Algebra-Gleichungen-Kubische Gleichungen
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Beispiel Nr: 19
$\begin{array}{l} \text{Gegeben:} ax^{3}+bx^{2}+cx+d=0
\\ \text{Gesucht:} \\ \text{Lösung der Gleichung} \\
\\ \\ \textbf{Gegeben:} \\ -5\frac{2}{5}x^3-37\frac{4}{5}x^2-75\frac{3}{5}x-43\frac{1}{5} =0\\ \\ \textbf{Rechnung:} \\\\-5\frac{2}{5}x^3-37\frac{4}{5}x^2-75\frac{3}{5}x-43\frac{1}{5}=0 \\\\ \text{Nullstelle für Polynmomdivision erraten:}-2\\
\,\small \begin{matrix} (-5\frac{2}{5}x^3&-37\frac{4}{5}x^2&-75\frac{3}{5}x&-43\frac{1}{5}&):( x +2 )=-5\frac{2}{5}x^2 -27x -21\frac{3}{5} \\
\,-(-5\frac{2}{5}x^3&-10\frac{4}{5}x^2) \\ \hline
&-27x^2&-75\frac{3}{5}x&-43\frac{1}{5}&\\
&-(-27x^2&-54x) \\ \hline
&&-21\frac{3}{5}x&-43\frac{1}{5}&\\
&&-(-21\frac{3}{5}x&-43\frac{1}{5}) \\ \hline
&&&0\\
\end{matrix} \\ \normalsize \\
\\
-5\frac{2}{5}x^{2}-27x-21\frac{3}{5} =0
\\
x_{1/2}=\displaystyle\frac{+27 \pm\sqrt{\left(-27\right)^{2}-4\cdot \left(-5\frac{2}{5}\right) \cdot \left(-21\frac{3}{5}\right)}}{2\cdot\left(-5\frac{2}{5}\right)}
\\
x_{1/2}=\displaystyle \frac{+27 \pm\sqrt{262\frac{11}{25}}}{-10\frac{4}{5}}
\\
x_{1/2}=\displaystyle \frac{27 \pm16\frac{1}{5}}{-10\frac{4}{5}}
\\
x_{1}=\displaystyle \frac{27 +16\frac{1}{5}}{-10\frac{4}{5}} \qquad x_{2}=\displaystyle \frac{27 -16\frac{1}{5}}{-10\frac{4}{5}}
\\
x_{1}=-4 \qquad x_{2}=-1
\\ \underline{x_1=-4; \quad1\text{-fache Nullstelle}} \\\underline{x_2=-2; \quad1\text{-fache Nullstelle}} \\\underline{x_3=-1; \quad1\text{-fache Nullstelle}} \\ \end{array}$