Why a Complete Template Matters
A denominator may contain several different factors, and some may occur more than once. Before finding any coefficients, a partial-fraction decomposition needs a complete list of the possible terms. If a required term is missing from the template, the template cannot represent every rational function with that denominator; a particular function can still fit only if the omitted term's coefficient happens to be zero.
This tutorial focuses on constructing that list when distinct linear factors, repeated linear factors, and repeated irreducible quadratic factors appear together. The task is to inventory the denominator carefully and then write a term for each required factor power.
Read the Denominator Before Writing Terms
For real partial fractions, a linear factor has the form \(x-r\), where \(r\) is real. An irreducible quadratic factor is a real quadratic that cannot be factored into real linear factors. It can be written, after dividing out its leading coefficient, as \(q(x)=x^2+px+s\), with \(p^2-4s<0\). Distinct factors in a denominator are counted once each, with their multiplicities recorded separately.
For instance, in \((x-2)^3(x+1)(x^2+4)^2\), the distinct factors are \(x-2\), \(x+1\), and \(x^2+4\). Their multiplicities are \(3\), \(1\), and \(2\). The factor \(x^2+4\) is irreducible over the reals because its discriminant is \(-16\).
If the rational function is improper, polynomial long division is performed first. The partial-fraction template applies to the proper rational remainder. If numerator and denominator have a common factor, cancel it before recording the factors and their multiplicities.
The Complete Mixed Template
Suppose the denominator is factored over the reals as
where the \(r_i\) are distinct real numbers, the \(q_j\) are distinct monic irreducible real quadratics, and all the displayed factors are distinct. Here \(m_i\) and \(n_j\) are positive integers. For a proper rational function with this denominator, the complete real partial-fraction form is
The coefficients \(A_{i,k}\), \(B_{j,k}\), and \(C_{j,k}\) are constants to be determined. A factor with multiplicity one contributes just its first-power term. A factor with multiplicity \(m\) contributes all \(m\) powers, not only the largest power.
The requirement that a quadratic numerator be linear is part of the standard real partial-fraction form. Thus, for \(q(x)=x^2+1\), the terms have numerators such as \(Bx+C\); for \(x-r\), the numerator is a constant. The coefficients may eventually turn out to be zero, but the template includes the possible terms before those coefficients are known.
Why Each Power Appears
The factor ladder can be understood by clearing the denominator. Suppose a denominator contains \((x-r)^m\), and let \(H(x)\) denote the product of all its other factors. A candidate term \(A_k/(x-r)^k\), after multiplication by the full denominator, contributes
As \(k\) ranges from \(1\) to \(m\), the powers of \(x-r\) in these contributions range from \(m-1\) down to \(0\). Including all the terms gives the decomposition the needed range of contributions at that factor. If one of the terms is omitted, the corresponding power is absent from the candidate expression.
For an irreducible quadratic \(q(x)\) with multiplicity \(n\), a term \((Bx+C)/q(x)^k\), after clearing the denominator, contributes \((Bx+C)q(x)^{n-k}\) times the other factors. The powers \(k=1,\ldots,n\) likewise provide a full ladder, with a linear numerator at each level. This explains the structure of the template; the coefficients are found only after the template has been written.
Worked Examples
Worked Example: A Denominator with All Three Factor Types
Write a complete partial-fraction form for a proper rational function with denominator \(D(x)=(x-1)^2(x+1)(x^2+1)^2\).
The factor \(x-1\) is linear with multiplicity \(2\), so it contributes terms with denominators \(x-1\) and \((x-1)^2\). The factor \(x+1\) is linear with multiplicity \(1\), so it contributes one term. The quadratic \(x^2+1\) has discriminant \(0^2-4(1)(1)=-4<0\), so it is irreducible over the reals; its multiplicity is \(2\), so it contributes two terms with linear numerators.
Check the template against the denominator: its five denominators are exactly the two powers of \(x-1\), the one power of \(x+1\), and the two powers of \(x^2+1\). Every factor power is represented, and each numerator has the appropriate form. No coefficients have been solved here; the result is the complete template.
Worked Example: Repeated Linear Factors with an Absent Coefficient
Consider \(R(x)=(3x^3-x^2-19x-1)/((x+2)^2(x-1)^2)\). Both distinct linear factors have multiplicity \(2\). Therefore the complete template must include both powers of each factor:
In this case, the coefficients are \(A=3\), \(B=1\), \(C=0\), and \(D=-2\). Verify the claim by multiplying each term by the common denominator \((x+2)^2(x-1)^2\). The four contributions are
Adding gives \(3x^3-x^2-19x-1\), the stated numerator. Thus the decomposition checks. The zero coefficient \(C\) does not make its term unnecessary when constructing the complete template: before solving, its value is not known.
Worked Example: A Repeated Irreducible Quadratic
Write and check a decomposition for \(R(x)=(3x^4-x^3+6x^2-6x+2)/((x-1)(x^2+1)^2)\). The linear factor occurs once, while the irreducible quadratic occurs twice. The complete form is
For this function, the coefficients are \(A=1\), \(B=2\), \(C=1\), \(D=3\), and \(E=-2\). To verify them, set \(Q=x^2+1\) and multiply the proposed decomposition by \((x-1)Q^2\). The three resulting numerator contributions are
Their sum is \(3x^4-x^3+6x^2-6x+2\), exactly the numerator of \(R\). This confirms both the coefficients and the denominator powers in the decomposition.
A Common Omission
A frequent mistake is to include only the highest power of a repeated factor. For example, for a denominator containing \((x-2)^3(x^2+1)^2\), the expression
is not a complete template. It omits the first and second powers of \(x-2\), as well as the first power of \(x^2+1\). The correct template is
The correction is to count the multiplicity of each distinct factor and write its complete ladder. Do not assume an omitted coefficient is zero before the coefficients have been determined.
A Reliable Construction Procedure
Cancel any common factors, and use polynomial division first if the function is improper.
Separate distinct linear factors from irreducible quadratic factors and record each multiplicity.
For a factor of multiplicity \(m\), include its first through \(m\)-th powers in the denominators.
Use a constant numerator for each linear-factor power and a linear numerator for each irreducible-quadratic power.
Compare every denominator term with the factor inventory; when coefficients are available, clear the denominator and compare the resulting numerator.
Why Careful Templates Save Work
A complete template prevents a coefficient calculation from being built on an impossible starting point. If an intermediate power has been left out, solving a system for the remaining coefficients cannot repair that omission. Conversely, including a term whose coefficient later proves to be zero is harmless: it keeps the setup complete and lets the algebra determine its value.
The essential habit is to treat each distinct factor independently. Record its multiplicity, write every power from one up to that multiplicity, and then combine the resulting lists. With mixed denominators, this systematic inventory is more reliable than trying to guess the decomposition by inspection.
Check Your Understanding
For each question, identify the factor multiplicities before writing terms.
- What terms belong in the template for a proper rational function with denominator \((x-3)^2(x+1)^3\)?
- What is the complete template for denominator \((x^2+4)^3\)?
- For denominator \((x-1)(x+2)^2(x^2+1)^2\), how many constant numerator coefficients and how many linear numerator coefficients appear?
- Why should a term be retained in the template even if its coefficient may eventually be zero?
- For an improper rational function, what must be done before applying the partial-fraction template?
