factor, such as \(x^2+1\), the numerator in the corresponding partial fraction must allow a linear expression, \(Bx+C\). A constant-only numerator can exclude decompositions that are needed to represent the original rational function.
A Cleared Identity Is Not the Original Rational Expression
There is an important distinction when substituting values. The original rational expression is defined only where its denominator is nonzero. But after multiplying an equation of rational expressions by a common denominator, the result is a polynomial identity. Values that made the original denominator zero can be substituted into that identity.
Here is why the cleared equation is an identity. Suppose two rational expressions are equal wherever their common denominator is nonzero. Multiplying by that denominator gives two polynomials that agree at every such value. There are infinitely many of these values. If their difference were a nonzero polynomial, it could have only finitely many roots. Therefore the difference is the zero polynomial, so the two polynomials agree for every value of \(x\), including values excluded from the original rational expressions.
This distinction makes substitutions at denominator roots especially useful: they often eliminate most terms and isolate a coefficient immediately.
Worked Example: Restoring a Missing Repeated-Factor Term
Worked Example: Restoring a Missing Repeated-Factor Term
Consider the rational function
A proposed decomposition includes only \(A/(x-1)\) and \(C/(x+2)\). Diagnose the omission and find the correct coefficients.
Step 1: Check the template. The factor \(x-1\) is repeated, with power \(2\). The complete form must include both powers:
Step 2: Clear denominators. Multiplying by \((x-1)^2(x+2)\) gives the polynomial identity
Step 3: Substitute useful values into the identity. At \(x=1\), the terms involving \(A\) and \(C\) vanish:
At \(x=-2\), the terms involving \(A\) and \(B\) vanish:
Now set \(x=0\) in the identity:
Thus \(-2A=-2\), so \(A=1\). The decomposition is
Step 4: Check by recombining. The right-hand side has common denominator \((x-1)^2(x+2)\) and numerator
This matches the original numerator. The omitted term was genuinely needed: if the \(B\)-term were absent, the cleared right-hand side would be \(A(x-1)(x+2)+C(x-1)^2\), which is zero at \(x=1\), while the left-hand side there is \(6\). No choices of \(A\) and \(C\) could make that incomplete identity true.
Worked Example: Correcting a Constant-Only Quadratic Numerator
Worked Example: Correcting a Constant-Only Quadratic Numerator
Consider
Suppose a proposed template uses \(A/(x-1)+C/(x^2+1)\). Check whether it is adequate, then find a valid decomposition.
Step 1: Use a linear numerator over the quadratic. Since \(x^2+1\) is irreducible over the real numbers, use the form
Step 2: Clear denominators and compare coefficients.
Expanding the right-hand side gives
Therefore the coefficient equations are
From the first equation, \(B=3-A\). From the second, \(C=B-3=-A\). Substituting \(C=-A\) into the third gives \(A-(-A)=2\), so \(A=1\). Then \(B=2\) and \(C=-1\). Hence
Step 3: Verify. Recombine the right-hand side:
The numerator matches, so the decomposition is verified. A constant-only quadratic numerator would have no \(x\)-term available in \((Bx+C)(x-1)\). In fact, setting \(B=0\) would require \(A=3\) from the \(x^2\)-coefficient and \(C=-3\) from the \(x\)-coefficient. Those values give constant term \(A-C=6\), not the required \(2\). The constant-only form cannot work here.
Worked Example: Finding and Fixing a Coefficient Arithmetic Error
Worked Example: Finding and Fixing a Coefficient Arithmetic Error
Suppose a proposed decomposition for
is being checked. Find the correct coefficients and identify a common sign error that could lead to an incorrect result.
Step 1: Write the form and clear denominators.
Step 2: Expand carefully. The product is
Thus the coefficient equations are
Step 3: Solve in sequence. The first equation gives \(B=3-A\). The second gives
Substitute this into the third equation:
Therefore \(A=2\), \(B=3-2=1\), and \(C=7-2(2)=3\). The decomposition is
Step 4: Verify by recombining.
The recombined numerator agrees with the original. A frequent error is to expand \((Bx+C)(x-2)\) with \(C+2B\) as the coefficient of \(x\). The correct middle term is \(Cx-2Bx=(C-2B)x\); the minus sign comes from multiplying \(Bx\) by \(-2\). Using \(C+2B=1\) would solve a different, incorrectly expanded equation.
A Reliable Diagnostic Routine
When a decomposition seems wrong, use the following sequence rather than changing coefficients by guesswork:
Check that the template includes every power of a repeated linear factor and a linear numerator over each irreducible quadratic.
Multiply the proposed decomposition by the common denominator to obtain a polynomial identity.
Expand each product, preserve signs, and match coefficients of equal powers of \(x\). Substitutions into the cleared identity can also simplify the equations.
Put the proposed terms over the original denominator and confirm that their numerator is exactly the original numerator.
For the substitution step, keep the domain distinction in view. A value that makes a denominator zero cannot be used in the original rational expressions. But after denominators have been cleared, the polynomial identity holds at that value as well. For instance, in the first worked example, substituting \(x=1\) into the original fraction is undefined, whereas substituting \(x=1\) into the cleared identity is valid and isolates \(B\).
Why These Checks Matter
Partial fractions are often used to prepare a rational function for integration. An omitted term can change the function itself, not merely its appearance, and an incorrect coefficient can lead to an incorrect integral. The template check catches structural omissions before solving; the polynomial identity catches coefficient errors; and recombination confirms the final equality directly.
Do not treat convenient substitutions as tests of the original expression when they make its denominator zero. Treat them as substitutions into the cleared polynomial identity, where they are legitimate. This careful distinction lets you use the most useful values without confusing a rational equation with the polynomial identity that follows from it.
Check Your Understanding
Use the template check, a cleared identity, and recombination where appropriate.
- What terms must be included for a repeated factor \((x+3)^3\)?
- Why can a constant-only numerator over \(x^2+4\) fail to represent a rational function?
- After clearing denominators, why may a zero of the original denominator be substituted into the resulting polynomial identity?
- In expanding \((Bx+C)(x-2)\), what is the coefficient of \(x\), and where does its sign come from?
- What calculation provides a direct final check of a proposed decomposition?
