What to Do First When f(x)/g(x) Is an Improper Fraction

Every rule in the four sections above assumes you're starting from a proper fraction (numerator's degree strictly less than denominator's). If you're handed an improper fraction instead — numerator's degree is greater than or equal to the denominator's — you cannot apply those rules directly; you first need to reduce the fraction to a polynomial part plus a genuinely proper remainder fraction, and only then decompose that remainder using Cases 1–4.

This reduction is exactly the division algorithm for polynomials: for any two polynomials $f(x)$ and $g(x) \neq 0$, there exist unique polynomials $q(x)$ and $r(x)$ such that

where either $r(x)=0$ or the degree of $r(x)$ is less than the degree of $g(x)$. Dividing both sides by $g(x)$ gives

and since $\dfrac{r(x)}{g(x)}$ is now a proper fraction, it can be resolved into partial fractions exactly as in the earlier sections. There are two flavors of this, depending on how much bigger the numerator's degree is:

• Degrees equal: $q(x)$ turns out to be just a constant (the ratio of the leading coefficients of $f$ and $g$), found by one quick division.
• Numerator's degree strictly greater: $q(x)$ is a genuine non-constant polynomial, found by ordinary long division of $f(x)$ by $g(x)$; the remainder $r(x)$ left over is what gets split into partial fractions.

Worked example (degrees equal). Resolve $\dfrac{2x^2+1}{x^2-1}$ into partial fractions.

The numerator and denominator both have degree 2, so this fraction is improper. Divide: $2x^2+1 = 2(x^2-1) + 3$, so

Now resolve the proper remainder using Case 1 (distinct linear factors): let $\dfrac{3}{(x-1)(x+1)} = \dfrac{A}{x-1}+\dfrac{B}{x+1}$, so $3 = A(x+1)+B(x-1)$. Putting $x=1$: $3=2A \Rightarrow A=\tfrac{3}{2}$. Putting $x=-1$: $3=-2B \Rightarrow B=-\tfrac{3}{2}$. Therefore …