Partial Fraction Decomposition (Linear Factors)

Definition A rational function is a quotient $\frac{P(x)}{Q(x)}$ of polynomials, where $Q$ is not the zero polynomial. It is proper if $\deg P < \deg Q$ and improper if $\deg P \ge \deg Q$. Partial fraction decomposition, as described below, applies only to proper rational functions.

Definition If $\frac{P}{Q}$ is improper, divide first. The division algorithm gives unique polynomials $S$ and $R$ with

Then $\frac{P(x)}{Q(x)} = S(x) + \frac{R(x)}{Q(x)}$ for every $x$ with $Q(x) \ne 0$, and only the proper part $\frac{R}{Q}$ is decomposed. Skipping this step cannot work, because a sum of partial fractions is always proper.

Definition Suppose the denominator factors as

with $c \ne 0$ and distinct real numbers $r_1, \ldots, r_n$, and suppose $\deg P < n$. Then there are unique real constants $A_1, \ldots, A_n$ with

for every $x$ other than $r_1, \ldots, r_n$. A factor written as $ax + b$ with $a \ne 0$ gets a term $\frac{A}{ax + b}$ in the same way.

Definition If $x - r$ divides $Q$ exactly $k$ times, so that $(x - r)^k$ is a factor of $Q$ but $(x - r)^{k + 1}$ is not, then the decomposition of a proper $\frac{P}{Q}$ contains the $k$ terms

one for every power from $1$ to $k$. A factor written as $(ax + b)^k$ with $a \ne 0$ gets the same $k$ terms with $ax + b$ in place of $x - r$. Together with the terms of the other factors this gives a unique decomposition, valid wherever $Q(x) \ne 0$.

Definition If a decomposition holds, multiplying both sides by $Q(x)$ gives two polynomials that agree at every $x$ with $Q(x) \ne 0$, which is all but finitely many numbers. Two polynomials that agree at infinitely many numbers are identical, because their difference would otherwise be a nonzero polynomial with infinitely many zeros. So the coefficients of each power of $x$ must match. The identity then holds at every $x$, including the zeros of $Q$, which is why substituting those zeros is allowed.

Definition If $\deg Q = n$, then after clearing denominators both sides have degree at most $n - 1$. Matching the coefficients of $1, x, \ldots, x^{n - 1}$ gives $n$ linear equations in the $n$ unknown constants, and by the uniqueness of the decomposition this system has exactly one solution. Solve it by substitution, elimination or matrices, as for any linear system. Substituting the zeros of $Q$ is a shortcut that gives equations with a single unknown.

Definition A partial fraction decomposition is an equality of functions on the domain of the original rational function, the set of all real $x$ with $Q(x) \ne 0$. At a zero of $Q$ the original function is undefined, even when every piece of the decomposition happens to be defined there. This can happen only when $P$ and $Q$ share a factor, as in rational functions with holes.