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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.

Mathejunior
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    Polynomials aren't "positive" or "negative". In this context, $-x^3-x+1$ is a perfectly acceptable remainder. – Angina Seng Jun 17 '18 at 04:38
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    if we make the quotient x^2−x+1 ...what is the degree of the "remainder" you get? – dxiv Jun 17 '18 at 04:39
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    Why shouldn't remainders be negative? And what does "negative" mean in terms of polynomials? You really shouldn't like the particulars of integers influence your thinking when they don't apply. – fleablood Jun 17 '18 at 04:40
  • @above(all), I get it, thanks! – Mathejunior Jun 17 '18 at 04:56
  • @Mathejunior Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here https://meta.stackexchange.com/questions/5234/how-does-accepting-an-answer-work – user Aug 04 '18 at 22:11
  • @gimusi Accepted, thank you! – Mathejunior Aug 12 '18 at 16:24

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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

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

user
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