How do I factor this completely? 4r^2 + 1 -4r ?
回答 (8)
2009-04-25 4:58 am
✔ 最佳答案
4r² + 1 - 4r rearrange into:4r² - 4r + 1
= 4r² - 4r + 1
= 4r² - 2r - 2r + 1 factor by grouping
= (4r² - 2r) - (2r + 1)
= (2r - 1)(2r - 1) same factors, so
= (2r - 1)²
2009-04-25 3:46 pm
(2r-1)^2
2009-04-25 12:52 pm
4r^2 + 1 - 4r
= 4r^2 - 4r + 1
= 4r^2 - 2r - 2r + 1
= (4r^2 - 2r) - (2r - 1)
= 2r(2r - 1) - 1(2r - 1)
= (2r - 1)(2r - 1)
= 4r^2 - 4r + 1
= 4r^2 - 2r - 2r + 1
= (4r^2 - 2r) - (2r - 1)
= 2r(2r - 1) - 1(2r - 1)
= (2r - 1)(2r - 1)
2009-04-25 11:52 am
4r² + 1 - 4r
= 4r² - 4r + 1
= (2r - 1)(2r - 1)
= (2r - 1)²
Done
= 4r² - 4r + 1
= (2r - 1)(2r - 1)
= (2r - 1)²
Done
2009-04-25 11:38 am
(2r-1)^2
2009-04-25 11:34 am
first change the order so it in decreasing order... 4r^2-4r+1=(2r-1)(2r-1) so (2r-1)^2
2009-04-25 11:32 am
(2r - 1)^2
2009-04-25 11:32 am
r+4 and r-1
收錄日期: 2021-05-01 12:18:32
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