Faktorisieren

Mathe 11
Entwickeln & Faktorisieren - Lösungen
2
Terme
2
1a) 4xy ⋅ (2x − 3y + 5z) = 4xy ⋅ 2x − 4xy ⋅ 3y + 4xy ⋅ 5z = 8x y − 12xy + 20xyz
1b) 3a ⋅ (3a − 6) ⋅ 2a = 3a ⋅ 2a ⋅ (3a − 6) = 6a 2 ⋅ (3a − 6) = 6a 2 ⋅ 3a − 6a 2 ⋅ 6 = 18a 3 − 36a 2
1c) (2x − 7) ⋅ (3x + 4) = 2x ⋅ 3 x + 2x ⋅ 4 − 7 ⋅ 3x − 7 ⋅ 4 = 6x 2 + 8 x − 21x − 28 = 6x 2 − 13x − 28
1d) ( x + 5) ⋅ ( x 2 − 5x + 3) = x ⋅ (x 2 − 5x + 3) + 5 ⋅ (x 2 − 5x + 3) = x 3 − 5x 2 + 3x + 5x 2 − 25x + 15 = x 3 − 22x + 15
2a) ( x + 3 y )2 = x 2 + 2 ⋅ x ⋅ 3 y + (3 y )2 = x 2 + 6 xy + 9 y 2
{ (a + b)2 = a 2 + 2ab + b 2 }
2b) ( x + 4 y ) ⋅ ( x − 4 y ) = x 2 − ( 4 y )2 = x 2 − 16 y 2
{ (a + b) ⋅ (a − b) = a 2 − b 2 }
2c) ( 4a − 5b)2 = ( 4a)2 − 2 ⋅ 4a ⋅ 5b + (5b)2 = 16a2 − 40ab + 25b2
{ (a − b)2 = a 2 − 2ab + b 2 }
2d) ( 1 x + 1 )2 = ( 1 x )2 + 2 ⋅ 1 x ⋅ 1 + ( 1 )2 = 1 x 2 + 1 x + 1
{ (a + b)2 = a 2 + 2ab + b 2 }
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4
2
2
4
4
4
4
16
3a) 15 xy − 9 x 2 + 6 xz = 3 x ⋅ 5 y − 3 x ⋅ 3 x + 3 x ⋅ 2z = 3 x ⋅ (5 y − 3 x + 2z )
3b) 21a 2b3 − 28ab 2 = 7ab 2 ⋅ 3ab − 7ab 2 ⋅ 4 = 7ab 2 ⋅ (3ab − 4)
3c) 12st 2 + 4st − 8s2 t = 4st ⋅ 3t + 4st ⋅ 1 − 4st ⋅ 2s = 4st ⋅ (3t + 1 − 2s)
3d) 2 x 2 − 1 xy 2 = 1 x ⋅ 2 x − 1 x ⋅ 1 y 2 = 1 x ⋅ (2x − 1 y 2 ) oder = 4 x 2 − 1 xy 2 = 1 x ⋅ ( 4x − y 2 )
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6
3
1
3
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3
6
2
6
6
2ax + 6ay + bx + 3by = 2a ⋅ ( x + 3y) + b ⋅ (x + 3y) = (2a + b) ⋅ (x + 3y)
4a)
oder
= 2ax + bx + 6ay + 3by = x ⋅ (2a + b) + 3y ⋅ (2a + b) = (x + 3y) ⋅ (2a + b)
3x 3 + 3x − 4x 2 − 4 = 3x ⋅ ( x 2 + 1) − 4 ⋅ ( x 2 + 1) = (3x − 4) ⋅ ( x 2 + 1)
4b)
oder = 3x 3 − 4x 2 + 3x − 4 = x 2 ⋅ (3x − 4) + 1⋅ (3x − 4) = ( x 2 + 1) ⋅ (3x − 4)
= 2a 3 + 5a 2 − 6a − 15 = a 2 ⋅ (2a + 5) − 3 ⋅ (2a + 5) = (a 2 − 3) ⋅ (2a + 5)
4c)
oder
= 2a 3 − 6a + 5a 2 − 15 = 2a ⋅ (a 2 − 3) + 5 ⋅ (a 2 − 3) = (2a + 5) ⋅ (a 2 − 3)
5a) x 2 + 6 xy + 9 y 2 = x 2 + 2 ⋅ x ⋅ 3 y + (3 y )2 = ( x + 3 y )2
{ siehe 2a) }
{ a 2 + 2ab + b 2 = (a + b)2 }
5b) 4a2 − 28ab + 49b 2 = (2a)2 − 2 ⋅ 2a ⋅ 7b + (7b)2 = (2a − 7b)2
{ a 2 − 2ab + b 2 = (a − b) 2 }
5c) 9 x 2 − 64 y 2 = (3 x )2 − (8 y )2 = (3 x + 8 y ) ⋅ (3 x − 8 y )
{ a 2 − b 2 = ( a + b) ⋅ ( a − b) }
5d) 4 x 2 − 1 = ( 2 x )2 − 12 = ( 2 x + 1) ⋅ ( 2 x − 1)
{ a 2 − b 2 = ( a + b) ⋅ ( a − b) }
25
5
6a) x 2 + 12xy +
6b) 9a 2 −
5
= x 2 + 2 ⋅ x ⋅ 6y +
+ 49b 2 = (3a)2 −
− 48x + 36 =
6c)
5
6d) 3y 2 + 18y +
= x 2 + 2 ⋅ x ⋅ 6y + (6y)2 = x 2 + 12xy + 36y 2
+ (7b) 2 = (3a) 2 − 2 ⋅ 3a ⋅ 7b + (7b)2 = 9a 2 − 42ab + 49b 2
− 2 ⋅ ? ⋅ 6 + 62 =
= 3 ⋅ (y 2 + 6y +
− 2 ⋅ 4x ⋅ 6 + 62 = (4x )2 − 2 ⋅ 4x ⋅ 6 + 62 = 16x 2 − 48x + 36
) = 3 ⋅ ( y 2 + 2 ⋅ y ⋅ 3 + 3 2 ) = 3 ⋅ (y 2 + 6y + 9) = 3y 2 + 18y + 27
7a) 3 x 2 − 12x + 12 = 3 ⋅ ( x 2 − 4 x + 4) = 3 ⋅ ( x − 2)2 7b) 8 x 3 + 24 x 2 + 18 x = 2x ⋅ ( 4 x 2 + 12 x + 9) = 2x ⋅ ( 2x + 3)2
7c) s3 t − st 3 = st ⋅ (s2 − t 2 ) = st ⋅ (s + t ) ⋅ (s − t )
7d) 5a 4 − 20 = 5 ⋅ (a 4 − 4) = 5 ⋅ (a 2 + 2) ⋅ (a 2 − 2)
7e*) 1 x 2 − 2xy + 5 y 2 = 1 x 2 − 10 xy + 25 y 2 = 1 ⋅ ( x 2 − 10 xy + 25 y 2 ) = 1 ⋅ ( x − 5 y )2
5
5
5
5
5
5
7f*) 6a 6 − 96b 4 = 6 ⋅ (a 6 − 16b 4 ) = 6 ⋅ [(a 3 )2 − ( 4b 2 )2 ] = 6 ⋅ (a 3 + 4b) ⋅ (a 3 − 4b)
8)
( x + 1)2 − (x − 1)2 = ( x 2 + 2x + 1) − ( x 2 − 2x + 1) = x 2 + 2x + 1 − x 2 + 2x − 1 = 4x
Man erhält das Vierfache der Zahl.
Backer '14