Factorise it : 14(3y-5z)^3 + 7(3y-5z)^2 and with details and simple steps?
回答 (4)
2010-01-04 3:04 pm
✔ 最佳答案
14(3y - 5z)^3 + 7(3y - 5z)^2= 7[2(3y - 5z)^3 + (3y - 5z)^2]
= 7(3y - 5z)^2[2(3y - 5z) + 1]
= 7(3y - 5z)^2[2(3y) - 2(5z) + 1]
= 7(3y - 5z)^2(6y - 10z + 1)
2010-01-04 2:24 pm
14(3y - 5z)^3 + 7(3y - 5z)^2
7(3y - 5z)^2 * [2(3y - 5z) + 1]
7(3y - 5z)^2 * (6y - 10z + 1)
7(3y - 5z)^2 * [2(3y - 5z) + 1]
7(3y - 5z)^2 * (6y - 10z + 1)
參考: Derive 6
2010-01-04 2:05 pm
14(3y - 5z)^3 + 7(3y - 5z)^2 = 7(3y - 5z)^2 [2(3y - 5z) + 1] = 7(3y - 5z)^2 (6y - 10z + 1)
參考: I have taught math for over 40 yr.
2010-01-04 2:04 pm
收錄日期: 2021-05-01 12:56:25
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20100104055516AAPNoEP