求下列各组复多项式的最高公因式:
(1)$x^3+(2+i)x^2+(3+2i)x+6$,$x^5+ix^4+10x^3+28x+21i$.
解:
\begin{align*}
x^5+ix^4+10x^3+28x+21i&=x^2(x^3+(2+i)x^2+(3+2i)x+6)+(-2x^4+(7-2i)x^3-6x^2+28x+21i)\\
-2x^4+(7-2i)x^3-6x^2+28x+21i&=-2x(x^3+(2+i)x^2+(3+2i)x+6)+(11x^3+4ix^2+40x+21i)\\
11x^3+4ix^2+40x+21i&=11(x^3+(2+i)x^2+(3+2i)x+6)+((-22-7i)x^2+(7-22i)x+21i-66)\\
\end{align*}
而
\begin{align*}
x^3+(2+i)x^2+(3+2i)x+6=\frac{-1}{22+7i}x((-22-7i)x^2+(7-22i)x+21i-66)+(2x^2+(3+2i+\frac{21i-66}{22+7i})x+6)
\end{align*}
下面看$(-22-7i)x^2+(7-22i)x+21i-66$与$2x^2+(3+2i+\frac{21i-66}{22+7i})x+6$.
\begin{align*}
(-22-7i)x^2+(7-22i)x+21i-66=\frac{-22-7i}{2}(2x^2+(3+2i-\frac{21i-66}{-22-7i})x+6)+42i
\end{align*}
综上，最高公因式竟然是1.
(2)$x^6+2x^4-4x^3-3x^2+8x-5$和$x^5+x^2-x+1$.
解:
\begin{align*}
x^6+2x^4-4x^3-3x^2+8x-5&=x(x^5+x^2-x+1)+(2x^4-5x^3-2x^2+7x-5)\\
x^5+x^2-x+1&=\frac{1}{2}x(2x^4-5x^3-2x^2+7x-5)+(\frac{5}{2}x^4+x^3-\frac{5}{2}x^2+\frac{3}{2}x+1)\\
2x^4-5x^3-2x^2+7x-5&=\frac{4}{5}(\frac{5}{2}x^4+x^3-\frac{5}{2}x^2+\frac{3}{2}x+1)-\frac{29}{5}(x^3-x+1)\\
2x^4-5x^3-2x^2+7x-5&=(2x-5)(x^3-x+1)
\end{align*}
因此两者的最大公因式为$x^3-x+1$.