Partial Fraction Decomposition

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Let

r(x) = \frac{b_0 + b_1x + \cdots + b_{n-1}x^{n-1}}{(c_1 + d_1x)(c_2 + d_2x)\cdots(c_n + d_nx)},

where d_i \neq 0 for all i = 1, 2, \dots, n. We assume the denominator consists of unique linear factors and there are no common factors between the numerator and the denominator.

Rewrite r(x) in the form

\hat{r}(x) = \frac{A_1}{c_1 + d_1x} + \frac{A_2}{c_2 + d_2x} + \cdots + \frac{A_n}{c_n + d_nx},

where A_1, A_2, \dots, A_n are real constants.

Let p(x) = b_0 + b_1x + \cdots + b_{n-1}x^{n-1} and q(x) = \prod_{k=1}^n(c_k + d_kx). Thus, r(x) = p(x)/q(x). Let us consider the linear factors of q(x).

Let q(x) = \prod_{k=1}^nl_k(x), where l_k(x) = c_k + d_kx. By inspection, l_k(-c_k/d_k) = 0 and l_i(-c_k/d_k) \neq 0 for all i \neq k since the factors are distinct.

If we equate r(x) = \hat{r}(x) and multiply both sides by the denominator q(x),

\begin{array}{rcl} r(x)q(x) & = & \hat{r}(x)q(x)\\ p(x) & = & A_1q(x)/l_1(x) + A_2q(x)/l_2(x) + \cdots + A_nq(x)/l_n(x)\\ & = & A_1\frac{\prod_{k=1}^nl_k(x)}{l_1(x)} + A_2\frac{\prod_{k=1}^nl_k(x)}{l_2(x)} + \cdots + A_n\frac{\prod_{k=1}^nl_k(x)}{l_n(x)}. \end{array}

Let’s define a special function

d_j(x) = \prod_{\begin{tiny}k \neq j\end{tiny}}l_k(x) = \prod_{\begin{tiny}k \neq j\end{tiny}}(c_k + d_kx)

that is the product of all but the j-th linear factor of q(x). Hence p(x) becomes

p(x) = A_1d_1(x) + A_2d_2(x) + \cdots + A_kd_k(x) + \cdots + A_nd_n(x)

Let x_k = -c_k/d_k. Thus,

p(x_k) = A_1d_1(x_k) + A_2d_2(x_k) + \cdots + A_kd_k(x_k) + \cdots + A_nd_n(x_k).

But d_i(x) contains the linear factor l_k(x) for every i \neq k. Hence only d_k(x_k) is nonzero and

p(x_k) = A_kd_k(x_k) \Rightarrow A_k = p(x_k)/d_k(x_k).

For example, rewrite

r(x) = \frac{2 + 4x + 6x^2}{(1 - x)(1 + x)(2 + 3x)}

in the form

\frac{A_1}{1 - x} + \frac{A_2}{1 + x} + \frac{A_3}{2 + 3x}.

We have

\begin{array}{rcl} p(x) & = & 2 + 4x + 6x^2, \\ d_1(x) & = & (1 + x)(2 + 3x), \\ d_2(x) & = & (1 - x)(2 + 3x), \rm{and} \\ d_3(x) & = & (1 - x)(1 + x). \end{array}

Hence x_1 = 1, x_2 = -1, and x_3 = -2/3. Thus, A_1 = p(1)/d_1(1) = 12/10 = 6/5, A_2 = p(-1)/d_2(-1) = 4/(-2) = -2, and A_3 = p(-2/3)/d_3(-2/3) = 2/(5/9) = 18/5.

We have

\begin{array}{rcl} r(x) & = & \frac{2 + 4x + 6x^2}{(1 - x)(1 + x)(2 + 3x)} \\ & = & \frac{6/5}{1 - x} - \frac{2}{1 + x} + \frac{18/5}{2 + 3x}. \end{array}

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