## Mathematics

Practical wisdom is only to be learned in the school of experience

# 高等代数多项式带余除法练习

用辗转相除法，得

$f(x)=q_1(x)g(x)+r_1(x)=1\cdot g(x)+[(1+t)x^2+(2-t)x+u]$

$g(x)=q_2(x)r_1(x)+r_2(x)=\bigg[\dfrac{1}{1+t}x+\dfrac{t-2}{(1+t)^2}\bigg]r_1(x)$

$+\bigg[\dfrac{(t^2+t-u)(1+t)+(t-2)^2}{(1+t)^2}x+\dfrac{u[(1+t)^2-(t-2)]}{(1+t)^2}\bigg]$

$\dfrac{(t^2+t-u)(1+t)+(t-2)^2}{(1+t)^2}=0$

$\dfrac{u[(1+t)^2-(t-2)]}{(1+t)^2}=0$