Division of Algebraic Expressions

Hi

 

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

 

[math]D(x)=d(x)q(x)+r(x)[/math]

 

 

In this example:

.......................If [math](2x^{119}+1)/(x^2-x+1)[/math] then figure out the reminder.

 

Therefore I can afirm that the reminder has the form of :[math]r(x)=ax+b[/math],

because [math]d(x)>r(x)[/math] and [math]d(x)=x^2-x+1[/math]

 

And the original problem could be written as:

..................[math]2x^{119}+1=(x^2-x+1)q(x)+ax+b[/math].........(I)

 

To this point I need two solution of:[math]d(x)=0[/math]

so I can find the values of a and b.

 

According to what I think:

...............If [math](2x^{119}+1) = D(x)[/math] and [math]d(x)=(x^2-x+1)[/math]

so in order to find the solution :[math] x^2-x+1=0[/math]......(II)

.............................[math](x+1)(x^2-x+1)=0[/math]

....................................[math]x^3+1=0 [/math]

..............................................[math]x^3=-1 [/math]........(III)

.................................. replacing (III) on [math]D(x)[/math]:

....................................[math]2(-1)x^2+1= -2x^2+1 [/math]

 

.................................but in (II) we'll have: [math] -x^2=-x+1[/math]

......................................so [math]-2x^2+1=-2x+3[/math]

 

........................................finally [math]-2x+3=r(x) [/math]

........................................[math]ax+b=-2x+3[/math]

 

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:

...................................[math]2x^{119}+1=(x^2-x+1)q(x)+ax+b.............(I)[/math]

........................................and if I try to fatorise the way I did:

..........................................[math]2x^{119}+1=(x^2-x+1)(x+1)(x+1)^{-1}q(x)+ax+b[/math]

...........................................[math]2x^{119}+1=(x^3+1)(x+1)^{-1}q(x)+ax+b[/math]

.................................................When I change x=-1

............................................... the expression: [math](x+1)^{-1}[/math] 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.

Edited August 19, 201114 yr by giorgio