Let $R(z) = \frac{P(z)}{Q(z)}$ be a fractional expression, i.e., a quotient of the polynomials $P(z)$ and $Q(z)$ such that $P(z)$ is not divisible by $Q(z)$ . Let's restrict to the case that the coefficients are real or complex numbers.

If the distinct complex zeros of the denominator are $b_1,\,b_2,\,\ldots,\,b_t$ with the multiplicities $\tau_1,\,\tau_2,\,\ldots,\,\tau_t$ ($t \ge 1$ ), and the numerator has not common zeros, then $R(z)$ can be decomposed uniquely as the sum $$R(z) = H(z)+ \sum_{j=1}^t\left(\frac{A_{j1}}{z-b_j}+\frac{A_{j2}}{(z-b_j)^2}+\ldots +\frac{A_{j\tau_j}}{(z-b_j)^{\tau_j}}\right),$$ where $H(z)$ is a polynomial and the $A_{jk}$ 's are certain complex numbers.

Let us now take the special case that all coefficients of $P(z)$ and $Q(z)$ are real. Then the imaginary (i.e. non-real) zeros of $Q(z)$ are pairwise complex conjugates, with same multiplicities, and the corresponding linear factors of $Q(z)$ may be pairwise multiplied to quadratic polynomials of the form $z^2\!+\!pz\!+\!q$ with real $p$ 's and $q$ 's and $p^2 < 4q$ . Hence the above decomposition leads to the unique decomposition of the form

Cf. the partial fractions of fractional numbers.

Example. $$\frac{-x^5\!+\!6x^4\!-\!7x^3\!+\!15x^2\!-\!4x\!+\!3} {(x\!-\!1)^3(x^2\!+\!1)^2} \,=\, -\frac{1}{x\!-\!1}\!+\!\frac{3}{(x\!-\!1)^3}\!+ \!\frac{x}{x^2\!+\!1}\!+\!\frac{2x\!-\!1}{(x^2\!+\!1)^2}$$