# Division of Algebraic Expressions

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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.

Edited by giorgio

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