Algebra-Gleichungen-Kubische Gleichungen

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