faktorisieren 32x^{10}-243y^{15}

Lösung

faktorisieren

32 x^{10} - 243 y^{15}

(2 x^{2} - 3 y^{3}) (16 x^{8} + 24 x^{6} y^{3} + 36 x^{4} y^{6} + 54 x^{2} y^{9} + 81 y^{12})

Schritte zur Lösung

32 x^{10} - 243 y^{15}

Schreibe

32 x^{10}

um:

(2 x^{2})^{5}

Schreibe

243 y^{15}

um:

(3 y^{3})^{5}

= (2 x^{2})^{5} - (3 y^{3})^{5}

Wende Regel zum Faktorisieren an:

x^{n} - y^{n} = (x - y) (x^{n - 1} + x^{n - 2} y + \dots + x y^{n - 2} y^{n - 1})

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

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

Vereinfache

(2 x^{2} - 3 y^{3}) ((2 x^{2})^{4} + 3 y^{3} (2 x^{2})^{3} + (3 y^{3})^{2} (2 x^{2})^{2} + (3 y^{3})^{3} \cdot 2 x^{2} + (3 y^{3})^{4}) : (2 x^{2} - 3 y^{3}) (16 x^{8} + 24 x^{6} y^{3} + 36 x^{4} y^{6} + 54 x^{2} y^{9} + 81 y^{12})

= (2 x^{2} - 3 y^{3}) (16 x^{8} + 24 x^{6} y^{3} + 36 x^{4} y^{6} + 54 x^{2} y^{9} + 81 y^{12})

f a c t or \frac{1}{2} x + 4

f a c t or d c + 3 c - a d - 3 a

f a c t or (5 x)^{2} + 4 (5 x) - 12

f a c t or 4 x^{2} + (- 8) x + (7)

f a c t or x - x^{b - 1} + x^{2}