faktorisieren 1/6 (2x^4+3x^3-16x-24)^2

Lösung

faktorisieren

\frac{1}{6} (2 x^{4} + 3 x^{3} - 16 x - 24)^{2}

\frac{1}{6} (2 x + 3)^{2} (x - 2)^{2} (x^{2} + 2 x + 4)^{2}

Schritte zur Lösung

\frac{1}{6} (2 x^{4} + 3 x^{3} - 16 x - 24)^{2}

Faktorisiere

(2 x^{4} + 3 x^{3} - 16 x - 24)^{2} : (2 x + 3)^{2} (x - 2)^{2} (x^{2} + 2 x + 4)^{2}

= \frac{1}{6} (2 x + 3)^{2} (x - 2)^{2} (x^{2} + 2 x + 4)^{2}

2 y + 3 > 9

(4 x - 3)^{2} = 14

∣ x - 5 ∣ + 1 \leq 2 x

x^{2} - 10 x + 16 < 0

5.4 x + 12 = 2 (2.7 x - 9)