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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..

Dont compare your life to others. You have no idea what their journey is all about. ^_^

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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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