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Given the lefthand side of the identity $\frac{3}{{x}^{2}\+1}\frac{5}{x4}\=\frac{5{x}^{2}\+3x17}{{x}^{3}4{x}^{2}\+x4}$, the righthand side is obtained by finding the common denominator $\left({x}^{2}\+1\right)\cdot \left(x4\right)$ and adding the two fractions. Given the righthand side, the lefthand side is called its partialfraction decomposition. The decomposition is properly applied to a rational function $f\left(x\right)\=u\left(x\right)\/v\left(x\right)$ for which the degree of $u$ is less than the degree of $v$. Otherwise, a long division must first be performed so that $f\left(x\right)\=q\left(x\right)\+r\left(x\right)\/v\left(x\right)$, where $q$ is the quotient, and $r$ is the remainder. The decomposition is then applied to the fraction $r\/v$, not $f$.
Decomposing a rational function into its partial fractions requires that the denominator of the fraction be factored into nothing worse than quadratics. Indeed, for each nonrepeated linear factor $x\+A$ in the denominator of the rational function, its partial fraction decomposition will contain a term of the form $a\/\left(x\+A\right)$; and for each nonrepeated quadratic factor ${x}^{2}\+Bxplus;C$, a term of the form $\left(bxplus;c\right)sol;\left({x}^{2}plus;Bxplus;C\right)$. If a factor $\mathrm{\σ}$ appears raised to the power $n$, then the decomposition must contain the sum $\sum _{k\=1}^{n}\frac{{\mathrm{\λ}}_{k}}{{\mathrm{\σ}}^{k}}$, where ${\mathrm{\λ}}_{k}$ is ${a}_{k}$ or ${b}_{k}x\+{c}_{k}$, depending on whether $\mathrm{\σ}$ is linear or quadratic. Table 6.4.1 illustrates the structure of the partialfraction decomposition in several explicit cases of factored denominators $v\left(x\right)$.
$\mathit{v}\left(\mathit{x}\right)$

Decomposition Template

$\left(x\+{A}_{1}\right)\left(x\+{A}_{2}\right)$

$\frac{{a}_{1}}{x\+{A}_{1}}\+\frac{{a}_{2}}{x\+{A}_{2}}$

$\left(x\+{A}_{1}\right){\left(x\+{A}_{2}\right)}^{3}$

$\frac{{a}_{1}}{x\+{A}_{1}}\+\frac{{a}_{2}}{x\+{A}_{2}}\+\frac{{a}_{3}}{{\left(x\+{A}_{2}\right)}^{2}}\+\frac{{a}_{4}}{{\left(x\+{A}_{2}\right)}^{3}}$

$\left({x}^{2}\+{B}_{1}x\+{C}_{1}\right)\left({x}^{2}\+{B}_{2}x\+{C}_{2}\right)$

$\frac{{b}_{1}x\+{c}_{1}}{{x}^{2}\+{B}_{1}x\+{C}_{1}}\+\frac{{b}_{2}x\+{c}_{2}}{{x}^{2}\+{B}_{2}x\+{C}_{2}}$

$\left(x\+A\right){\left({x}^{2}\+Bxplus;C\right)}^{2}$

$\frac{a}{x\+A}\+\frac{{b}_{1}x\+{c}_{1}}{{x}^{2}\+Bxplus;C}plus;\frac{{b}_{2}xplus;{c}_{2}}{{\left({x}^{2}plus;Bxplus;C\right)}^{2}}$

Table 6.4.1 Decomposition templates for different denominators of $f\=u\/v$



To a linear factor (simpler than the quadratic) there corresponds a partial fraction with a "simple" numerator, namely, a constant; to a quadratic factor (messier than the linear), a "messy" numerator, namely, the linear expression $bxplus;c$. This observation is the basis for the author's own mnemonic: "Simplesimple; messymessy, but watch out for the repeats." Simple (linear) factors require simple numerators, but messy (quadratic) factors require "messy" numerators. The repeatedfactor rule is the one giving students most difficulty, and the irony in the mnemonic is the caution against "repeats" in the midst of a sentence that has multiple repeats.
A final issue to consider is the number field over which the denominator is factored. Working over the reals, the factor ${x}^{2}\+1$ is irreducible, and would be considered a quadratic factor. But working over the complex numbers, every polynomial is the product of linear factors, even if some of the linear factors are repeated. Controls engineers typically work over the complex field, so all factors would be considered linear. In an integral calculus course, a quadratic factor that was irreducible over the reals would typically remain a quadratic factor.
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