faktorisieren 16d^8-8d^4+1

Lösung

faktorisieren

16 d^{8} - 8 d^{4} + 1

(2 d^{2} + 1)^{2} (2 d + 1)^{2} (2 d - 1)^{2}

Schritte zur Lösung

16 d^{8} - 8 d^{4} + 1

Angenommen

u = d^{2}

= 16 u^{4} - 8 u^{2} + 1

Faktorisiere

16 u^{4} - 8 u^{2} + 1 : (2 u + 1) (2 u - 1) (2 u + 1) (2 u - 1)

= (2 u + 1) (2 u - 1) (2 u + 1) (2 u - 1)

Setze in

u = d^{2}

ein

= (2 d^{2} + 1) (2 d^{2} - 1) (2 d^{2} + 1) (2 d^{2} - 1)

Fasse zusammen

= (2 d^{2} + 1)^{2} (2 d^{2} - 1)^{2}

Wende Exponentenregel an:

a^{m} b^{m} = (ab)^{m},

angenommen

a \geq 0, b \geq 0

(2 d^{2} + 1)^{2} (2 d^{2} - 1)^{2} = ((2 d^{2} + 1) (2 d^{2} - 1))^{2}

= ((2 d^{2} + 1) (2 d^{2} - 1))^{2}

Faktorisiere

(2 d^{2} + 1) (2 d^{2} - 1) : (2 d^{2} + 1) (2 d + 1) (2 d - 1)

= ((2 d^{2} + 1) (2 d + 1) (2 d - 1))^{2}

Wende Exponentenregel an:

(a \cdot b)^{n} = a^{n} b^{n}

= (2 d^{2} + 1)^{2} (2 d + 1)^{2} (2 d - 1)^{2}

f a c t or 24 x + 18 y + 39

\frac{9}{2} (2 + x) \leq \frac{1}{3} (5 x + 14) \leq 11 - \frac{3 x}{2}

s im pl i f y 6 x y^{0}

f a c t or 4 x^{2} + 3 x y - 10 y^{2}

- 7 + 2 x \leq 5