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The simplest way to effect the long division in Maple is to invoke a partialfraction decomposition. As a step in the decomposition, Maple carries out the long division.
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Controldrag the rational function.

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Context Panel:
Conversions≻Partial Fractions≻$x$


$\frac{{x}^{4}7{x}^{2}plus;5x8}{{x}^{2}2xplus;3}$$\to$${{x}}^{{2}}{\+}{2}{}{x}{}{6}{\+}\frac{{}{13}{}{x}{\+}{10}}{{{x}}^{{2}}{}{2}{}{x}{\+}{3}}$



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The quotient (${x}^{2}\+2x6$) and remainder ($13xplus;10$) can be read from the decomposition. Since the zeros of the divisor ${x}^{2}2xplus;3$ are the complex numbers $1\pm i\sqrt{2}$, the divisor cannot be factored over the reals, and the fraction "remainder/divisor" cannot be further decomposed by the technique of partial fractions.
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An alternate approach to determining the quotient and remainder, shown in Table 6.4.1(b), makes use of the quo command.
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Controldrag the rational function.

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Context Panel: Numerator (Denominator)

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Context Panel: Assign to a Name≻$n$ (d)


$\frac{{x}^{4}7{x}^{2}plus;5x8}{{x}^{2}2xplus;3}$${}$$\stackrel{\text{numerator}}{\to}$${{x}}^{{4}}{}{7}{}{{x}}^{{2}}{\+}{5}{}{x}{}{8}$$\stackrel{\text{assign to a name}}{\to}$${n}$${}$

$\frac{{x}^{4}7{x}^{2}plus;5x8}{{x}^{2}2xplus;3}$$\stackrel{\text{denominator}}{\to}$${{x}}^{{2}}{}{2}{}{x}{\+}{3}$$\stackrel{\text{assign to a name}}{\to}$${d}$

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Context Panel: Evaluate and Display Inline


$\mathrm{quo}\left(n\,d\,x\,\'r\'\right)$ = ${{x}}^{{2}}{\+}{2}{}{x}{}{6}$${}$

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Context Panel: Evaluate and Display Inline


$r$ = ${}{13}{}{x}{\+}{10}$${}$

Table 6.4.1(b) Use of the quo command to find the quotient and remainder



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The letter to which the remainder is assigned does not have to be "r", but it needs to have the single quotes around it in the quo command. If this name is already assigned and is unquoted, then a syntax error will result. The quotes serve to unassign the name, allowing the quo command to attach a new remainder to the name.
A final alternative for obtaining the partialfraction decomposition of a rational function is the convert command with the option parfrac.
$\mathrm{convert}\left(\frac{{x}^{4}7{x}^{2}plus;5x8}{{x}^{2}2xplus;3}comma;\mathrm{parfrac}comma;x\right)$ = ${{x}}^{{2}}{\+}{2}{}{x}{}{6}{\+}\frac{{}{13}{}{x}{\+}{10}}{{{x}}^{{2}}{}{2}{}{x}{\+}{3}}$${}$



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