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Determine the quotient and the remainder of the division:

($1$).of $f\in \mathbb K[x]$ by $x^2-a$ in $\mathbb K[x],$Where $\mathbb K$ is a field.

($2$).of $x^m-1$ by $x^n-1$ in $\mathbb Z[x],$for $m,n\in\mathbb N^*$ .

Results used

Division Algorithm for $\mathbb F[x]$:Let $\mathbb F$ be a field and let $f(x)$ and $g(x)\mathbb F[x]$ with $g (x)\neq 0.$Then thete exists unique polynomials $q(x),r(x) \in \mathbb F[x]$ such that $f(x)=g(x)q(x)+r(x)$ and either $r(x)=0$ or $deg(r(x))<deg(g(x)).$

Remainder theorem:Let $\mathbb F$ be a field,$a\in \mathbb F,$ and $f(x)\in \mathbb F[x].$Then $f(a)$ is the remainder in the division of $f(x)$ by $x-a.$

Solution of $(1)$

By Remainder theorem,remainder,$r(x)=f(\pm\sqrt a)$.

Let $q(x)$ be the quotient,then by division algorithm,$f(x)=(x^2-a)q(x)+r(x)\implies f(x)=(x^2-a)q(x)+f(\pm\sqrt a)\implies q(x)=\frac{f(x)-f(\pm\sqrt a)}{x^2-a}.$

Hence quotient and remainder are $r(x)=f(\pm\sqrt a)$,$q(x)=\frac{f(x)-f(\pm\sqrt a)}{x^2-a}$

Is it correct?

Solution of $(2)$

On comparing with the division algorithm, we have

$f(x)=x^m-1,g(x)=x^n-1$ Now by remainder theorem ,we have $r(x)=f(1^{1/n})=f(1)=0$.

Let $q(x)$ be the quotient,then by division algorithm,$f(x)=(x^2-a)q(x)+r(x)\implies f(x)=(x^n-1)q(x)+f(1^{1/n})\implies q(x)=\frac{x^m-1-0}{x^n-1}=\frac{x^m-1}{x^n-1}.$

Hence quotient and remainder are $r(x)=f(1^{1/n})=f(1)=0$,$q(x)=\frac{x^m-1-0}{x^n-1}=\frac{x^m-1}{x^n-1}$

How do we guarantee that $r(x),q(x)\in \mathbb Z[x]?$

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  • $\begingroup$ For (1), it seems to me that the remainder is not what you say: for instance when $f(x)=x^2+x-a$, obviously we have $r(x)=x$. $\endgroup$ – René Gy Jan 27 at 12:15
  • $\begingroup$ Let $f(x)=q(x)(x^2-a)+r(x)$ by division algorithm. Since $\text{Deg}(x^2-a)=2$, we must have $\text{Deg}(r(x))$=1,i.e.,$r(x)=cx+d$. Let $x=0,1,-1,\sqrt{a},-\sqrt{a}$, then you will get many equations. It is not hard to find out $c$ and $d$. For (2), $\mathbb{Z}$ is not a field, so you can not use remainder theorem. $\endgroup$ – zongxiang yi Mar 14 at 9:57

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