$y=\frac{2x^3+2x^2-32x+40}{x^3+2x^2-13x+10}=\frac{2{\left(x-2\right)}^2\cdot \left(x+5\right)}{\left(x-1\right)\left(x-2\right)\left(x+5\right)}=\frac{2\left(x-2\right)}{\left(x-1\right)}$ .
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Zähler $2x^3+2x^2-32x+40=0$, $x_1=2$: .
$\begin{array}{ccccc}\hfill \hfill & \hfill 2\hfill & \hfill 2\hfill & \hfill -32\hfill & \hfill 40\hfill \\ \hfill 2\hfill & \hfill \hfill & \hfill 4\hfill & \hfill 12\hfill & \hfill -40\hfill \\ & & & & \\ \hfill \hfill & \hfill 2\hfill & \hfill 6\hfill & \hfill -20\hfill & \hfill 0\hfill \\ \hfill \hfill \end{array}$ .
$2x^2+6x-20$, $x_2=2$: .
$\begin{array}{cccc}\hfill \hfill & \hfill 2\hfill & \hfill 6\hfill & \hfill -20\hfill \\ \hfill 2\hfill & \hfill \hfill & \hfill 4\hfill & \hfill 20\hfill \\ & & & \\ \hfill \hfill & \hfill 2\hfill & \hfill 10\hfill & \hfill 0\hfill \\ \hfill \hfill \end{array}$ .
$2\left(x+5\right)=0$, $x_3=-5$: .

Nenner: $x^3+2x^2-13x+10=0$, $x_1=1$: .
$\begin{array}{ccccc}\hfill \hfill & \hfill 1\hfill & \hfill 2\hfill & \hfill -13\hfill & \hfill 10\hfill \\ \hfill 1\hfill & \hfill \hfill & \hfill 1\hfill & \hfill 3\hfill & \hfill 10\hfill \\ & & & & \\ \hfill \hfill & \hfill 1\hfill & \hfill 3\hfill & \hfill -10\hfill & \hfill 0\hfill \\ \hfill \hfill \end{array}$ .
$x^2+3x-10=0$, $x_2=2,x_3=-5$ .

$\Rightarrow$ Beseitigung der Definitionslücken ($x=2,-5$) .
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$y=\frac{2x^3+2x^2-32x+40}{x^3+2x^2-13x+10}=\frac{2{\left(x-2\right)}^2\left(2+5\right)}{\left(x-1\right)\left(x-2\right)\left(x+5\right)}$ $=\frac{2\left(x-2\right)}{\left(x-1\right)}$ .
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