x^3+3x^2+3x+1 = x^2+2x
--------------------
x+1
10x^2 + 23x - 18 = 5x -2
----------------------
2x+3
2006-11-29 00:04:52 · 11 answers · asked by 梅子怪李 2 in Science & Mathematics ➔ Mathematics
Aaaarrghh!!! then what is?! (thank you ^^)
2006-11-29 17:32:07 · update #1
I'm dividing not adding -.-
2006-11-29 17:49:16 · update #2
No. Try again...
2006-11-29 00:39:15 · answer #1 · answered by Askhole Ninja 3 · 0⤊ 0⤋
x^3+3x^2+3x+1 = x^2+2x
x^3+3x^2-x^2+3x-2x+1 = 0
x^3+2x^2+x+1=0
x^2(x+2)+(x+1)=0
hence your answer is wrong if you solve this you are likely to get an imaginary number for x
10x^2 + 23x - 18 = 5x -2
10x^2 + 23x - 18 - 5x +2=0
10x^2+18x-16=0
5 x^2+9x-8=0
use x=(b^2-4ac/2a)^1/2
to get x
x=(24.1)^1/2=4.90
2006-11-29 00:22:40 · answer #2 · answered by riya s 2 · 0⤊ 0⤋
(x^3+3x^2+3x+1) / (x+1) =
(x+1)^3 / (x+1) =
(x+1)^2, which does not equal (x^2 + 2x)
2006-11-29 00:08:49 · answer #3 · answered by fcas80 7 · 0⤊ 0⤋
no
when you add powers, they must be the same, unlike multiplying and dividing. for your question
x^3+3x^2+3x+1 = x^2+2x
move all to one side: x^3+3x^2+3x+1-x^2-2x
=x^3+2x^2+x+1
=(x+1)^3
2006-11-29 02:25:46 · answer #4 · answered by arumisan 2 · 0⤊ 0⤋
x^3+3x^2+3x+1 = (x+1)^3
2006-11-29 00:21:23 · answer #5 · answered by James Chan 4 · 0⤊ 0⤋
You are dividing by (x+1). If x equals -1, you would be dividing by zero, which you cannot do. So, any answer would at least need a "for x not equal to minus one".
Similar for the second question.
The answer to your question: no, it is not correct.
2006-11-29 01:44:49 · answer #6 · answered by Anonymous · 0⤊ 0⤋
No. You add the 16 to the other side of the equation to get: 11t = 36+16 11t = 52 t = 52/11 t = 4.72 (repeating)
2016-05-23 01:34:25 · answer #7 · answered by ? 4 · 0⤊ 0⤋
My brain hurts trying to explain the answer since it has been so long since I did this, but I did not see how that would be the correct answer to the questions.
2006-11-29 00:16:56 · answer #8 · answered by Chris 2 · 0⤊ 0⤋
Sorry, me is only in form 2. I am just learning algebra. Sorry I cant help u.
2006-11-29 01:11:24 · answer #9 · answered by ErC 4 · 0⤊ 0⤋
no
2006-11-29 00:13:39 · answer #10 · answered by raju p 1 · 0⤊ 1⤋