faktorisieren 125(x+2)^3+64(x-5)^3

Lösung

faktorisieren

125 (x + 2)^{3} + 64 (x - 5)^{3}

7 (9 x - 10) (3 x^{2} + 100)

Schritte zur Lösung

125 (x + 2)^{3} + 64 (x - 5)^{3}

Schreibe

125 (x + 2)^{3} + 64 (x - 5)^{3}

um:

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

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

Wende Formel zur Summe von dritten Potenzen an:

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

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

= (4 (x - 5) + 5 (x + 2)) (4^{2} (x - 5)^{2} - 5 \cdot 4 (x - 5) (x + 2) + 5^{2} (x + 2)^{2})

Fasse zusammen

= (4 (x - 5) + 5 (x + 2)) (16 (x - 5)^{2} - 20 (x - 5) (x + 2) + 25 (x + 2)^{2})

Faktorisiere

25 (x + 2)^{2} - 20 (x - 5) (x + 2) + 16 (x - 5)^{2} : 7 (3 x^{2} + 100)

= 7 (4 (x - 5) + 5 (x + 2)) (3 x^{2} + 100)

Multipliziere aus

4 (x - 5) + 5 (x + 2) : 9 x - 10

= 7 (9 x - 10) (3 x^{2} + 100)

s im pl i f y (\frac{1}{4})^{x}

x (x - 4) = 5

f a c t or 25 c^{2} - 20 c + 4

\frac{X}{4} + \frac{X}{3} \leq 90

s im pl i f y \frac{5}{2 5 - 6}