Why Repeated Factors Need a Full Set of Terms
A repeated factor in a denominator does not call for just one fraction. If \((x-a)^m\) occurs in the denominator, its partial-fraction template includes a separate term for each denominator power: \((x-a)\), \((x-a)^2\), and so on through \((x-a)^m\).
The reason is algebraic. After denominators are cleared, these terms correspond to different powers of \(x-a\) in a polynomial identity. A general polynomial needs all those powers available, even though a particular rational function may have a zero coefficient for one or more of them.
Multiplicity and the Repeated-Factor Template
A denominator contains the repeated linear factor \((x-a)^m\) when \((x-a)^m\) divides it. The multiplicity is \(m\) if the denominator is not divisible by \((x-a)^{m+1}\). For example, \((x-2)^3(x+1)\) contains \(x-2\) with multiplicity three.
For a proper rational function whose denominator has a factor \((x-a)^m\), the terms associated with that factor have the form below. Other distinct denominator factors have their own terms; they do not change the powers required for this repeated factor.
Here \(A_1,\ldots,A_m\) are constants to be determined. This is a template: it lists the terms that may be needed before the coefficients are known. If the coefficient calculation gives \(A_j=0\), that term contributes nothing for the particular function. That does not make the general template unnecessary.
The Polynomial Identity Behind the Template
First consider a fraction with denominator \((x-a)^m\) and a numerator \(R(x)\) of degree less than \(m\). To express it using the repeated-factor terms, combine those terms over the common denominator. The numerator becomes
Thus finding the coefficients means representing \(R(x)\) as a polynomial in powers of \(x-a\). Those powers run from \((x-a)^{m-1}\) down to the constant power \((x-a)^0=1\). A polynomial of degree at most \(m-1\) can have any of these powers, so the template must allow for each one.
Justification. Set \(t=x-a\), so \(x=t+a\). Then \(R(t+a)\) is a polynomial in \(t\) of degree less than \(m\). Collecting its constant, linear, quadratic, and higher powers gives a representation \(c_0+c_1t+\cdots+c_{m-1}t^{m-1}\). Substituting \(t=x-a\) gives the stated expression.
For uniqueness, suppose two such expressions represent the same polynomial. Subtracting them gives a polynomial in \(t\) that is zero for every \(t\). A polynomial is the zero polynomial only when each of its coefficients is zero. Therefore the two lists of coefficients agree. Dividing the expression for \(R(x)\) by \((x-a)^m\) gives the partial-fraction terms: the coefficient of \((x-a)^{m-j}\) becomes the numerator of the term with denominator \((x-a)^j\).
This explains why the complete template is necessary: leaving out one denominator power would leave out the corresponding power of \(x-a\) from the numerator identity. A general numerator may contain that power.
Worked Examples
Worked Example: Expanding a Numerator Over a Triple Factor
Write \(\frac{2(x-1)^2-3(x-1)+4}{(x-1)^3}\) as repeated-factor terms.
The denominator contains \((x-1)^3\), so the template includes denominators \((x-1)\), \((x-1)^2\), and \((x-1)^3\). The numerator is already expressed in powers of \(x-1\). Dividing each term by \((x-1)^3\) gives
Each of the three coefficients is nonzero. To check, put the terms over the common denominator \((x-1)^3\): their numerator is \(2(x-1)^2-3(x-1)+4\), exactly the original numerator. Thus all three powers are needed for this particular fraction.
Worked Example: Matching Coefficients with Another Factor
Find the partial-fraction coefficients for \(\frac{5(x-1)^3+5(x-1)^2+9(x-1)+9}{(x-1)^3(x+2)}\).
Let \(t=x-1\). Then \(x+2=t+3\), and the fraction becomes \(\frac{5t^3+5t^2+9t+9}{t^3(t+3)}\). Since the repeated factor is \(t^3\), include all three associated terms, as well as a term for the distinct factor \(t+3\):
Multiply both sides by \(t^3(t+3)\). The polynomial identity to solve is
Expand each product and collect like powers:
Therefore the coefficients of \(t^3,t^2,t,\) and \(1\), in that order, give
From \(3C=9\), \(C=3\). Then \(3B+3=9\), so \(B=2\). Next, \(3A+2=5\), so \(A=1\). Finally, \(1+D=5\), so \(D=4\). Returning to \(x\), the result is
To check, combine the right-hand side over \(t^3(t+3)\). Its numerator is \(t^2(t+3)+2t(t+3)+3(t+3)+4t^3\), which expands to \(5t^3+5t^2+9t+9\), the original numerator in terms of \(t\). The identity is verified.
Worked Example: A Coefficient Can Be Zero
Express \(\frac{(x+2)^2+5}{(x+2)^3}\) using repeated-factor terms and determine whether all coefficients are nonzero.
The factor \((x+2)^3\) has multiplicity three, so start with all three terms. The numerator is \((x+2)^2+0(x+2)+5\). Dividing each part by \((x+2)^3\) gives
The coefficient of \(1/(x+2)^2\) is zero. Combining the terms over the common denominator gives numerator \((x+2)^2+0(x+2)+5\), as required. The zero coefficient shows why “every power is needed” describes the general template, not a claim that every particular decomposition has every coefficient nonzero.
What Goes Wrong When a Power Is Omitted
Suppose a denominator contains \((x-a)^3\), but a proposed template includes only terms with denominators \((x-a)\) and \((x-a)^3\). After clearing the common denominator, the available numerator powers are \((x-a)^2\) and \(1\). There is no term that can produce a multiple of \(x-a\). Therefore a numerator containing a nonzero linear term in \(x-a\) cannot be represented by that shortened template.
For example, set \(t=x-1\). The fraction \(t/t^3=1/t^2\) cannot be represented using only \(A/t+C/t^3\). Clearing denominators would require \(t=At^2+C\). At \(t=0\), this identity requires \(C=0\). It would then say \(t=At^2\) for every \(t\), which is impossible: for \(t=1\), it requires \(A=1\), while for \(t=2\), it requires \(A=1/2\). Including the missing term resolves the problem: \(1/t^2\) has coefficient \(1\) on that term.
Why This Matters
The repeated-factor template is not an arbitrary convention. It is chosen so that, after clearing denominators, the numerator can match a general polynomial of the required degree. Each denominator power supplies a different power of \(x-a\) in that identity. Omitting one term restricts which numerators the decomposition can represent.
This reasoning also provides a practical check while setting up a decomposition: if a factor has multiplicity \(m\), count \(m\) possible terms for it. For instance, multiplicity four calls for denominators \((x-a)\), \((x-a)^2\), \((x-a)^3\), and \((x-a)^4\). Coefficient solving may later show that one or more of them vanish, but the complete template makes no assumption in advance about which powers the numerator will require.
Check Your Understanding
Use the polynomial identity behind the template to answer these questions.
- For a factor \((x-3)^4\), which four denominator powers belong in its partial-fraction template?
- Why do the terms for \((x-a)^m\) produce powers from \(1\) through \((x-a)^{m-1}\) in the numerator after clearing denominators?
- Can a particular decomposition have a zero coefficient for one of the template terms? What does that imply about the general template?
- Why can a numerator containing a nonzero multiple of \(x-a\) not be represented by a triple-factor template that omits \(1/(x-a)^2\)?
- For \(\frac{3(x+1)^2-2(x+1)+7}{(x+1)^3}\), write the repeated-factor terms directly by dividing each numerator term by the denominator.
