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= x(5x - 1) + 5 (5x - 1)
= (x(5x) - x(1)) + (5(5x) - 5(1))
= 5(x)(x) - x + (5)(5)x - 5
= 5x² - x + 25x - 5
= 5x² + (-x + 25x) - 5
= 5x² + (-1 + 25)x - 5
= 5x² + 24x - 5
c.(7 - 2x) × (2x - 7) = ...
= 7(2x - 7) - 2x (2x - 7)
= (7(2x) - 7(7)) + (-2x(2x) - 2x(-7))
= (7)(2)x - 49 + (-2)(2)(x)(x) + (-2)(-7)x
= 14x - 49 - 4x² + 14x
= - 4x² + 14x + 14x - 49
= - 4x² + (14 + 14)x - 49
= - 4x² + 28x - 49
Soal nomor 2
Tentukan nilai r pada persamaan bentuk aljabar (2x + 3y)(px + qy) = rx2 + 23xy + 12y2
Jawab:
(2x + 3y)(px + qy) = rx2 + 23xy + 12y2
2x × (px + qy) + 3y × (px + qy) = rx2 + 23xy + 12y2
2px2 + 2qxy + 3pxy + 3qy2 = rx2 + 23xy + 12y2
(2p)x2 + (2q + 3p)xy + (3q)y2 = rx2 + 23xy + 12y2
Jadi 2p = r
2q + 3p = 23
3q = 12
q = 12/3
q = 4
2q + 3p = 23
2 (4) + 3p = 23
8 + 3p = 23
3p = 23 – 8
3p = 15
p = 15/3
p = 5
2p = r
2 (5) = r
r = 10
Jadi nilai r adalah 10