faktorisieren y^5+m^5b^5

Lösung

faktorisieren

y^{5} + m^{5} b^{5}

(y + bm) (y^{4} - b y^{3} m + b^{2} y^{2} m^{2} - b^{3} y m^{3} + b^{4} m^{4})

Schritte zur Lösung

y^{5} + m^{5} b^{5}

Schreibe

m^{5} b^{5}

um:

(mb)^{5}

= y^{5} + (mb)^{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})

n is odd

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

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

Vereinfache

(y + mb) (y^{4} - y^{3} mb + y^{2} (mb)^{2} - y (mb)^{3} + (mb)^{4}) : (y + bm) (y^{4} - b y^{3} m + b^{2} y^{2} m^{2} - b^{3} y m^{3} + b^{4} m^{4})

= (y + bm) (y^{4} - b y^{3} m + b^{2} y^{2} m^{2} - b^{3} y m^{3} + b^{4} m^{4})

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

f a c t or 2 π^{2} + 2 π r h

f a c t or 2 π (sin (x))

f (x) = sin (x) + cos (x)

f a c t or 2 x + π x