What is the waiting-to-be-factored form of 4n^2+30n+30?
回答 (5)
2008-11-29 4:04 pm
✔ 最佳答案
2(2n^2 + 15n + 15)
2008-11-30 2:14 am
4n^2 + 30n + 30
= 2(2n^2 + 15n + 15n)
= 2(2n^2 + 15n + 15n)
2008-11-30 12:23 am
4(n+1.188262309)(n+6.311737692)=
=(4n+7.5)(n+6.311737692)
=(4n+7.5)(n+6.311737692)
2008-11-30 12:23 am
4n^2+30n+30
= 2 (2n^2 + 15n + 15)
= 4 [n+ ((15-â105)/4)][n + ((15 +â105)/4 ]
where (- 15 ± â105)/4 are roots of the equation 2n^2 + 15n + 15 = 0
= 2 (2n^2 + 15n + 15)
= 4 [n+ ((15-â105)/4)][n + ((15 +â105)/4 ]
where (- 15 ± â105)/4 are roots of the equation 2n^2 + 15n + 15 = 0
2008-11-30 12:08 am
4n^2+30n+30
first factor out a 2
2 (2n^2 + 15n + 15)
That's as far as you can take it, the Trinomial part cannot be factored so you need to use the quadratic formula to find the factors
first factor out a 2
2 (2n^2 + 15n + 15)
That's as far as you can take it, the Trinomial part cannot be factored so you need to use the quadratic formula to find the factors
收錄日期: 2021-05-01 11:33:14
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20081129080155AA2jQsX