2x^3+7x^2-3x-3=0

Lösung

2 x^{3} + 7 x^{2} - 3 x - 3 = 0

x = - \frac{1}{2}, x = \frac{- 3 + 21}{2}, x = \frac{- 3 - 21}{2}

Dezimale

x = - 0.5, x = 0.79128 \dots, x = - 3.79128 \dots

Schritte zur Lösung

2 x^{3} + 7 x^{2} - 3 x - 3 = 0

Faktorisiere

2 x^{3} + 7 x^{2} - 3 x - 3 : (2 x + 1) (x^{2} + 3 x - 3)

(2 x + 1) (x^{2} + 3 x - 3) = 0

Anwendung des Nullfaktorprinzips: Wenn

ab = 0

dann

a = 0

oder

b = 0

2 x + 1 = 0 or x^{2} + 3 x - 3 = 0

Löse

2 x + 1 = 0 : x = - \frac{1}{2}

Löse

x^{2} + 3 x - 3 = 0 : x = \frac{- 3 + 21}{2}, x = \frac{- 3 - 21}{2}

Die Lösungen sind

x = - \frac{1}{2}, x = \frac{- 3 + 21}{2}, x = \frac{- 3 - 21}{2}

x^{4} - 2 x^{3} + 6 x^{2} + 22 x + 13 = 0

x^{4} - 8 x^{3} + 24 x^{2} - 32 x + 15 = 0

x^{3} - 4 x^{2} - 11 x + 44 = 0

(x - 2)^{3} (x - 1)^{2} x = 0

x^{3} - 3 x - 1.8 = 0