Why Quadratic Factors Need Their Own Template
When a real rational function has a denominator that factors into linear and irreducible quadratic factors, its partial-fraction form must account for each factor. The quadratic factors require a different template from the linear factors: each quadratic term has a numerator with two unknown coefficients.
The word irreducible here means irreducible over the real numbers. Such a quadratic does not factor into real linear factors. Recognizing this distinction matters because the shape of a partial-fraction term depends on the type of factor in the denominator.
For example, \(x^2+1\) is irreducible over the reals because its discriminant is \(0^2-4(1)(1)=-4\), which is negative. By contrast, \(x^2-1\) is not irreducible: it factors as \((x-1)(x+1)\).
The Template for One Quadratic Factor
Suppose a proper rational function has an irreducible quadratic factor \(q(x)\) in its denominator, and that this factor occurs once. Its partial-fraction contribution has the form
where \(A\) and \(B\) are constants to be determined. The numerator is written as a general polynomial of degree less than two, so both its constant and \(x\)-terms are represented. The coefficients may turn out to be zero in a particular decomposition, but they are included in the template before the coefficients are known.
For two distinct irreducible quadratic factors \(q_1(x)\) and \(q_2(x)\), each occurring once, the corresponding part of the template is
Each factor gets its own pair of constants. The coefficients for one quadratic are not shared with those for another.
Worked Example: Two Distinct Irreducible Quadratics
Consider a denominator \((x^2+1)(x^2+4)\). Both factors are irreducible over the reals: their discriminants are \(-4\) and \(-16\). The denominator has two distinct quadratic factors, each with multiplicity one, so the template is
To check how this form recombines, take \(A=2\), \(B=1\), \(C=-1\), and \(D=3\). Using the common denominator gives
Thus the displayed template terms recombine to the stated rational function. The numerator has degree \(3\), while the denominator has degree \(4\), so this is proper.
Repeated Quadratic Factors
If an irreducible quadratic \(q(x)\) occurs to a power greater than one, the template includes a term for every positive power from \(q(x)^1\) through the highest power present. For multiplicity \(m\), the full contribution is
The denominator powers are \(q(x),q(x)^2,\ldots,q(x)^m\). Each power has its own numerator and its own two constants. The repeated-factor template is not just one fraction over \(q(x)^m\), and it is not a single numerator reused over several powers.
Worked Example: A Quadratic Repeated Three Times
For denominator \((x^2+1)^3\), the template must include three terms:
Here is a concrete check using \(A=1\), \(B=2\), \(C=-1\), \(D=3\), \(E=2\), and \(F=-1\). The common denominator is \((x^2+1)^3\), so the combined numerator is
Therefore the recombined rational function is
The numerator has degree \(5\), less than the denominator's degree \(6\). Recombining confirms that all three terms in the repeated-quadratic template contribute with the indicated powers.
Quadratics Alongside Other Factors
A denominator can contain both quadratic factors and other factors. In the full partial-fraction form, each irreducible quadratic contributes the quadratic terms described here, while linear factors contribute their own terms. The presence of linear factors does not change the quadratic template.
For instance, if a denominator contains a repeated irreducible quadratic \(q(x)^2\) and a distinct irreducible quadratic \(r(x)\), the quadratic portion of the template is
The two terms over \(q(x)\) reflect its multiplicity two; the single term over \(r(x)\) reflects its multiplicity one. If the denominator also contains linear factors, their terms are added separately.
Worked Example: Distinct Quadratics With Different Leading Coefficients
Let \(q(x)=2x^2+2x+3\) and \(r(x)=x^2-2x+5\). Their discriminants are
so both are irreducible over the reals. For the denominator \(q(x)r(x)\), the template is
Choose \(A=1\), \(B=1\), \(C=2\), and \(D=-1\) to check the form. Recombining over the common denominator gives
The numerator degree is \(3\), and the denominator degree is \(4\), so the rational function is proper. The check also shows that the template works without requiring the quadratic factors to be monic.
Why the Powers Appear: Clearing the Denominator
The structure of the template can be checked by multiplying through by the full denominator. This removes the fractions and turns a proposed decomposition into a polynomial identity. For a repeated quadratic \(q(x)^m\), multiplying one of its template terms by \(q(x)^m\) leaves a polynomial factor:
This identity explains why each denominator power has a distinct role after clearing denominators. The term over \(q(x)^j\) is multiplied by \(q(x)^{m-j}\); it does not disappear into a single term over the highest power. When other factors are present, their factors also appear in the corresponding cleared-denominator terms.
Separate real linear factors from irreducible quadratic factors, and record the multiplicity of each factor.
For a quadratic of multiplicity \(m\), include a separate numerator over every power from one through \(m\).
Write the numerator over each quadratic power as \(A_jx+B_j\), without assuming in advance that either coefficient is zero.
Use the common denominator and verify that the numerator produced matches the rational function being represented.
A Common Template Error
A frequent mistake is to use a constant numerator over an irreducible quadratic, as if the quadratic behaved like a linear factor. For example, one might incorrectly write
This identity would require \(x+1=A\) for every real \(x\), which is impossible for a constant \(A\). For instance, setting \(x=0\) would require \(A=1\), while setting \(x=1\) would require \(A=2\). The correction is to use the quadratic template \((Ax+B)/(x^2+1)\). In this example, \(A=1\) and \(B=1\), and the numerator then agrees exactly.
A second error is omitting a power when the quadratic is repeated. If the denominator contains \(q(x)^3\), writing only a term over \(q(x)^3\) does not give the full repeated-quadratic template. The template includes terms over \(q(x)\), \(q(x)^2\), and \(q(x)^3\), each with its own numerator.
Using the Template Carefully
A template specifies the allowed form of the partial fractions; it does not, by itself, determine the coefficients. Keep these tasks separate. First identify the real factors and their multiplicities, then write all required terms with fresh unknown constants. Only after the template is complete should you determine the constants and check the resulting identity.
The most reliable final check is to put the terms over a common denominator and expand the numerator. If that numerator equals the numerator of the original rational function, the proposed decomposition is verified. Checking factor types, multiplicities, and numerator constants before doing coefficient calculations can prevent a great deal of unnecessary work.
Check Your Understanding
For each question, focus on the structure of the template and on how it can be checked.
- What condition on the discriminant makes a real quadratic irreducible over the reals?
- Write the template for a denominator containing two distinct irreducible quadratics, each with multiplicity one.
- Write all the quadratic terms required when an irreducible quadratic \(q(x)\) occurs to the fourth power.
- For a denominator \(q(x)^2r(x)\), where \(q\) and \(r\) are distinct irreducible quadratics, how many quadratic terms are required?
- What polynomial identity results from multiplying a term over \(q(x)^j\) by \(q(x)^m\)?
