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I have previously done division of polynomials in one variable, where I would do the following:

To compute $$\frac{x^3+4x^2+x-2}{x+1}$$

I would write $$\begin{align}x^3+4x^2+x-2&=(x+1)(ax^2+bx+c)\\ &= ax^3 + bx^2+cx+ax^2+bx+c\\ &= ax^3 + (a+b)x^2+(b+c)x+c \end{align}$$

Equating coefficients gives us the following systems: $$\begin{align}a&=1\\ a+b &= 4\\ b+c &= 1\\ c &= -2 \end{align}$$

From this we can see that $$\begin{align}a&=1\\ b &= 3\\ c &= -2 \end{align}$$

Therefore we conclude that $$\frac{x^3+4x^2+x-2}{x+1} = x^2+3x-2$$

How would I go about calculating something similar but in more than one variable, say $$\frac{x^3-2x^2-xy+2y}{x-2}$$

I want to be able to write this as $$x^3-2x^2-xy+2y=(x-2)A$$ where $A$ is some polynomial wih unknown coefficients

My question is, what form should this $A$ take?

[Note: This came about as part of this question]

Community
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lioness99a
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    Treat $y$ as just any other constant and run the algorithm as usual. $A$ will just be a polynomial in $x$ and $y$. – Josh Mar 01 '17 at 15:01

2 Answers2

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$A\,$ should take the form $\,x^2+bx+cy\,$ where $\,b,c\,$ could either be constants or involve $\,y$

But for this problem, you can just do it directly,

$$x^3-2x^2-xy+2y= x^2(x-2)-y(x-2)=(x-2)(x^2-y)$$

If $\,y\,$ is a perfect square, then

$$x^3-2x^2-xy+2y=(x-2)(x+\sqrt y)(x-\sqrt y)$$

Mengchun Zhang
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Note that division in the polynomial ring $R[x,y]$, is like to division in $(R[y])[x]$. So you may put $$A=x^2+ay^2+bxy. $$ Note that since the degree of $y$ in the polynomial $x^3-2x^2-xy+2y$ is $1$, so immediately $a=0$.