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I'm trying to compute $$\frac{x^3-2x^2-xyz+2xy+yz}{x-2}$$

I have written this in the form $$\begin{align}x^3-2x^2-xyz+2xy+yz&=(x-2)(x^2+bx+cy+dz)\\ x^3-2x^2-(yz+2y)x+yz&= x^3 + (b-2)x^2 + (cy+dz-2b)x+(-2cy-2dz) \end{align}$$

This leaves me with the following equations $$\begin{align}b-2&=-2\\ cy+dz-2b&=yz+2y \\ -2cy-2dz &= yz \end{align}$$

From this we can see that $b=0$

Therefore we are left with $$\begin{align}cy+dz&=yz+2y \\ cy +dz&=-\frac{yz}{2} \end{align}$$

I'm now stuck, how can I find values for $c$ and $d$ from this? If I have hit a dead-end, feel free to suggest a different method for solving this equation

lioness99a
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1 Answers1

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Let $F(x)={x}^{3}-xyz-2\,{x}^{2}+2\,xy+yz$. Then by Bézout's theorem we have $F(x)=(x-2) Q(x)+F(2).$ Now $F(2)=4y-yz$.

$$ F(x)-F(2)={x}^{3}-xyz-2\,{x}^{2}+2\,xy+2\,yz-4\,y=\left( x-2 \right) \left( {x}^{2}-yz+2\,y \right). $$ Thus $$ \frac{{x}^{3}-xyz-2\,{x}^{2}+2\,xy+yz}{x-2}={x}^{2}-yz+2\,y+\frac{4y-yz}{x-2} $$

Leox
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