faktorisieren 6(2p+q)^2-5(2p+q)-25

Lösung

faktorisieren

6 (2 p + q)^{2} - 5 (2 p + q) - 25

(3 (2 p + q) + 5) (2 (2 p + q) - 5)

Schritte zur Lösung

6 (2 p + q)^{2} - 5 (2 p + q) - 25

Zerlege die Ausdrücke in Gruppen

= (6 (2 p + q)^{2} + 10 (2 p + q)) + (- 15 (2 p + q) - 25)

Klammere

2 (2 p + q)

aus

6 (2 p + q)^{2} + 10 (2 p + q)

aus

: 2 (2 p + q) (3 (2 p + q) + 5)

Klammere

- 5

aus

- 15 (2 p + q) - 25

aus

: - 5 (3 (2 p + q) + 5)

= 2 (2 p + q) (3 (2 p + q) + 5) - 5 (3 (2 p + q) + 5)

Klammere gleiche Terme aus

3 (2 p + q) + 5

= (3 (2 p + q) + 5) (2 (2 p + q) - 5)

f a c t or 6 t^{2} - 8 + 40 cos (π t)

f a c t or (m - n)^{2} + 5 (m - n) - 24

f a c t or 12 (a - b)^{2} + 7 (a - b) - 12

f a c t or \infty^{2} - \infty

f a c t or + 9 + x y - x - 9 y^{2}