From 2 identities and factoriz

👨🏻‍🏫 學術台數學

tsz hin

2010-01-05 2:36 am

1) 6(a-1)b^3-(a-1)b^2

how to factorize this??
Give me the steps!!!The whole steps please...

更新1:

6(a-1)b^3-(a-1)b^2 = (a-1)(6b^3-b^2) = b^2(a-1)(6b-1) 點解(a-1)(6b^3-b^2)會變左做b^2(a-1)(6b-1)...有冇steps.....

回答 (3)

知識就是力量

2010-01-05 2:37 am

✔ 最佳答案

6(a-1)b^3-(a-1)b^2
= (a-1)(6b^3-b^2)
= b^2(a-1)(6b-1)

2010-01-04 18:38:39 補充：
Any answer?
If you have, please tell me whether it's wrong or right.

2010-01-04 19:20:37 補充：
係6b^3-b^2 入面各term抽 b^2
所以變了 (a-1)*b^2*(6b-1)
即是b^2(a-1)(6b-1)

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魔龍帝齊格益力多

2010-01-09 10:47 pm

6(a-1)b^3-(a-1)b^2
=(a-1)(6b3-b2)
=b2(a-1)(6b-1)

2010-01-09 14:52:56 補充：
我把6同b3乘左(6b3),因為6b3同b2都要乘(a-1),so(6b3-b2),

2010-01-09 15:16:07 補充：
6b3-b2), 仲有得抽

2010-01-09 15:16:59 補充：
同(ab+bc)=b(a+c)道理一樣

2010-01-11 15:06:02 補充：
你可把b^2(6b-1)乘左去,你就會明白.

參考: me

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用戶7639

2010-01-05 2:59 am

1) 6(a-1)b3-(a-1)b2
=6(a-1)(b2)(b)-(a-1)b2
=(a-1)(b2)(6b-1)

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收錄日期: 2021-04-23 20:39:22
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