faktorisieren x^6-12x^4+48x^2-64

Lösung

faktorisieren

x^{6} - 12 x^{4} + 48 x^{2} - 64

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

Schritte zur Lösung

x^{6} - 12 x^{4} + 48 x^{2} - 64

Angenommen

u = x^{2}

= u^{3} - 12 u^{2} + 48 u - 64

Faktorisiere

u^{3} - 12 u^{2} + 48 u - 64 : (u - 4) (u - 4) (u - 4)

= (u - 4) (u - 4) (u - 4)

Setze in

u = x^{2}

ein

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

Faktorisiere

x^{2} - 4 : (x + 2) (x - 2)

= (x + 2) (x - 2) (x^{2} - 4) (x^{2} - 4)

Faktorisiere

x^{2} - 4 : (x + 2) (x - 2)

= (x + 2) (x - 2) (x + 2) (x - 2) (x^{2} - 4)

Faktorisiere

x^{2} - 4 : (x + 2) (x - 2)

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

Fasse zusammen

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

f a c t or x^{2 + 14 x + 49}

f a c t or x^{2} - 17 + 72

f a c t or (x^{2} y^{5}) 2

f a c t or x^{3} - 7 x^{3} + 5 x - 35

f a c t or (- 1) - π