factorize〖3(x+1)〗^2-7(x+1)+2
回答 (3)
2014-10-02 11:13 pm
Method1：
3 (x+1)² - 7(x+1) + 2
= [3(x+1) - 1][ (x+1) - 2]
= (3x+3-1)(x+1-2)
= (3x+2)(x-1)
Method2：
3 (x+1)² - 7(x+1) + 2
= 3 (x² + 2x + 1) - 7x - 7 + 2
= 3x² + 6x + 3 - 7x - 5
= 3x² - x - 2
= (3x+2)(x-1)
3 (x+1)² - 7(x+1) + 2
= [3(x+1) - 1][ (x+1) - 2]
= (3x+3-1)(x+1-2)
= (3x+2)(x-1)
Method2：
3 (x+1)² - 7(x+1) + 2
= 3 (x² + 2x + 1) - 7x - 7 + 2
= 3x² + 6x + 3 - 7x - 5
= 3x² - x - 2
= (3x+2)(x-1)
2014-10-02 11:11 pm
[3(x＋1)]^2 - 7(x＋1)＋2
＝ (3x＋3)^2 - 7x－7＋2
＝9x^2 ＋ 18x＋9 - 7x－5
＝9x^2 ＋11x ＋4
＝ (3x＋3)^2 - 7x－7＋2
＝9x^2 ＋ 18x＋9 - 7x－5
＝9x^2 ＋11x ＋4
收錄日期: 2021-04-15 16:56:22
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20141002000051KK00072