Division of $f(x):= x^6+x^4+x^3-x^2-1$ by $g(x):=x^4+x^3-1$ seems not so correct

Question

Let's say $f(x):= x^6+x^4+x^3-x^2-1$ and $g(x):=x^4+x^3-1$. Given that degree $(f(x))$ $>$ degree $(g(x))$, we can divide $f$ by $g$. The issue here is, continuous division of polynomial (starting from the higher powers) we have learnt in junior classes, at the end of the day, we get the remainder as $(-x^3-x+1)$ (quotient being $x^2-x+2$) but remainders can't be negative right? So, if we make the quotient $x^2-x+1$, I think the remainder obtained will be more suitable.

Answer 1

The division is correct indeed we have that

$$(x^4+x^3-1)(x^2-x+2)=x^6-x^5+2x^4+x^5-x^4+2x^3-x^2+x-2=$$ $$=x^6+x^4+2x^3-x^2+x-2=(x^6+x^4+x^3-x^2-1)+(x^3+x-1)$$

and therefore

$$x^6+x^4+x^3-x^2-1=(x^4+x^3-1)(x^2-x+2)+(-x^3-x+1)$$