question: factor 4y^ -1
更新1:
SO easy, know that I see it! I just do not have an eye for algebra. It's like a puzzle to me and I just can't find that corner piece. thanks again for all your help!
I need help factoring this algebra problem?
回答 (5)
2009-10-03 10:17 am
✔ 最佳答案
Answer:Given, 4y^2 - 1
=> (2^2)(y^2) - (1^2)
=> (2y)^2 - (1)^2
WKT, a^2 - b^2 = (a-b) (a+b)
So, (2y)^2 - (1)^2 = (2y - 1)(2y +1)
2009-10-03 9:26 am
(2y+1)(2y-1)
2009-10-03 11:09 am
a^2 - b^2 ≡ (a + b)(a - b)
4y^2 - 1
= (2y)^2 - 1^2
= (2y + 1)(2y - 1)
4y^2 - 1
= (2y)^2 - 1^2
= (2y + 1)(2y - 1)
2009-10-03 11:01 am
4y^2 - 1 = ( 2y - 1 ) ( 2y + 1 )
2009-10-03 9:42 am
question: factor 4y^ -1 =
= 4y^2 + 2y - 2y - 1
= 2y(2y + 1) - (2y + 1)
= (2y + 1)(2y - 1)
A N S W E R
= 4y^2 + 2y - 2y - 1
= 2y(2y + 1) - (2y + 1)
= (2y + 1)(2y - 1)
A N S W E R
收錄日期: 2021-05-01 12:46:34
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