Why the Numerator Has a Degree Bound
A partial-fraction decomposition breaks a rational function into simpler terms associated with its denominator factors. Each term has a numerator, and that numerator is not chosen with arbitrary degree. Its degree is restricted by the factor in the denominator.
For a real linear factor, the numerator is a constant. For an irreducible quadratic factor, the numerator is linear or constant. This is not a special rule invented for quadratic expressions: it follows from the general requirement that each partial-fraction term be a proper rational function.
Understanding this degree bound helps distinguish a valid decomposition template from an expression that still contains a polynomial part. It also explains why the quadratic numerator has two possible coefficients, rather than three.
Irreducible Quadratics and Proper Terms
A quadratic polynomial \(q(x)=ax^2+bx+c\), where \(a\ne 0\), is irreducible over the real numbers if it cannot be factored into real linear factors. Equivalently, its discriminant satisfies \(b^2-4ac<0\). Its degree is \(2\).
A rational function \(P(x)/Q(x)\), with \(Q\) not the zero polynomial, is proper when the degree of \(P\) is less than the degree of \(Q\). A partial-fraction term associated with a denominator factor is written as a proper rational term: its numerator has degree strictly less than the degree of that factor.
The zero polynomial is also allowed as a numerator; it simply makes that term vanish. A constant numerator is included by taking \(A=0\). Thus the general numerator is \(Ax+B\), not necessarily a polynomial of degree exactly one.
Why the Bound Follows from Polynomial Division
The key fact is the polynomial division theorem. For any polynomial \(N(x)\) and any nonzero polynomial \(q(x)\), there are unique polynomials \(S(x)\) and \(R(x)\) such that \(N(x)=S(x)q(x)+R(x)\), where either \(R=0\) or \(\deg R<\deg q\). For a quadratic \(q\), the remainder therefore has degree at most one.
Write \(N=S q+R\), with \(R=0\) or \(\deg R<2\).
Divide the identity by \(q\) to obtain \(\frac{N}{q}=S+\frac{R}{q}\).
Since \(\deg R<2\), write \(R=Ax+B\), allowing either coefficient to be zero.
This proves the degree bound: any numerator over a quadratic can be divided into a polynomial part and a remainder of degree less than two. The proper rational term uses the remainder, so its numerator is linear or constant.
The restriction is also necessary for a term to be proper. If \(N\) has degree at least two and \(q\) has degree two, then \(N/q\) is not a proper rational term. Polynomial division identifies exactly what must be removed before the remaining fraction is proper.
Worked Example: The General Form for a Quadratic Factor
Worked Example: The General Form for a Quadratic Factor
Consider a rational function with denominator \((x-2)(x^2+4)\). The factor \(x-2\) has degree one, and \(x^2+4\) is an irreducible quadratic because its discriminant is \(0^2-4(1)(4)=-16<0\).
The numerator over the linear factor must have degree less than one, so it is a constant \(A\). The numerator over the quadratic factor must have degree less than two, so it is \(Bx+C\). The appropriate form is therefore
To check that these terms combine over the original denominator, multiply each term by the missing factor:
Expand the numerator:
The combined numerator has degree at most two, while the denominator has degree three, so the resulting rational function is proper. The form uses a constant above the linear factor and a linear-or-constant polynomial above the quadratic factor, as required. This example establishes the structure of the terms; determining \(A\), \(B\), and \(C\) is a separate coefficient question.
Worked Example: Reducing a Quadratic Numerator
Worked Example: Reducing a Quadratic Numerator
Consider \(\frac{3x^2+2x+1}{x^2+2x+5}\). First confirm that the denominator is an irreducible quadratic. Its discriminant is \(2^2-4(1)(5)=4-20=-16<0\).
The numerator has degree two, equal to the denominator's degree, so this fraction is not proper. Divide the numerator by the denominator. The leading terms give the quotient \(3\), since \(3x^2/x^2=3\). Multiply the denominator by \(3\), then subtract:
Thus the division gives quotient \(3\) and remainder \(-4x-14\), whose degree is one. The expression becomes
Check by recombining:
The recombined numerator is the original numerator. The remainder term has a linear numerator, as the degree bound requires; the constant \(3\) is the separate polynomial part.
Worked Example: Checking a Decomposition's Numerator Degrees
Worked Example: Checking a Decomposition's Numerator Degrees
Consider the expression \(\frac{2}{x+1}+\frac{3x-2}{x^2+1}\). The factor \(x^2+1\) is irreducible over the reals because its discriminant is \(0^2-4(1)(1)=-4<0\). Its degree is two, and the numerator \(3x-2\) has degree one, which is below two.
Combine the terms using the common denominator \((x+1)(x^2+1)\):
Expand both parts of the numerator:
Add them and simplify:
Therefore, the combined expression is \(\frac{5x^2+x}{(x+1)(x^2+1)}\). Its numerator has degree two and its denominator degree three, so it is proper. Recombining confirms the expression, and the numerator over the irreducible quadratic remains linear.
What the Degree Bound Does—and Does Not—Say
The bound is determined by the irreducible factor associated with a term, not by the degree of the entire denominator of the original rational function. In the first example, the whole denominator had degree three, but the term over \(x^2+4\) used a numerator of degree less than two because its associated factor had degree two.
Likewise, the bound does not say that every quadratic-factor term must have a nonzero \(x\)-coefficient. A numerator \(Bx+C\) includes a constant numerator when \(B=0\), and it can represent a zero term when both coefficients are zero.
For a repeated quadratic factor, the relevant degree bound is still based on the irreducible quadratic itself, rather than treating the whole power as a single irreducible factor. The detailed arrangement of terms for repeated factors is a distinct issue; the principle here is that each numerator is constrained by the degree of its irreducible denominator factor.
Common Mistake: Allowing a Quadratic Numerator
A frequent error is to write \(\frac{Ax^2+Bx+C}{x^2+1}\) as a standard proper partial-fraction term over \(x^2+1\). This assigns a numerator of degree two to a denominator factor of degree two, so the term is not proper. The error can be seen directly by dividing the numerator by the denominator.
Worked Example: Correcting an Overly High-Degree Numerator
Suppose someone proposes \(\frac{2x^2+3x+5}{x^2+1}\) as a single proper term over the irreducible quadratic \(x^2+1\). Divide \(2x^2+3x+5\) by \(x^2+1\). The leading-term quotient is \(2\). Subtract the product of the denominator and quotient:
Therefore the correct separation is
Check by recombining:
The corrected fraction has a linear numerator over the quadratic. The original expression was algebraically valid, but it had not been written as a proper partial-fraction term because its polynomial part had not been separated.
Why This Rule Matters
The degree bound makes partial-fraction forms precise and finite. Each irreducible quadratic contributes only two possible numerator coefficients, \(A\) and \(B\), rather than an unrestricted polynomial. That gives the decomposition a standard structure and prevents polynomial parts from being hidden inside proper rational terms.
When checking a proposed form, identify the denominator factor for each term and compare degrees. For a linear factor, a proper numerator is constant. For an irreducible quadratic, it is linear or constant. If a proposed numerator has degree at least that of its factor, use polynomial division to separate the polynomial part before treating the remaining fraction as proper.
Check Your Understanding
Use the degree of each denominator factor to justify your answers.
- Why does an irreducible quadratic factor require a numerator of degree less than two?
- What is the general numerator form over \(x^2+6x+10\), and why is this quadratic irreducible over the reals?
- Is \(\frac{4x^2-x+7}{x^2+3}\) a proper rational term? If not, what operation separates its polynomial part?
- In a term with denominator factor \(x^2+1\), is a constant numerator allowed? Explain using the general form.
- When a quadratic factor appears in the denominator of a larger rational function, is its numerator degree bound determined by the whole denominator or by that quadratic factor?
