# Determine the quotient and the remainder of the division:

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))

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 $$r(x),q(x)\in \mathbb Z[x]?$$

• 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 $r(x)=x$. – René Gy Jan 27 at 12:15
• 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. – zongxiang yi Mar 14 at 9:57