Why Numerator Degrees Are Restricted
A partial-fraction template is not formed by placing an arbitrary polynomial over each factor of a denominator. Each numerator has a specific degree bound. For a linear factor, the numerator is constant; for an irreducible quadratic factor, it is linear or constant. The same rule applies at every power of a repeated factor.
These restrictions are not just a convenient convention. Polynomial division shows how any higher-degree numerator over a factor can be rewritten using lower powers of that factor, with a bounded-degree remainder. For a proper rational function, the polynomial terms produced along the way disappear into the polynomial part, which must be zero.
The Standard Form and Its Degree Bound
Work over the real numbers and suppose a denominator is factored into powers of distinct monic irreducible polynomials:
Here, each \(m_i\) is a positive integer, and no two \(q_i\) are the same factor. Over the real numbers, irreducible factors have degree one or two. A standard complete partial-fraction template has, for each factor \(q_i\), one term for every power from \(1\) through \(m_i\):
Thus, if \(q_i(x)\) is linear, \(A_{i,k}(x)\) must be a constant. If \(q_i(x)\) is an irreducible quadratic, \(A_{i,k}(x)\) may be a linear polynomial \(ax+b\), including the cases \(a=0\) or \(b=0\). A repeated factor does not change the bound: the numerators over \(q_i\), \(q_i^2\), and every higher power still have degree less than \(\deg q_i\).
This rule concerns the degree of each numerator compared with its own factor. It does not say that the numerator of the original rational function must have lower degree than every factor. The original numerator is compared with the degree of the entire denominator to decide whether the rational function is proper. If it is improper, polynomial division first separates its polynomial part.
Why Polynomial Division Gives the Bound
Let \(q(x)\) be a nonconstant polynomial of degree \(d\), and let \(B(x)\) be any polynomial. The polynomial division algorithm gives polynomials \(C(x)\) and \(R(x)\) such that
Dividing by \(q(x)^k\), where \(k\) is a positive integer, gives
The remainder term now has the required numerator bound. If \(k>1\), the quotient term still has a denominator involving \(q\), but its power has decreased by one. Divide its numerator by \(q\) again and repeat. Each repetition lowers the denominator power, so the process ends after finitely many steps.
Here is the resulting identity, with each remainder having degree less than \(d\):
The polynomial \(P(x)\) may be zero. If the original rational function is proper, then its polynomial part is zero after all the factors have been reduced in this way. To see why, each bounded-degree fraction \(R_j(x)/q(x)^j\) tends to zero as \(x\) becomes arbitrarily large: its numerator has smaller degree than its denominator. The sum of these fractions therefore tends to zero. A proper rational function also tends to zero. Their difference is the polynomial part, so that polynomial tends to zero. The only polynomial that tends to zero as \(x\) grows without bound is the zero polynomial.
For an improper rational function, a nonzero polynomial part can remain. This is why the degree bounds belong to the partial-fraction terms, not necessarily to every term in the full expression after division.
Worked Example: Checking a Mixed-Factor Template
Worked Example: Checking a Mixed-Factor Template
Consider the rational function
The factor \(x-2\) is linear and is repeated twice. The factor \(x^2+1\) is an irreducible quadratic over the real numbers. The bounded-degree rule therefore requires two constant numerators for the repeated linear factor and one linear-or-constant numerator for the quadratic factor. The template is
One valid decomposition is
Check it by combining terms over the common denominator. The three numerator contributions are
Adding these contributions gives
This reproduces the original numerator. As a numerical check, set \(x=0\), which is not a pole. The original fraction is \(4/4=1\). The decomposition gives
The template has the required bounds: \(A\) and \(B\) are constants, and \(Cx+D=x+1\) has degree one, which is less than the degree two of \(x^2+1\).
Worked Example: Repeated Irreducible Quadratic
Worked Example: Repeated Irreducible Quadratic
For a denominator \((x^2+1)^2\), the factor \(x^2+1\) has degree two and is repeated twice. The template has a numerator of degree less than two over each power:
For instance, consider this expression in that form:
Combining the terms over the common denominator gives the numerator
Thus the combined fraction is
Check at \(x=1\). The combined fraction is \(6/4=3/2\). The original two terms give \(3/2+0=3/2\), so the values agree. Both displayed numerators satisfy the rule: \(2x+1\) and \(x-1\) have degree one, below the quadratic factor's degree two.
Worked Example: Reducing an Overlong Numerator
Worked Example: Reducing an Overlong Numerator
Consider the improper rational function
A numerator of degree two placed directly over the repeated linear factor would not obey the standard bounded-degree template. First divide \(x^2+1\) by \(x-1\):
Therefore,
The numerator \(x+1\) over \(x-1\) is still too high in degree, so divide again:
Substitution gives the standard form
Check by recombining the right-hand side over \((x-1)^2\):
For a numerical check, set \(x=2\). The original expression is \(5/1=5\), and the reduced form gives \(1+2+2=5\). The constant numerators over the repeated linear factor now have degree zero, less than the factor's degree one. The separate constant \(1\) is the polynomial part.
A Common Template Mistake
A frequent error is to write an unrestricted polynomial over a factor and treat the result as a standard partial-fraction form. For example, the equality
is algebraically correct, but it is not yet in standard form: the numerator \(x+1\) has degree one, which is not less than the degree one of \(x-1\). Calling this a complete bounded-degree template is the mistake. Divide once more, \(x+1=(x-1)+2\), to obtain
The correction separates the polynomial part and leaves constant numerators over both powers of the linear factor. An equality can be true while its proposed form is still not the standard partial-fraction form.
How to Check a Proposed Template
Before determining any coefficients, check the denominator factors and the form of the proposed numerators. This catches structural errors before any coefficient calculations begin.
Identify each distinct irreducible linear or quadratic factor and its multiplicity.
A factor \(q^m\) requires terms over \(q,q^2,\ldots,q^m\).
Require degree zero for a linear factor and degree at most one for an irreducible quadratic.
If the rational function is improper, first account for polynomial division; do not force that polynomial part into the partial-fraction numerators.
The degree restriction matters in practice because it keeps the template precise and prevents unnecessary coefficients. It also reflects the way polynomial division reduces higher-degree numerators to remainders. Once the template is structurally correct, coefficient methods can be applied to determine its constants and linear numerators.
Check Your Understanding
For each question, use the factor degree to justify the required numerator form.
- What numerator form is allowed over \((3x-1)^4\) at each power?
- Write the numerator form required over each power of the irreducible quadratic \(x^2+4\).
- Why does the degree bound remain unchanged for the term over \(q^5\) when \(q\) is an irreducible quadratic?
- Use polynomial division to explain what happens to \(B(x)/q(x)^k\) when \(\deg B\geq\deg q\).
- Why may an improper rational function have a polynomial part even though each partial-fraction numerator obeys the degree bound?
