faktorisieren 2x^3-5x^2+x+2

Lösung

faktorisieren

2 x^{3} - 5 x^{2} + x + 2

(x - 1) (2 x + 1) (x - 2)

Schritte zur Lösung

2 x^{3} - 5 x^{2} + x + 2

Wende den rationalen Nullstellentest an

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

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

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

Faktorisiere

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

= (x - 1) (2 x + 1) (x - 2)

x^{\frac{5}{4}}

x - (2 x + 1) = 8 - (3 x + 3)

y^{2} - 5 y + 6 = 0

1 + 8 x - 14 = 19

- 10 x > - 7