Why the Choice of x Matters
When a rational function has repeated denominator factors, its partial-fraction decomposition includes several unknown coefficients. Clearing denominators produces a polynomial identity, and comparing coefficients is one way to find those unknowns. But expanding and matching every power can create a larger system than necessary.
A useful alternative is to evaluate the cleared identity at selected values of \(x\). A root of a denominator factor may make several terms vanish at once. Other values, chosen for simple arithmetic, can then give equations for the coefficients that remain. The aim is not to avoid the coefficient system altogether; it is to make that system easier to solve.
Start with the Cleared Identity
Consider a rational function whose denominator contains a repeated factor. A typical decomposition might include terms such as \(A/(x-a)\) and \(B/(x-a)^2\), along with terms from other factors. Multiplying the proposed decomposition by the full denominator eliminates the fractions and gives a polynomial identity involving the unknown coefficients.
This last point is essential. The original rational expression is undefined at a denominator root, so we do not substitute that value into the original fractions. We substitute it into the cleared polynomial identity, where every term is defined.
At a root, a factor can make terms vanish. For example, if the cleared identity contains a term multiplied by \((x-a)^2\), setting \(x=a\) makes that term zero. The remaining equation may determine one coefficient immediately, but it does not necessarily determine all the coefficients. Additional values of \(x\), or coefficient comparison, may still be required.
A Justification for the Method
The reason these substitutions are legitimate follows from the definition of a polynomial identity. Suppose two polynomials \(P(x)\) and \(Q(x)\) are equal for every \(x\) in the domain of an original rational identity after denominators have been cleared. There are infinitely many such real values of \(x\), since a denominator has only finitely many real roots. The difference \(P(x)-Q(x)\) is therefore a polynomial that is zero for infinitely many values. A nonzero polynomial has only finitely many roots, so \(P-Q\) must be the zero polynomial. Hence \(P(x)=Q(x)\) for every real \(x\), including the roots of the original denominator.
In practice, this means we can evaluate the cleared identity at a factor’s root to eliminate terms. Then we choose other values that keep the arithmetic manageable. A good choice produces a simpler equation; it does not change the identity or the answer.
Worked Example: One Repeated Linear Factor
Worked Example: One Repeated Linear Factor
Find \(A\), \(B\), and \(C\) in the decomposition
Multiply both sides by \((x-1)^2(x+2)\). The cleared identity is
Set \(x=1\). The terms containing \(x-1\) vanish, leaving
Now set \(x=-2\). The terms containing \(x+2\) vanish, so
These two root substitutions have determined \(B\) and \(C\), but the coefficient \(A\) remains. Choose \(x=0\), which makes the arithmetic simple:
Thus the proposed decomposition is
Check by recombining the right-hand side over the common denominator. Its numerator is
This matches the original numerator, so the decomposition checks.
When Root Substitutions Are Not Enough
Repeated factors often produce a useful first equation at a root, but the remaining unknowns may still be linked. Choose further values that make the cleared identity easy to evaluate. Values such as \(0\), \(1\), or \(-1\) are often convenient, provided they are not excluded by the original expression when considering the rational equation. In the cleared polynomial identity, any value is valid; in practice, values that keep the arithmetic simple are preferable.
It is also useful to plan which equations are needed before substituting. If there are four unknowns and the root evaluations provide only two independent equations, two more independent equations are needed. They may come from convenient substitutions or from comparing coefficients. Repeating substitutions without checking whether they add new information can waste effort.
Worked Example: A Repeated Linear Factor and a Quadratic
Worked Example: A Repeated Linear Factor and a Quadratic
Find the coefficients in
After multiplying by the denominator, obtain
The repeated linear factor has root \(x=-1\). Substitution gives
There are three coefficients left. Set \(x=0\):
Set \(x=1\). The left side is \(4+6+6=16\), so
Subtract \(A+D=4\) from \(A+C+D=3\) to obtain \(C=-1\). Finally, set \(x=-2\). The left side is \(16-12+6=10\), and the identity gives
Using \(D=4-A\) from \(A+D=4\), solve:
The decomposition is therefore
To check, recombine the numerators:
Expand each term:
The numerator agrees with the original one, verifying the result.
Worked Example: Two Repeated Linear Factors
Worked Example: Two Repeated Linear Factors
Determine the coefficients in
Clearing denominators gives
Set \(x=2\). Only the term involving \(B\) remains on the right:
Set \(x=-1\). Only the term involving \(D\) remains:
Now use \(x=0\). The left side is \(15\), giving
Use \(x=1\). The left side is \(5-8-16+15=-4\):
From \(-A+2C=4\), \(A=2C-4\). Substitute this into the second equation:
Then \(A=2(3)-4=2\). The full set of coefficients is \(A=2\), \(B=-1\), \(C=3\), and \(D=2\).
Check by recombining. The numerator from these coefficients is
Expanding the four terms gives
This is the original numerator, so the decomposition is verified.
Common Pitfall: Substituting Before Clearing Denominators
A tempting but invalid step is to put \(x=a\) directly into a rational equation whose denominator contains \(x-a\). For example, at \(x=1\), the expression \(A/(x-1)\) is undefined. It is not valid to substitute \(x=1\) into the original partial-fraction equation and then treat the resulting fractions as ordinary numbers.
The correction is to multiply by the common denominator first. In the first example, the cleared equation was
This polynomial identity is defined at \(x=1\). Substituting there gives \(9=3B\), so \(B=3\). The value at the root is useful precisely because we evaluate the cleared identity, not the undefined rational expressions.
A second mistake is to assume that root substitutions always determine every coefficient. They often eliminate many terms, but a decomposition with several unknowns may still need equations from other values of \(x\) or from coefficient comparison. Check how many unknowns remain and whether the equations obtained are independent before deciding that the system is solved.
A Practical Strategy
Include every coefficient required by the denominator factors, including the full set of terms for repeated factors.
Form a polynomial identity before evaluating at any denominator root.
Substitute roots that make groups of terms vanish, and solve the resulting equations.
Use values such as zero or small integers to obtain equations for the remaining coefficients.
Solve the remaining equations, then recombine the terms and check that the numerator is recovered.
Strategic substitution is useful because it turns the structure of the denominator into a guide for simplifying the coefficient system. It does not replace careful bookkeeping: every substitution must be applied to the cleared identity, every unknown must be determined, and the final decomposition should be checked by recombination.
Check Your Understanding
Answer these questions using the cleared-identity method.
- Why is it valid to substitute a root of the denominator into a cleared polynomial identity, even though the original rational expression is undefined there?
- For a denominator containing \((x-3)^2\), which value of \(x\) is a natural first choice, and what makes that substitution useful?
- If root substitutions determine two coefficients but leave three unknowns, what should you do next?
- Why is direct substitution into the original rational equation at a denominator root invalid?
- After finding the coefficients, what calculation checks that the decomposition reproduces the original rational function?
