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a.(x+2) (x-3) = x² - 3x + 2x - 6 = x² - x - 6
b.(2x-3) (3x-4) = 6x² - 8x - 9x + 12 = 6x² - 17x + 12
c.(4K+1)² = 16k² + 4k + 4k + 1 = 16k² + 8k + 1
d.(2+a) (a²-2a+3) ⇒ dimisalkan sebagai angka 3 karena di soal tak lengkap
= a³ + 2a² - 2a² - 4a + 3a + 6
= a³ - a + 6