Why Repeated Factors Need a Template
A rational function can often be rewritten as a sum of simpler fractions. When a denominator contains a repeated linear factor, the form of that sum must account for the repetition. For a factor raised to the third power, for example, the template includes terms over the first, second, and third powers—not just one fraction over the full factor.
Writing the template is a structural step: it records which terms are allowed before any coefficients are determined. This tutorial focuses on that form, not on calculating its unknown constants.
When the Template Applies
A rational function is a quotient \(P(x)/Q(x)\) of polynomials, with \(Q(x)\) not the zero polynomial. It is proper when the degree of \(P\) is less than the degree of \(Q\). Partial-fraction templates apply directly to proper rational functions after the denominator has been factored into the relevant factors.
For a denominator made entirely of real linear factors, write its distinct factors as \(x-a_1,\ldots,x-a_r\), with multiplicities \(m_1,\ldots,m_r\). The factors must be distinct, so \(a_i\ne a_j\) when \(i\ne j\). The standard partial-fraction decomposition theorem says that a proper rational function with this denominator can be expressed using the following form:
Here every \(A_{i,j}\) is a constant. They are unknown until a later coefficient-solving step. If the denominator includes other kinds of factors, additional types of terms are needed; the formula above describes the case where it factors into real linear factors.
Why the Numerators Are Constants
The general partial-fraction rule assigns to a power of an irreducible factor a numerator polynomial whose degree is less than the degree of that factor. A linear factor has degree \(1\). Thus its numerator must have degree less than \(1\), which means it is a constant (including the possibility of zero). This explains why the repeated-linear-factor terms use constants rather than polynomials in \(x\).
The exponent on the denominator determines which term is being written. A factor with multiplicity \(m\) contributes a term for each exponent from \(1\) through \(m\). The coefficients for these terms are separate constants; they are not assumed to be equal.
Building a Template
To construct a template, first check whether the rational function is proper. Then factor the denominator and record each distinct linear factor and its multiplicity. For each factor, write a constant over every power from the first through the power appearing in the denominator. Finally, combine the lists for all factors.
If the numerator degree is at least the denominator degree, divide first and keep the resulting polynomial part.
Write the denominator as a product of distinct linear factors, each raised to its multiplicity.
For a factor with multiplicity \(m\), include its powers \(1\) through \(m\), each with its own constant numerator.
Multiply the proposed sum by the common denominator. The result must be a polynomial identity with the original numerator on the other side.
Worked Templates
Worked Example: One Factor Repeated Three Times
Write the template for \(\dfrac{2x+1}{(x-2)^3}\). The numerator has degree \(1\), while the denominator has degree \(3\), so this is proper. The only distinct linear factor is \(x-2\), with multiplicity \(3\). Therefore the template is
To check the form, multiply both sides by \((x-2)^3\). The resulting identity must be
Expanding the right-hand side gives
Thus the template is consistent precisely when this expanded polynomial equals \(2x+1\). The check confirms that each term has the correct denominator power and that clearing denominators produces a polynomial identity. No coefficient values have been determined here.
Worked Example: A Repeated Factor and a Distinct Factor
Construct the template for \(\dfrac{x^2+1}{(x+1)^2(x-3)}\). The denominator has degree \(3\), and the numerator has degree \(2\), so the fraction is proper. The factor \(x+1\) has multiplicity \(2\); the distinct factor \(x-3\) has multiplicity \(1\). Include two terms for the repeated factor and one for the other factor:
Multiplying through by the full denominator gives
Expand the products separately: \((x+1)(x-3)=x^2-2x-3\), and \((x+1)^2=x^2+2x+1\). Therefore the right-hand side becomes
This verifies the template by showing exactly the polynomial identity it represents. In particular, the repeated factor contributes both its first- and second-power terms, while the factor occurring once contributes only its first-power term.
Worked Example: Different Multiplicities and a Scaled Linear Factor
Consider \(\dfrac{x^4+2x+1}{(2x-5)^3(x+2)^2}\). The denominator degree is \(3+2=5\), and the numerator degree is \(4\), so the function is proper. The factor \(2x-5\) has multiplicity \(3\), and \(x+2\) has multiplicity \(2\). Both are linear, including the scaled factor \(2x-5\). The complete template is
To check every denominator power, multiply by \((2x-5)^3(x+2)^2\). The required identity is
Each term follows by canceling its own denominator factor against the common denominator: for example, the \(A_2\) term leaves one factor \(2x-5\) and two factors \(x+2\). The highest degree on the right is at most \(4\), matching the degree of the numerator. The displayed identity is the check on the form; finding the constants is a separate task.
If the Fraction Is Improper
If the numerator degree is at least the denominator degree, do polynomial division before writing the partial-fraction terms. The quotient is the polynomial part; the remainder over the original denominator is proper and receives the template.
Worked Example: Divide Before Writing the Template
Consider \(\dfrac{x^3+1}{(x-1)^2}\). Since \((x-1)^2=x^2-2x+1\), divide \(x^3+0x^2+0x+1\) by \(x^2-2x+1\). The first quotient term is \(x\), and
Subtracting gives \(2x^2-x+1\). The next quotient term is \(2\), and
Subtracting again gives the remainder \(3x-1\). Thus
The remainder fraction is proper because its numerator degree is \(1\), less than the denominator degree \(2\). Since \(x-1\) has multiplicity \(2\), its template is
For the check, multiplying the proper-fraction part by \((x-1)^2\) gives the identity \(3x-1=A(x-1)+B\). The division itself checks because \((x+2)(x-1)^2+(3x-1)=(x+2)(x^2-2x+1)+3x-1=x^3+1\). Both the quotient and the remainder are accounted for.
A Common Incomplete Template
A frequent error is to include only the highest power of a repeated factor. For instance, writing
is not the complete template for a denominator factor of multiplicity \(3\). The corrected form is
The correction is to list every power from \(1\) through the multiplicity, with an independent constant for each term. Do not omit the intermediate powers, and do not use one shared constant for all of them.
Why This Form Matters
A correct template organizes the algebra before coefficient calculations begin. It identifies the allowed terms and provides a direct check: after clearing denominators, the decomposition must produce a polynomial identity. A missing power or an incorrect multiplicity changes that identity and means the proposed form is incomplete.
Keep three checks in view: the fraction must be proper (or first be divided), every factor and multiplicity must be read from the factored denominator, and each repeated linear factor must contribute one constant-numerator term for every power up to its multiplicity.
Check Your Understanding
For each question, focus on constructing or checking the form; do not solve for the unknown constants.
- Write the complete template for \(\dfrac{3x+2}{(x+4)^4}\).
- Write the template for \(\dfrac{x^2-1}{(x-2)^2(x+3)^3}\). Is the fraction proper?
- For a factor \((5x+1)^2\), how many terms appear in its partial-fraction template, and what are their denominator powers?
- Why does a term associated with a linear factor have a constant numerator?
- What must be done before writing partial fractions if the numerator degree is greater than or equal to the denominator degree?
