faktorisieren 243x^{10}-y^5

Lösung

faktorisieren

243 x^{10} - y^{5}

(3 x^{2} - y) (81 x^{8} + 27 x^{6} y + 9 x^{4} y^{2} + 3 x^{2} y^{3} + y^{4})

Schritte zur Lösung

243 x^{10} - y^{5}

Schreibe

243 x^{10}

um:

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

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

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

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

Vereinfache

(3 x^{2} - y) ((3 x^{2})^{4} + (3 x^{2})^{3} y + (3 x^{2})^{2} y^{2} + 3 x^{2} y^{3} + y^{4}) : (3 x^{2} - y) (81 x^{8} + 27 x^{6} y + 9 x^{4} y^{2} + 3 x^{2} y^{3} + y^{4})

= (3 x^{2} - y) (81 x^{8} + 27 x^{6} y + 9 x^{4} y^{2} + 3 x^{2} y^{3} + y^{4})

f a c t or (3 x + 10) - 2^{2}

f a c t or ab x - ab y + pq x - pq y

f a c t or r m + y m + r n + y n

\frac{d}{d t} ((x) (t^{2} e^{t}))

f a c t or (x + y)^{3} + 3 x y (1 - x - y) - 1