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#1 2006-11-09 02:37:30
- unique
- Member
- Registered: 2006-10-04
- Posts: 419
divide
divide
x^3 + 2x - 1 by x + 3
so
x^3 + 2x - 1
---------------
x + 3
x^3 + 3x - 2
the way i did it ..it seems to me that i did it wrong
can you correct me on this?
Desi
Raat Key Rani !
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#2 2006-11-09 17:51:46
- haruka-san
- Member
- Registered: 2006-11-08
- Posts: 205
Re: divide
x^3 is it read as x raised to 3??
you are dividing polynomials.. right?
i think that is not the right way to divide it..
oh.. i have to get my pencil..
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#3 2006-11-09 19:17:23
- Devantè
- Real Member
- Registered: 2006-07-14
- Posts: 6,400
Re: divide
Yes, x^3 = x cubed. 
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#4 2006-11-10 03:25:33
- unique
- Member
- Registered: 2006-10-04
- Posts: 419
Re: divide
ok
x+3 / x^3 + 2x - 1
x - 1
-------------------
x + 3 | x^3 + 2x - 1
x^3 + 3x
----------------
5x - 1
-1x - 3
----------------
+1x + 3
--------------
6x + (-2)
now what do i do? and please explain it well to me how to do it if i did this wrong!
Desi
Raat Key Rani !
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#5 2006-11-10 06:51:28
- mathsyperson
- Moderator
- Registered: 2005-06-22
- Posts: 4,900
Re: divide
You're doing it as if the dividend has an x² in it, when it actually has an x³.
x² -3x +11 r -34.
-------------------------
x+3 | x³ +2x -1
x³ +3x²
--------------------
-3x² +2x -1
-3x² -9x
--------------------
11x -1
11x +33
----------
-34
So I get that (x³ +2x -1)/(x+3) = x² -3x +11, with a remainder of -34.
A quick check is to put x=-3 into the dividend and see if we get the remainder.
(-3)³ +2(-3) -1 = -27-6-1 = -34, so the answer is probably right.
Why did the vector cross the road?
It wanted to be normal.
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