Why Repeated Factors Need a Full Ladder
A denominator containing a repeated linear factor does not call for just one partial-fraction term. If the factor has multiplicity \(m\), the complete pattern includes a term for each of its powers, beginning with the first power and ending with the \(m\)th.
This detail matters whether the repeated factor appears by itself or alongside other factors. Leaving out a power can make the proposed decomposition incapable of representing the original rational function. The goal here is to recognize and write the complete pattern; the coefficients can then be determined by the usual algebraic methods for partial fractions.
The Complete Pattern
First suppose a proper rational function has denominator \((x-r)^mS(x)\), where \(S(r)\ne 0\). The condition \(S(r)\ne 0\) says that \(S\) contains no additional factor \(x-r\). In a complete partial-fraction decomposition, the terms associated with this repeated factor are:
Here \(A_1,\ldots,A_m\) are constants. The order is important: the list includes every denominator power from \(1\) through \(m\). For example, if the factor is \((x-r)^3\), its terms are \(A_1/(x-r)\), \(A_2/(x-r)^2\), and \(A_3/(x-r)^3\). The coefficients are allowed to be zero; writing the complete pattern does not claim that every coefficient must be nonzero.
If other factors occur in the denominator, the decomposition also has terms associated with those factors. This tutorial’s pattern concerns the terms contributed by each repeated linear factor. When there is more than one such factor, write a complete ladder for each one.
A linear factor may be presented in a non-monic form, such as \(ax+b\), where \(a\ne 0\). Its root is \(r=-b/a\), so \(ax+b=a(x-r)\). The corresponding powers differ from \((x-r)^j\) only by constant factors, which can be absorbed into the unknown coefficients. Thus the standard pattern is written using \(x-r\).
Why Every Power Appears
The complete ladder can be understood by looking near the root \(r\). Write the rational function as \(H(x)/(x-r)^m\), where \(H(x)=P(x)/S(x)\) and \(S(r)\ne 0\). Since \(H\) is smooth near \(r\), its Taylor expansion through degree \(m-1\) has the form shown below, with \(G(x)\) bounded near \(r\).
Dividing by \((x-r)^m\) gives:
The terms that can be singular at \(r\) therefore have precisely the powers \(1\) through \(m\). Their coefficients may vanish for a particular numerator, but the full pattern is needed to allow all possible numerators. The remaining part, \(G(x)\), is bounded near \(r\); it is not another term in the ladder associated with a pole at \(r\).
Worked Example: A Repeated Factor and a Simple Factor
Worked Example: Decomposing a Fraction with a Repeated Linear Factor
Write the partial-fraction form for \(1/((x-1)^2(x+2))\), determine its coefficients, and verify the result.
The factor \(x-1\) has multiplicity \(2\), so it contributes a term over the first power and a term over the second power. The factor \(x+2\) is simple and contributes one term. Set up:
Multiply both sides by \((x-1)^2(x+2)\):
Substitute \(x=1\). The terms containing \(x-1\) vanish, leaving \(1=3B\), so \(B=1/3\). Substitute \(x=-2\). The terms containing \(x+2\) vanish, leaving \(1=9C\), so \(C=1/9\).
To find \(A\), compare the coefficients of \(x^2\) after expanding the identity. The \(x^2\) coefficient on the right is \(A+C\), while the left side has coefficient \(0\). Therefore \(A+C=0\), and \(A=-1/9\).
For a check, expand the right side of the multiplied identity using these values:
The \(x^2\) coefficients sum to \(-1/9+1/9=0\). The \(x\) coefficients sum to \(-1/9+1/3-2/9=0\). The constant terms sum to \(2/9+2/3+1/9=1\). Thus the identity holds, and dividing by the common denominator verifies the decomposition.
Worked Example: A Factor of Multiplicity Three
Worked Example: Listing and Verifying Three Powers
For the rational function \((2x^2-5x+6)/(x-1)^3\), write the complete decomposition pattern and verify its coefficients.
The denominator has one linear factor, \(x-1\), with multiplicity \(3\). The required pattern is therefore:
Multiply through by \((x-1)^3\):
Set \(x=1\). The left side becomes \(2-5+6=3\), and the right side becomes \(C\). Thus \(C=3\). Expand the identity:
Matching the \(x^2\) coefficients gives \(A=2\). Matching the \(x\) coefficients gives \(-2A+B=-5\), so \(-4+B=-5\) and \(B=-1\). The constant terms then check: \(A-B+C=2-(-1)+3=6\).
Consequently, the decomposition is:
Recombining the right side over the common denominator gives numerator \(2(x-1)^2-(x-1)+3\). Expanding yields \(2x^2-4x+2-x+1+3=2x^2-5x+6\), which is the original numerator. This verifies the decomposition.
Worked Example: A Repeated Factor Alongside Another Linear Factor
Worked Example: A Repeated Factor at Zero
Decompose \(1/[x(x-1)^2]\). Identify the repeated-factor terms and check the result by recombining.
The factor \(x-1\) is repeated twice, so its terms must include both powers. The factor \(x\) is simple. Thus write:
Multiplying by \(x(x-1)^2\) gives \(1=A(x-1)^2+Bx(x-1)+Cx\). Substituting \(x=0\) gives \(1=A\), so \(A=1\). Substituting \(x=1\) gives \(1=C\), so \(C=1\). Comparing \(x^2\) coefficients gives \(A+B=0\); since \(A=1\), \(B=-1\).
The resulting decomposition is:
To verify it, combine the right side over \(x(x-1)^2\). Its numerator is \((x-1)^2-x(x-1)+x\). Expanding gives \(x^2-2x+1-x^2+x+x=1\), the original numerator. The decomposition is correct.
A Common Mistake: Leaving Out a Lower Power
A frequent error is to include only the term with the highest denominator power for a repeated factor. For instance, one might incorrectly propose that the fraction in the first worked example has the form \(B/(x-1)^2+C/(x+2)\), omitting the term over \(x-1\).
After multiplying by the denominator, that incomplete proposal would require:
Expanding gives \(1=Cx^2+(B-2C)x+(2B+C)\). For this to equal the constant polynomial \(1\), the \(x^2\) coefficient requires \(C=0\). The \(x\) coefficient then requires \(B=0\). But those values make the constant term \(2B+C=0\), not \(1\). No choices of \(B\) and \(C\) can satisfy the identity. The missing \(A/(x-1)\) term is necessary.
Checklist for Writing the Terms
For example, \((x+4)^3\) has root \(r=-4\) and multiplicity \(3\).
For \((x+4)^3\), include terms over \(x+4\), \((x+4)^2\), and \((x+4)^3\).
A distinct simple linear factor receives its own term; another repeated linear factor receives its own complete ladder.
Multiply by the denominator and verify that the resulting polynomial numerator matches the original.
For an improper rational function, the partial fractions apply to its proper remainder after polynomial division. If numerator and denominator have a common factor, cancel it first when appropriate, since cancellation can change which factors remain in the denominator. These steps determine which factors and multiplicities the decomposition must use.
The practical value of the full pattern is reliability: once every denominator factor and multiplicity has been identified, the form of the decomposition is fixed before the coefficient calculations begin. This prevents a missing term from being mistaken for an arithmetic error later.
Check Your Understanding
For each question, identify the complete set of terms associated with the repeated linear factor. Then use the examples and the derivation above to explain your reasoning.
- What terms are required for a denominator containing \((x-5)^4\)?
- What is the full repeated-factor ladder for \((2x+1)^3\), written using the monic factor determined by its root?
- If the denominator is \(x^2(x-3)^2\), which terms come from the factor \(x^2\), and which come from \((x-3)^2\)?
- Why can the coefficient of one of the listed terms be zero without changing the required pattern?
- What goes wrong if the \(1/(x-r)\) term is omitted from the pattern for a repeated factor \((x-r)^m\)?
