faktorisieren 32m^{15}-243n^{10}

Lösung

faktorisieren

32 m^{15} - 243 n^{10}

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

Schritte zur Lösung

32 m^{15} - 243 n^{10}

Schreibe

32 m^{15}

um:

(2 m^{3})^{5}

Schreibe

243 n^{10}

um:

(3 n^{2})^{5}

= (2 m^{3})^{5} - (3 n^{2})^{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 m^{3})^{5} - (3 n^{2})^{5} = (2 m^{3} - 3 n^{2}) ((2 m^{3})^{4} + (2 m^{3})^{3} 3 n^{2} + (2 m^{3})^{2} (3 n^{2})^{2} + 2 m^{3} (3 n^{2})^{3} + (3 n^{2})^{4})

= (2 m^{3} - 3 n^{2}) ((2 m^{3})^{4} + (2 m^{3})^{3} 3 n^{2} + (2 m^{3})^{2} (3 n^{2})^{2} + 2 m^{3} (3 n^{2})^{3} + (3 n^{2})^{4})

Vereinfache

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

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

f a c t or 15 e^{2} + 28 e + 5

f a c t or f (π) - g (π)

f a c t or (3 x - 5) (3 x - 5)

f a c t or 4 0^{2} - 2 0^{2}

f (X) = (X)^{2}