WP-MA: Klasse 9 – Umformen von Summen in Produkte				erarbeitet von R. Bothe
| Aufgabenübersicht - WP | Aufgabe |
Lösungswege:
Forme die folgenden Summen in Produkte um
a)	14x^3 + 7x^2 + 21 x	=	7x × (2x^2 + x + 3)
b)	9y^2 + 3y^2x - 6yx^2	=	3y × (3y + xy - 2x^2)
c)	3 × (a + b) + a^2 × (a + b)	=	(a + b) × (3 + a^2)
d)	(3x - 2)^2 - 3ax + 2a	=	(3x - 2)^2 - a (3x - 2)
		=	(3x - 2) × (3x - 2 - a)
e)	9x^3 + 6x^2 + 21x +14	=	3x^2 × (3x + 2) + 7 × (3x + 2)
		=	(3x + 2) × (3x^2 + 7)
f)	9m^2 - 12m + 4	=	(3m - 2)^2
g)	e^2 + 8ex + 16x^2	=	(e + 4x)^2
h)	k^2 - 9	=	(k + 3)(k - 3)
i)	k^4- 16	=	(k^2 + 4)(k^2 - 4)
		=	(k^2 + 4)(k + 2)(k - 2)
j)	9x^2 - 4y^2	=	(3x + 2y)(3x - 2y)
k)	2x^2 - 20x +50	=	2(x^2 - 10x + 25)
		=	2(x - 5)^2
l)	b^2 × (a + b ) - a^2 × (a + b)	=	(a + b)(b^2 - a^2)
		=	(a + b)(b + a)(b - a)
		=	(a + b)^2× (b - a)
		=	- (a + b)^2× (a - b)
m)	(x + 3 )^2 + 4x^3 + 12x^2	=	(x + 3)^2 +4x^2(x + 3)
		=	(x + 3)(x + 3 + 4x^2)
		=	(x + 3)(4x^2 + x + 3)
n)	x^2 × (x - 3) - 4x +12	=	x^2 (x - 3) - 4(x - 3)
		=	(x - 3)(x^2 - 4)
		=	(x - 3)(x + 2)(x - 2)
o)	x^4 - 4x^3 - 2x^2 + 8x	=	x^3(x - 4) - 2x(x - 4)
		=	(x - 4)(x^3 - 2x)
		=	(x - 4) × x × (x^2 - 2)
		=	x(x - 4)(x + Ö2)(x - Ö2)
p)	x^2 - 6x - 16	=	(x^2 - 6x + 9) - 9 - 16
		=	(x - 3)^2 - 25
		=	(x - 3 + 5)(x - 3 - 5)
		=	(x + 2)(x - 8)
q)	x^3 + 4x^2 - 5x	=	x × (x^2 + 4x - 5)
		=	x × [(x^2 + 4x + 4) - 4 - 5]
		=	x × [(x + 2)^2 - 9]
		=	x × (x + 2 + 3)(x + 2 - 3)
		=	x × (x + 5)(x - 1)
r)	x^4 + 5x^2 + 6	=	x^4 + 3x^2 + 2x^2 + 6
		=	x^2 × (x^2 + 3) + 2 × (x^2 + 3)
		=	(x^2 + 3)(x^2 + 2)
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