faktorisieren 8x^6+y^9z^3

Lösung

faktorisieren

8 x^{6} + y^{9} z^{3}

(2 x^{2} + y^{3} z) (4 x^{4} - 2 x^{2} y^{3} z + y^{6} z^{2})

Schritte zur Lösung

8 x^{6} + y^{9} z^{3}

Schreibe

8 x^{6}

um:

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

Schreibe

y^{9} z^{3}

um:

(y^{3} z)^{3}

= (2 x^{2})^{3} + (y^{3} z)^{3}

Wende Formel zur Summe von dritten Potenzen an:

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

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

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

Vereinfache

(2 x^{2} + y^{3} z) ((2 x^{2})^{2} - y^{3} z \cdot 2 x^{2} + (y^{3} z)^{2}) : (2 x^{2} + y^{3} z) (4 x^{4} - 2 x^{2} y^{3} z + y^{6} z^{2})

= (2 x^{2} + y^{3} z) (4 x^{4} - 2 x^{2} y^{3} z + y^{6} z^{2})

s im pl i f y (x + y)^{(x + y)}

s im pl i f y 82

f a c t or \frac{x ^{2} + 2 x}{x + 2}

f a c t or (3 x + 12)^{'} + 15

f a c t or 2 a + 2 b + a x + b x