Why Coefficients Require More Than One Substitution
A repeated linear factor changes how coefficients are determined in a partial-fraction decomposition. For distinct linear factors, substituting a root into the cleared equation can isolate one coefficient for each factor. With a repeated factor, one such substitution isolates the coefficient of its highest reciprocal power, but the other coefficients generally remain to be found.
The practical method is to clear denominators first. The result is a polynomial identity: one equation that must hold for every value of the variable. Substituting the repeated root gives a useful equation, and matching coefficients of like powers supplies the rest.
The Setup
Suppose a proper rational function has a denominator containing the repeated factor \((x-a)^m\), where \(m\) is a positive integer. The terms in its partial-fraction decomposition associated with this factor have the form
Here \(A_1,\ldots,A_m\) are constants to be determined. The decomposition may also have terms associated with other factors in the denominator. Once the complete decomposition form is specified, multiplying both sides by the full denominator removes all fractions and produces a polynomial identity.
For example, if the denominator is \((x-a)^m Q(x)\), where \(Q(a)\ne 0\), the contribution from \(A_m/(x-a)^m\) becomes \(A_mQ(x)\) after clearing denominators. At \(x=a\), that contribution is \(A_mQ(a)\). Contributions from terms with smaller powers of \(x-a\) still have a factor of \(x-a\) and therefore become zero there. Terms associated with other denominator factors also vanish there, provided those factors do not themselves vanish at \(a\).
A Reliable Solving Process
The following steps apply when the partial-fraction form is already known. They are for finding its constants, not for deciding which terms belong in the form.
Include each term supplied by the repeated-factor template and any other terms in the given denominator’s decomposition.
Write the resulting polynomial identity, keeping track of the factors that remain in each term.
This can isolate the coefficient of the highest reciprocal power associated with that factor.
Use the polynomial identity to solve for the remaining constants. Substitute any known constants into the equations before solving.
Substitute the constants into the cleared identity and confirm that its numerator is the original numerator.
Why Root Substitution Works
Consider a decomposition containing the terms \(A_1/(x-a),\ldots,A_m/(x-a)^m\), with the other denominator factors collected in \(Q(x)\). After clearing the full denominator \((x-a)^mQ(x)\), the contribution from \(A_j/(x-a)^j\) is
If \(j<m\), then \(m-j\) is positive, so this contribution is zero at \(x=a\). If \(j=m\), the contribution is \(A_mQ(x)\), whose value at \(x=a\) is \(A_mQ(a)\). Thus, when the other denominator factors do not vanish at \(a\), substitution into the cleared identity gives an equation that isolates \(A_m\).
This substitution does not usually determine \(A_1,\ldots,A_{m-1}\). To find them, expand the remaining identity and equate coefficients. This is valid because two polynomials that agree for all values away from the finitely many denominator roots agree at infinitely many values; their difference is therefore the zero polynomial, so its coefficients are all zero.
Worked Examples
Worked Example: A Repeated Factor and a Distinct Linear Factor
Find the constants in
Multiply both sides by \((x+1)(x-2)^2\). This gives the polynomial identity
Set \(x=2\). The terms containing \(x-2\) vanish:
Now expand the right-hand side. Since \((x-2)^2=x^2-4x+4\) and \((x+1)(x-2)=x^2-x-2\), the identity becomes
Equating coefficients gives
Substitute \(C=3\). The first two equations become \(A+B=1\) and \(4A+B=7\). Subtract the first from the second: \(3A=6\), so \(A=2\). Then \(B=1-A=-1\). The constant equation checks: \(4(2)-2(-1)+3=8+2+3=13\).
Therefore the decomposition is
To check, recombine the right-hand side’s numerator: \(2(x-2)^2-(x+1)(x-2)+3(x+1)\). Expanding gives \(2x^2-8x+8-x^2+x+2+3x+3=x^2-4x+13\), as required.
Worked Example: A Repeated Factor of Power Three
Find the constants in
Clear the denominator to obtain
Set \(x=1\). All terms except the one containing \(D\) vanish:
For the other constants, use \((x-1)^3=x^3-3x^2+3x-1\), \((x+2)(x-1)^2=x^3-3x+2\), and \((x+2)(x-1)=x^2+x-2\). Matching coefficients in the cleared identity gives
Use \(D=4\). From the first two equations, \(B=3-A\) and \(C=-4+3A\). Substituting into the third gives
Therefore \(B=3-1=2\) and \(C=-4+3(1)=-1\). The constant equation checks: \(-1+2(2)-2(-1)+2(4)=-1+4+2+8=13\).
The decomposition is
Recombining gives the numerator \((x-1)^3+2(x+2)(x-1)^2-(x+2)(x-1)+4(x+2)\). Expanding the four terms gives \(x^3-3x^2+3x-1+2x^3-6x+4-x^2-x+2+4x+8=3x^3-4x^2+13\), which verifies the result.
Worked Example: A Denominator with Only a Repeated Factor
Find the constants in
Multiplying through by \((x+1)^3\) gives
Substitute the repeated root \(x=-1\):
Expand and match coefficients:
Thus \(A=2\), and \(2(2)+B=-3\) gives \(B=-7\). The constant equation checks: \(2-7+12=7\). Hence
Recombining gives \(2(x+1)^2-7(x+1)+12=2x^2+4x+2-7x-7+12=2x^2-3x+7\). The numerator agrees with the original.
A Common Pitfall: Substituting Before Clearing Denominators
In the first example, substituting \(x=2\) into the original rational function would produce a denominator of zero. A calculation such as \(9/[3(0)]=C\) is not valid: division by zero is undefined, and the original function is not defined at \(x=2\).
The correction is to first form the cleared polynomial identity. At \(x=2\), that identity gives \(9=3C\), so \(C=3\). The distinction matters: root substitution is applied to the polynomial identity, not to the original rational expression at its pole.
Another common error is to stop after finding \(C\). In the first example, the root substitution gives \(C=3\), but it does not supply \(A\) and \(B\). Those constants come from the remaining coefficient equations.
Using the Method Effectively
Substitution at a repeated root is most useful as the first step, not as a replacement for coefficient matching. It often reduces the number of unknowns in the later equations and can prevent unnecessary algebra. After substituting, expand carefully, group terms by powers of \(x\), and check every resulting equation against the original numerator.
Keep the roles of the two techniques distinct:
| Technique | What it does |
|---|---|
| Substitute the repeated root into the cleared identity | Often isolates the coefficient of the highest reciprocal power for that factor |
| Match polynomial coefficients | Determines the remaining constants from the full identity |
| Recombine the terms | Checks that the proposed constants reproduce the original numerator |
If the rational function is not proper, polynomial division may be needed before using a partial-fraction form for the proper part. The coefficient calculations here begin once the appropriate decomposition form has been written.
Check Your Understanding
Use a cleared polynomial identity in each problem. Check your answers by recombining the terms.
- For \(\frac{5x+1}{(x-2)^2}=\frac{A}{x-2}+\frac{B}{(x-2)^2}\), what does substituting \(x=2\) into the cleared identity give?
- For \(\frac{x^2+2}{x(x+1)^2}=\frac{A}{x}+\frac{B}{x+1}+\frac{C}{(x+1)^2}\), find \(A\), \(B\), and \(C\).
- Why is it invalid to substitute the repeated root into the original rational function before clearing denominators?
- After substituting the repeated root into a cleared identity, why might some coefficients still be unknown?
- What final algebraic check confirms that a proposed decomposition is correct?
