Division of Algebraic Expressions

Science Forums: Division of Algebraic Expressions - Science Forums

Welcome to ScienceForums.Net!

Welcome to ScienceForums.Net! We welcome science discussion at all levels — from beginners to researchers, covering topics from biology to computer science, and much more. Registration is fast and free, and allows you to post on the forums, so register now and join the discussions!

After you've registered, come in and introduce yourself, or visit the forum index. If you need any help registering, posting, or if you just have some questions about our site, please feel free to contact us at staff at scienceforums dot net.

Start new topics and reply to others
Subscribe to topics and forums to get automatic updates
Create a ScienceForums.Net Blog!

Page 1 of 1

You cannot start a new topic
You cannot reply to this topic

Division of Algebraic Expressions Algebraic division Rate Topic:

#1 19 August 2011 - 07:39 PM giorgio

Lepton

Hi

It is well knowm by the theorem of the reminder that:

$D(x)=d(x)q(x)+r(x)$

In this example:
.......................If $(2x^{119}+1)/(x^2-x+1)$ then figure out the reminder.

Therefore I can afirm that the reminder has the form of : $r(x)=ax+b$ ,
because $d(x)>r(x)$ and $d(x)=x^2-x+1$

And the original problem could be written as:
.................. $2x^{119}+1=(x^2-x+1)q(x)+ax+b$ .........(I)

To this point I need two solution of: $d(x)=0$
so I can find the values of a and b.

According to what I think:
...............If $(2x^{119}+1) = D(x)$ and $d(x)=(x^2-x+1)$
so in order to find the solution : $x^2-x+1=0$ ......(II)
............................. $(x+1)(x^2-x+1)=0$
.................................... $x^3+1=0$
.............................................. $x^3=-1$ ........(III)
.................................. replacing (III) on $D(x)$ :
.................................... $2(-1)x^2+1= -2x^2+1$

.................................but in (II) we'll have: $-x^2=-x+1$
......................................so $-2x^2+1=-2x+3$

........................................finally $-2x+3=r(x)$
........................................ $ax+b=-2x+3$

I just found the above anwer substituting values for another, but really would like to know
if I can find it in the form (I) below:
................................... $2x^{119}+1=(x^2-x+1)q(x)+ax+b.............(I)$
........................................and if I try to fatorise the way I did:
.......................................... $2x^{119}+1=(x^2-x+1)(x+1)(x+1)^{-1}q(x)+ax+b$
........................................... $2x^{119}+1=(x^3+1)(x+1)^{-1}q(x)+ax+b$
.................................................When I change x=-1
............................................... the expression: $(x+1)^{-1}$ has no meaning.

I will appretiate if anyone can remark me if I'm making or if I made a mistake in my logic(method)

Thanks in advance, and excuse me if my coding is wrond.

This post has been edited by giorgio: 19 August 2011 - 07:44 PM