標題: 一個中三 Factorization 問題 [打印本頁] 作者: anthony 時間: 17/9/2005 06:58 PM 標題: 一個中三 Factorization 問題 如果 x^3-8=(x-2)(x^2+2x+4)，那麼x^3+x-10等如多少？ P.S.我看過書的答案，x^3+x-10應該等如(x-2)(x^2+2x+5)，但我不懂這一題數的步驟，麻煩大家盡量用中三程度的方法來解釋。作者: 法皇拿破崙 時間: 17/9/2005 07:06 PM X^3-8 = (x-2)(x^2 + 2x +4) X^3+ x – 10 = (X^3-8) + (x –2) <--- you have to use the above equation, so you need to rewrite like this = (x-2)(x^2 + 2x +4) + (x –2) = (x-2) (x^2 + 2x +4+1) <---take out the common factor(x-2) = ( x-2) (x^2 + 2x +4+5)
[ Last edited by 法皇拿破崙 on 2005-9-17 at 07:11 PM ]作者: 細薯 時間: 18/9/2005 09:04 PM
Originally posted by 法皇拿破崙 at 2005-9-17 07:06 PM: = (x-2) (x^2 + 2x +4+1) <---take out the common factor(x-2) = ( x-2) (x^2 + 2x+4+5)