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There are few things which I can't quite grasp while trying to find Taylor polynomial of composite functions and their products. This will be lenghty for such an easy topic so i apologize in advance. I will use this problem as an example:

Find Taylor polynomial centered at $0$ of the fourth degree of the function:$$f(x) = frac1+x+x^21-x+x^2$$

I work with remainder in Peano form, defined to be $omega(x)= fracR_n(x-a)^n$ and $R_n = f(x)-P_n(x)$, where $P_n$ is Taylor polynomial of n-th degree and $a$ is where the polynomial is centered at. By Taylor theorem we know for $P_n$ this holds:
$lim_xto aomega(x) = 0$ and any given polynomial with this property is Taylor polynomial.

So since they're basically two multiplied functions I tried to make use of the Taylor expansion for $(1+x)^alpha$ like this:$$(1+x+x^2)(1+(-x+x^2))^-1=(1+x+x^2)bigg(sum_k=0^nalphachoosek(-x+x^2)^k+omega(-x+x^2)(-x+x^2)^nbigg)$$
Where in $omega(-x+x^2)$ the argument of $omega(x)$ tends to zero as x tends to zero, so by limit of composite function: $lim_xto0omega(-x+x^2)=0$. The remainder is multiplyied by $(-x+x^2)^n$. (Sorry for that notation. They use it at our uni and I'm not sure whether it's standard or not)

What degree should I plug into the sum? The problem asks for polynomial of degree $4$ but it's multiplied by polynomial of second degree. My initial thought was that I can expand the sum to second degree and I will get polynomial of 4th degree. It obviously didn't work. Can someone explain why I can't do this? The only explanation I came up with was that the remainder won't satisfy this limit:
$$lim_xtoinftyfracR_4x^4$$ so by Taylor theorem I can't tell if it's Taylor polynomial or not.

How does this multiplication of remainder change it? Is it okay to multiply it like this? From the expansion above:$$omega(-x+x^2)(-x+x^2)^n+omega(-x+x^2)(-x^2+x^3)^n+omega(-x+x^2)(-x^3+x^4)^n$$
Does this whole term need to tend to zero when divided by $x^4$ by Taylor theorem? That would explain why I need to plug at least $n=4$, for $n<4$ the $omega(-x+x^2)(-x+x^2)^n$ would be $frac00$.

To get the Taylor expansion around $x=0$, you should expand the terms of the form $(-x+x^2)^k$ as a sum of powers of $x$, adding equal powers up to order $4$. There are easier ways to get the Taylor expansion.

1. 
   Long division. You divide $1+x+x^2$ by $1-x+x^2$ as polynomials, but in increasing orders of powers. The first term in the quotient will be $1$. Then compute
   $$
   1+x+x^2-1times(1-x+x^2)=2,x.
   $$
   Dividing $2,x$ by $1-x+x^2$, we find that the next term in the expansion will be $2,x$. Compute now
   $$
   2,x-(2,x)times(1+x+x^2)=-2,x^2-2,x^3.
   $$
   Keep on going until you get to $x^4$.


2. 
   Undetermined coefficients. Let
   $$
   frac1+x+x^21-x+x^2=a_0+a_1,x+a_2,x^2+a_3,x^3+a_4,x^4+dots
   $$
   Then
   $$
   1+x+x^2=(1-x+x^2)times(a_0+a_1,x+a_2,x^2+a_3,x^3+a_4,x^4+dots)tag*
   $$
   Multiply the right hand side keeping only powers $le4$. I wild it up to $x^2$:
   $$
   a_0+(a_1-a_0)x+(a_2-a_1+a_0)x^2+dots
   $$
   Identifying coefficients on both sides of (*) we get the equations
   beginalign
   1&=a_0\
   1&=a_1-a_0\
   1&=a_2-a_1+a_0\
   0&=dots
   endalign
   This allows to computebthe values of the coefficients $a_k$ recursively.