faktorisieren 1/27 x^3-8/125

Lösung

faktorisieren

\frac{1}{27} x^{3} - \frac{8}{125}

(\frac{x}{3} - \frac{2}{5}) (\frac{x ^{2}}{9} + \frac{2 x}{15} + \frac{4}{25})

Schritte zur Lösung

\frac{1}{27} x^{3} - \frac{8}{125}

Faktorisiere

\frac{x ^{3}}{27} - \frac{8}{125} : (\frac{x}{3} - \frac{2}{5}) ((\frac{1}{3})^{2} x^{2} + \frac{2}{5} \cdot \frac{1}{3} x + (\frac{2}{5})^{2})

= (\frac{x}{3} - \frac{2}{5}) ((\frac{1}{3})^{2} x^{2} + \frac{2}{5} \cdot \frac{1}{3} x + (\frac{2}{5})^{2})

Vereinfache

(\frac{x}{3} - \frac{2}{5}) ((\frac{1}{3})^{2} x^{2} + \frac{2}{5} \cdot \frac{1}{3} x + (\frac{2}{5})^{2}) : (\frac{x}{3} - \frac{2}{5}) (\frac{x ^{2}}{9} + \frac{2 x}{15} + \frac{4}{25})

= (\frac{x}{3} - \frac{2}{5}) (\frac{x ^{2}}{9} + \frac{2 x}{15} + \frac{4}{25})

s im pl i f y 2 5^{\frac{3}{2}}

f a c t or x^{3} - 3 x^{2} - 16 x + 48

∣ x - 9 ∣ = 8

x^{2} - 7 x + 31 = 0

2 x^{2} + 4 x + 4 = 0