Perkalian Bentuk Aljabar

~ 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 Bentuk Bentuk Bentuk Bentuk Bentuk Bentuk

(x + a)(x + b) (a + b)2 (a – b)(a + b) (a + b)n (a + b)(a2 – ab + b2) (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 jumla h

hasil 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)

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 – b 2  (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– b

b b a

a a– b a

a

ILUSTRASI a– b

a– b

b

a a+ b a

a a– b a

b a– b

d. Bentuk (a + b)n 1

 Perhatikan SEGITIGA PASCAL 1 1 1

1 1 1

3 4

5

2

3

1 1

6 4 1 10 10 5 1 15 20 15 6 1

 (a + b)0 = 1 1 6  (a + b)1 = a + b  (a + b)2 = a2 + 2ab + b2  (a + b)3 = a3 + 3a2b + 3ab2 + b3

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