factor by grouping 14r^2-45r+14?
回答 (4)
2008-07-04 6:18 pm
✔ 最佳答案
Hi,Your problem as written does not factor.
14r² - 45r + 14
However, if it was supposed to be 14r² - 45r - 14, then the factors are:
(2r - 7)(7r + 2) <==ANSWER
I hope that helps!! :-)
2008-07-04 6:28 pm
14r^2-45r+14 = 14(r^2 -(45/14)r + 1)
= 14[r^2 -(45/14)r + 45^2/28^2 -45^2/28^2 + 1]
= 14[(r - 45/28)^2 - 1241/784]
= 14[(r -45/28 - sqrt(1241)/28][(r -45/28 + sqrt(1241)/28]
= 14[r^2 -(45/14)r + 45^2/28^2 -45^2/28^2 + 1]
= 14[(r - 45/28)^2 - 1241/784]
= 14[(r -45/28 - sqrt(1241)/28][(r -45/28 + sqrt(1241)/28]
2008-07-04 6:21 pm
stop using the word "grouping" I thought "grouping" refer to expressions with 4 terms and you group together 2 at a time.
You can factor this if it is 14r^2 - 45r - 14
then it is (7r-2)(2r+7)
You can factor this if it is 14r^2 - 45r - 14
then it is (7r-2)(2r+7)
2008-07-04 6:05 pm
14r^2 - 45r + 14
= cannot be factored
= cannot be factored
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