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Find polynomials $q(x)$ and $r(x)$ such that $f(x)=g(x)q(x)+r(x)$ where $r(x)=0$ or $\deg r(x)<\deg g(x)$ provided that $f(x)=2x^4+x^2-x+1$ and $g(x)=3x^2+2$ in $\Bbb Z[x]$.

The problem I'm having with this is that I can't think of any possible function in $\Bbb Z[x]$ that satisfies either of the conditions that the remainder $r(x)$ is $0$ or $\mbox{deg}(r(x))<\mbox{deg}(g(x))$.

Kamil Jarosz
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Jon Snow
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  • Yes, but that is not possible in $\Bbb Z[x]$ – Jon Snow Jul 16 '15 at 01:48
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    Do long division on f and g, that is divide f by g until you have a remainder, the remainder is either 0 or a linear factor ax+b which is order 1, one less than g. As a start, subtract 2/3 * x**2 of g from f. – user247608 Jul 16 '15 at 01:49
  • Does the statement also require that $q(x)$ and $r(x)$ be in $\Bbb Z[x]$? – coldnumber Jul 16 '15 at 01:53
  • @cockneywanker, If you subtract $\frac23x^2$ of $g$ from $f$, 1) That'll be part of your remainder, and 2) you'll have a $\frac23x^2$ lying around, which is not in $\Bbb Z[x]$ – Jon Snow Jul 16 '15 at 01:54
  • @coldnumber, yes, that's part of the division algorithm – Jon Snow Jul 16 '15 at 01:55

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The problem is $\mathbb{Z}[x]$ is not a PID, so it's not an Euclidean domain and you can't use the division algorithm in general.

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