Why repeated quadratic factors need a complete template
A rational function may have a denominator containing the same irreducible quadratic factor more than once. The partial-fraction form must account for each power of that factor, not merely its highest power. This matters whether the goal is to rewrite the rational function for algebraic analysis or to prepare it for integration.
For example, if the denominator contains \((x^2+1)^3\), a complete template includes terms over \(x^2+1\), \((x^2+1)^2\), and \((x^2+1)^3\). Each term has a numerator of degree at most one. The coefficients in those numerators may turn out to be zero, but the terms belong in the template before the coefficients are determined.
Definitions and the template
A real quadratic \(q(x)=ax^2+bx+c\), where \(a\ne 0\), is irreducible over the real numbers if it cannot be factored into real linear factors. For a quadratic, this is equivalent to its discriminant satisfying \(b^2-4ac<0\).
A rational function \(P(x)/Q(x)\) is proper when \(\deg P<\deg Q\). The partial-fraction template below applies to a proper rational function whose denominator consists of powers of distinct irreducible real quadratics, with no linear factors. If the rational function is improper, first use polynomial long division and apply the template to the proper remainder.
For one repeated quadratic \(q(x)\) of multiplicity \(m\), this reduces to one term over each power from \(q\) through \(q^m\). The numerator over each power is linear or constant: it has the form \(Ax+B\). This is the standard partial-fraction rule that the numerator degree must be less than the degree of the corresponding irreducible factor. Since a quadratic has degree two, its numerator has degree at most one.
How the coefficients are determined
Once the template is written, multiply both sides by the original denominator. The result is a polynomial identity: two polynomials that agree for every \(x\) must have equal coefficients of corresponding powers of \(x\). This gives equations for the unknown numerator coefficients.
For example, for one quadratic \(q(x)\) repeated twice, clearing denominators gives
The equivalence follows by multiplying both sides by \(q(x)^2\), wherever the original rational functions are defined. The polynomial identity can then be solved by comparing coefficients. Afterward, recombining the terms provides a direct check.
Worked Examples
Worked Example: A quadratic repeated twice
Decompose \(\frac{x^3+2x^2+3x+4}{(x^2+1)^2}\). The factor \(x^2+1\) is irreducible over the reals, since its discriminant is \(0^2-4(1)(1)=-4<0\). Because it is repeated twice, write the full template
Multiply through by \((x^2+1)^2\):
Expand the right-hand side, keeping each coefficient visible:
Compare coefficients of \(x^3,x^2,x,\) and \(1\):
Thus \(A=1\), \(B=2\), \(C=3-A=2\), and \(D=4-B=2\). Therefore,
Check by recombining. The numerator on the right, over the common denominator, is
This is the original numerator, so the decomposition checks.
Worked Example: A quadratic repeated three times
Decompose \(\frac{x^5+x^4+4x^3+x^2+6x+2}{(x^2+1)^3}\). Since the factor has multiplicity three, include all three denominator powers:
Clear denominators. The first numerator is multiplied by \((x^2+1)^2\), the second by \(x^2+1\), and the third is unchanged:
Using \((x^2+1)^2=x^4+2x^2+1\), expand each product:
Therefore the coefficients of \(x^5,x^4,x^3,x^2,x,\) and \(1\) give
Substitution in order gives \(A=1\), \(B=1\), \(C=4-2A=2\), \(D=1-2B=-1\), \(E=6-A-C=3\), and \(F=2-B-D=2\). Hence
To check, recombine over the common denominator. The numerator becomes
The three expansions are \(x^5+x^4+2x^3+2x^2+x+1\), \(2x^3-x^2+2x-1\), and \(3x+2\). Adding them gives \(x^5+x^4+4x^3+x^2+6x+2\), as required.
Worked Example: Two distinct irreducible quadratics, one repeated
Consider the proper rational function
The factors \(x^2+1\) and \(x^2+4\) are distinct irreducible real quadratics: their discriminants are \(-4\) and \(-16\). The first factor is repeated twice; the second occurs once. The complete template is
In this example the coefficients are \(A=1\), \(B=1\), \(C=2\), \(D=-1\), \(E=3\), and \(F=2\). Verify them by combining the terms over the common denominator. The numerator is
Compute the products separately:
Adding the coefficients gives \(4x^5+3x^4+13x^3+8x^2+15x+2\), exactly the numerator in the original function. Thus the decomposition is
Why every power appears
The template includes every power because a numerator over the highest power alone cannot, in general, represent all proper numerators. For a quadratic \(q(x)\) repeated \(m\) times, the full template has separate linear numerators over \(q,q^2,\ldots,q^m\). Clearing denominators turns these terms into contributions multiplied by \(q^{m-1},q^{m-2},\ldots,1\), respectively. Omitting a power removes one of these available contributions, so the remaining terms need not be able to match the desired numerator.
There is a useful divisibility check on the top-power term. If one tried to represent \(1/q(x)^m\) using only terms with denominators \(q,q^2,\ldots,q^{m-1}\), then after multiplying by \(q^m\), the right-hand side would be a sum of terms each divisible by \(q(x)\). It would therefore be divisible by \(q(x)\). The left-hand side would be \(1\), which is not divisible by a quadratic \(q(x)\). That representation is impossible. The term over \(q^m\) is necessary in the general template, although its numerator can be zero for a particular function.
A common mistake
A frequent error is to include only the term over the highest power. For instance, one might try to write
Multiplying by \((x^2+1)^2\) would require the polynomial identity \(x^3+2x^2+3x+4=Ax+B\). This cannot hold: the left-hand side has degree three, while the right-hand side has degree at most one. The correction is to include the missing term over \(x^2+1\), as in the complete template. Another error is to use only constant numerators. That also restricts the form unnecessarily: a term over an irreducible quadratic may require a nonzero coefficient of \(x\), as the worked examples demonstrate.
Check Your Understanding
Use the template and coefficient-comparison method to answer these questions.
- Write the complete partial-fraction template for a proper rational function with denominator \((x^2+3x+5)^4\).
- For \(q(x)=x^2+4\), what is the discriminant, and what does its sign tell you about factoring over the reals?
- Why must a template for a quadratic repeated three times include terms over \(q(x)\), \(q(x)^2\), and \(q(x)^3\)?
- When clearing denominators in a template with terms over \(q(x)\) and \(q(x)^2\), what factors multiply the two numerators?
- How can recombining the terms check a proposed partial-fraction decomposition?
