Mathematics

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高等代数多项式带余除法练习

设$f(x)=x^3+(1+t)x^2+2x+2u,g(x)=x^3+tx+u$的最大公因式是一个二次多项式,求$t,u$的值.

 用辗转相除法,得

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

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

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

由题,$r_2(x)=0$,即

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

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

在实数域上解得$u=0,t=-4$.

在复数域上解得$u=0,t=-4$;$u=0,t=\dfrac{1}{2}(1+\sqrt{3}\,\mathrm{i})$;$u=0,t=\dfrac{1}{2}(1-\sqrt{3}\,\mathrm{i})$;$u=-7-\sqrt{11}\,\mathrm{i},t=\dfrac{1}{2}(-1+\sqrt{11}\,\mathrm{i})$;$u=-7+\sqrt{11}\,\mathrm{i},t=\dfrac{1}{2}(-1-\sqrt{11}\,\mathrm{i})$.

 

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