~ BENTUK ALJABAR ~
1. Perkalian suku banyak dengan konstanta Tiap suku dikalikan konstanta Contoh
5(a + b + c) = ..... Jawab: 5(a + b + c) = 5a + 5b + 5c
-2(3p – 2q + r) = ..... Jawab: -2(3p – 2q + r) = -2(3p) – (-2)(2q) + (-2)r
= -6p + 4q – 2r
3(-4a2b + 2ab2 – 5ab) = ..... Jawab: 3(-4a2b + 2ab2 – 5ab) = -12a2b + 6ab2 – 15ab
2. Perkalian suku banyak dengan suku satu Tiap suku dikalikan suku satu itu. Contoh:
a(a + b + c) = ..... Jawab: a(a + b + c) = a2 + ab + ac
-5p(2p2 – p + 1) = ..... Jawab: -5p(2p2 – p + 1) = -10p3 + 5p2 – 5p
2x2yz(-5x2 + 2xy -3xz + y2 – 5yz + z2) = ..... Jawab: 2x2yz(-5x2 + 2xy -3xz + y2 – 5yz + z2) =
-10x4yz + 4x3y2z – 6x3yz2 + 2x2y3z – 10x2y2z2 + 2x2yz3
3. Perkalian suku banyak dengan suku banyak Tiap suku pada suku pertama dikalikan
dengan tiap suku pada suku banyak kedua. Contoh:
(x – 2y)(2x + 3y) = ..... Jawab: (x – 2y)(2x + 3y) = x(2x + 3y) – 2y(2x + 3y)
= 2x2 + 3xy – 4xy – 6y2
= 2x2 – xy – 6y2
3. Perkalian suku banyak dengan suku banyak Contoh:
(a + b + 2c)(3a – 4b + c) = ..... Jawab: (a + b + 2c)(3a – 4b + c)
= a(3a – 4b + c) + b(3a – 4b + c) + 2c(3a – 4b + c)= 3a2 – 4ab + ac + 3ab – 4b2 + bc + 6ac – 8bc + 2c2
= 3a2 – ab + 7ac – 4b2 – 7bc + 2c2
4. Perkalian Istimewa
a. Bentuk (x + a)(x + b)b. Bentuk (a + b)2
c. Bentuk (a – b)(a + b)d. Bentuk (a + b)n
e. Bentuk (a + b)(a2 – ab + b2)f. Bentuk (a – b)(a2 + ab + b2)
a. Bentuk (x + a)(x + b)
(x + a)(x + b) = x(x + b) + a(x + b) = x2 + bx + ax + ab = x2 + ax + bx + ab = x2 + (a + b)x + ab
(x + a)(x + b) = x2 + (a + b)x + ab
jumlahhasil kali
a. (x + a)(x + b) = x2 + (a + b)x + ab
Contoh: (x + 5)(x + 2) = ..... Jawab: (x + 5)(x + 2) = x2 + (5 + 2)x + (5)(2)
= x2 + 7x + 10
(a – 6)(a + 3) = ..... Jawab: (a – 6)(a + 3) = a2 + (-6 + 3)a + (-6)(3)
= a2 – 3a – 18
a. (x + a)(x + b) = x2 + (a + b)x + ab
Contoh: (3p – 2)(3p – 3) = ..... J: (3p – 2)(3p – 3) = (3p)2 + (-2 + (-3))(3p) + (-2)(-3)
= 9p2 + (-5)(3p) + 6 = 9p2 – 15x + 6
(-5y + 2)(-5y – 7) = ..... J: (-5y + 2)(-5y – 7) = (-5y)2 + (2 + (-7))(-5y) + (2)(-7)
= 25y2 + (-5)(-5y) – 14 = 25y2 + 25y – 14
b. Bentuk (a + b)2 dan (a – b)2
(a + b)2 = (a + b)(a + b) = a(a + b) + b(a + b) = a2 + ab + ab + b2
= a2 + 2ab + b2
(a + b) = a2 + 2ab + b2
(a – b)2 = a2 + 2a(-b) + (-b)2 = a2 – 2ab + b2
(a – b)2 = a2 – 2ab + b2
b. (a + b)2 = a2 + 2ab + b2
Contoh: (x + 5)2 = ..... Jawab: (x + 5)2 = x2 + 2(x)(5) + 52
= x2 + 10x + 25
(2p + 3q)2 = ..... Jawab: (2p + 3q)2 = (2p)2 + 2(2p)(3q) + (3q)2
= 4p2 + 12pq – 9q2
b. (a – b)2 = a2 – 2ab + b2
Contoh: (a – 4)2 = ..... Jawab: (a – 4)2 = a2 – 2(a)(4) + (4)2
= a2 – 8a + 16
(5ab – 3c)2 = ..... Jawab: (5ab – 3c4)2 = (5ab)2 – 2(5ab)(3c4) + (3c4)2
= 25a2b2 – 30abc4 – 9c8
c. Bentuk (a – b)(a + b)
(a – b)(a + b) = a(a + b) – b(a + b) = a2 + ab – ab – b2
= a2 – b2
(a – b)(a + b) = a2 – b2
c. (a – b)(a + b) = a2 – b2
Contoh: (x – 5)(x + 5) = x2 – 52
= x2 – 25
(2p + 3q)(2p – 3q) = (2p)2 – (3q)2
= 4p2 – 9q2
(-5ab + 7c)(-5ab – 7c) = (-5ab)2 – (7c)2
= 25a2b2 – 49c2
ILUSTRASI
a
a
b
b
a
a
a – b
a – b
ILUSTRASI
a
a
a – b
a – b
a – b
a – b
a
b
a
b
a + b
d. Bentuk (a + b)n
Perhatikan SEGITIGA PASCAL
(a + b)0 = 1 (a + b)1 = a + b (a + b)2 = a2 + 2ab + b2
(a + b)3 = a3 + 3a2b + 3ab2 + b3
11 1
1 2 11 3 3 1
1 4 6 4 11 5 10 10 5 1
1 6 15 20 15 6 1
d. Bentuk (a + b)n
Contoh: (2p + q)3 = ..... Jawab: (2p + q)3 = (2p)3 + 3(2p)2(q) + 3(2p)(q)3 + q3
= 8p3 + 12p2q + 6pq3 + q3
(x + y)4 = ..... Jawab: (x + y)4 = x4 + 4x3y + 6x2y2 + 4xy3 + y4
e. Bentuk (a + b)(a2 – ab + b2)
(a + b)(a2 – ab + b2) = a(a2 – ab + b2) + b(a2 – ab + b2) =a3 – a2b + ab2 + a2b – ab2 + b3 =a3 + b3
Jadi, (a + b)(a2 – ab + b2) = a3 + b3
dan (a – b)(a2 + ab + b2) = a3 – b3
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