Templates For Mixed Linear And Quadratic Factors

Why Mixed Denominators Need a Complete Form

A rational function can have a denominator containing several kinds of factors at once. Some factors may be linear, some may be irreducible quadratics, and some may be repeated. To decompose such a function correctly, every factor and every repeated power must be represented in the partial-fraction form.

The form is determined by the denominator, not by guessing which terms seem useful. A missing power or an overly large numerator can make the proposed form incomplete or redundant. The first step is therefore to factor the denominator over the real numbers and record the multiplicity of each factor.

The Form for Each Kind of Factor

Each distinct linear factor contributes one constant numerator term. If that factor is repeated, the form includes a term for every power, from the first power up to the full multiplicity. Thus, a factor \((x-r)^m\) contributes terms with denominators \(x-r,(x-r)^2,\ldots,(x-r)^m\).

An irreducible quadratic over the real numbers is a quadratic polynomial with no real roots. Each power of such a factor contributes a linear numerator: a constant plus a multiple of \(x\). If \(Q(x)\) is an irreducible quadratic and its multiplicity is \(n\), terms are included over \(Q(x),Q(x)^2,\ldots,Q(x)^n\).

For example, suppose the denominator has the factorization

where \(r\ne s\), \(Q\) is irreducible over the real numbers, and these factors are pairwise distinct. The complete form is

The letters \(A_j,B_j,C_j,E_j\) represent real constants to be determined. A coefficient is allowed to be zero; its possible value of zero does not mean that the corresponding term can be omitted from the general form in advance.

Worked Example: Distinct Linear Factors and a Quadratic

Worked Example: Repeated Linear and Quadratic Factors

Worked Example: Several Distinct and Repeated Factors

Why the Numerator Degrees and Powers Are Required

The degree of a numerator over a factor is limited by the degree of that factor. A linear factor has degree one, so the numerator over any of its powers is a constant. An irreducible quadratic has degree two, so the numerator over any of its powers has degree at most one. These bounds are essential: larger numerators would not be the standard partial-fraction form.

Repeated factors require every power because a rational expression may have contributions associated with each successive power. For example, a denominator containing \((x-r)^3\) requires terms over \(x-r\), \((x-r)^2\), and \((x-r)^3\). A denominator containing \(Q(x)^2\) requires terms over \(Q(x)\) and \(Q(x)^2\). The form includes all of these before any coefficients are solved.

Why the Complete Form Exists and Is Unique

The following argument explains why the stated forms are sufficient, not just plausible. Suppose the denominator factors into distinct monic irreducible real polynomials \(p_i(x)\), each with degree \(d_i\) and multiplicity \(e_i\). The linear and irreducible quadratic cases have \(d_i=1\) and \(d_i=2\), respectively. Write \(D(x)=\prod_i p_i(x)^{e_i}\), and let \(d=\deg D\).

For each power \(p_i^j\), allow a numerator of degree less than \(d_i\). The resulting candidate terms are \(N_{ij}(x)/p_i(x)^j\), where \(\deg N_{ij}<d_i\). Clearing the denominator converts their sum into a polynomial:

Each summand has degree at most \(d-1\), since its degree is at most \((d_i-1)+(d-jd_i)\le d-1\). Thus every such form produces a numerator of degree less than \(d\).

There are \(\sum_i e_i d_i=d\) freely chosen real coefficients in all the numerators. The space of polynomials of degree less than \(d\) also has dimension \(d\). It remains to show that different choices of coefficients cannot produce the same polynomial. This is the uniqueness step.

This proves that the linear map from the \(d\) coefficients in the proposed form to the \(d\) coefficients of a polynomial of degree less than \(d\) is injective. An injective linear map between finite-dimensional spaces of equal dimension is also surjective. Consequently, every proper numerator of degree less than \(d\) has exactly one decomposition of the stated form.

A Common Mistake

A frequent error is to include only the highest power of a repeated factor. For instance, with denominator \((x-1)^2(x^2+4)^2\), writing only

is incomplete. It omits the terms over \(x-1\) and \(x^2+4\). The correct form includes both powers of each repeated factor:

The correction is to list the powers systematically before writing any numerators. For a factor of multiplicity \(m\), write all powers \(1\) through \(m\); then assign the appropriate numerator degree to each term.