faktorisieren y^6+7y^3-8

Lösung

faktorisieren

y^{6} + 7 y^{3} - 8

(y - 1) (y^{2} + y + 1) (y + 2) (y^{2} - 2 y + 4)

Schritte zur Lösung

y^{6} + 7 y^{3} - 8

Wende den rationalen Nullstellentest an

= (y - 1) \frac{y ^{6} + 7 y ^{3} - 8}{y - 1}

\frac{y ^{6} + 7 y ^{3} - 8}{y - 1} = y^{5} + y^{4} + y^{3} + 8 y^{2} + 8 y + 8

= (y - 1) (y^{5} + y^{4} + y^{3} + 8 y^{2} + 8 y + 8)

Faktorisiere

y^{5} + y^{4} + y^{3} + 8 y^{2} + 8 y + 8 : (y^{2} + y + 1) (y + 2) (y^{2} - 2 y + 2^{2})

= (y - 1) (y^{2} + y + 1) (y + 2) (y^{2} - 2 y + 2^{2})

2^{2} = 4

= (y - 1) (y^{2} + y + 1) (y + 2) (y^{2} - 2 y + 4)

f a c t or 2 x^{2} + 5 x + 14 x + 35

f a c t or 81 y^{4} - 1

f a c t or 9 y^{2} - 24 y + 16

5 (x + 2) + 4 x = 6 x + 1

s im pl i f y \frac{- 12}{- 2}