【用戶】Jiunndar Yeh
【年級】幼兒園下
【評論內容】(23x + a)(bx + c) = 23bx2 + (ab + 23c)x + ac = 161x2 + 152x + 12比較得 23b = 161 ⇒ b = 161/23 = 7又得 ab + 23c = 7a + 23c = 152 ...式子(1) ac = 12 ⇒ (a, c) = (1, 12), (2, 6), (3, 4), (4, 3), (6, 2), (12,1)因式子(1)中可發現7a與23c必須同時為奇數或同時為偶數才可能使其和為偶數於是可排除(a, c) = (1, 12), (3, 4), (4, 3), (12,1)等選擇 (因為代進去式子(1)可發現其為(奇數)+(偶數) 或 (偶數) + (奇數) = (奇數),不可能為152(偶數))故將(a, c) = (2, 6), (6, 2)代入式子(1)後,可發現當a = 2, b = 6時,式子(1)才會成立故得a + b + c = 2 + 7 + 6 = 15 ...選項(B)正確[此時161x2 + 152x + 12 = (23x + 2)(7x + 6)符合條件]