考研高等代数真题分类汇编04

2024-04-09 19:20:31 阅读数 623

在实数域上将多项式 ${f(x)=x^{5}+32}$ 分解为不可约多项式的乘积.

解答:为了方便, 记 ${x=2 t}$ , 则
${ f(x)=x^{5}+32=32\left(t^{5}+1\right) . }$
若 ${t=r(\cos \theta+\mathrm{i} \sin \theta)}$ 满足 ${t^{5}=1}$ , 则有
${ r^{5}(\cos 5 \theta+\mathrm{i} \sin 5 \theta)=1=\cos \pi+\mathrm{i} \sin \pi . }$
由此可知 ${r^{5}=1}$ , 且 ${5 \theta=\pi+2 k \pi}$ , 其中 ${k}$ 为整数, 即有 ${r=1, \theta=\dfrac{(2 k+1) \pi}{5}}$ , 现在记
${ \omega_{k}=\cos \frac{(2 k+1) \pi}{5}+\mathrm{i} \sin \frac{(2 k+1) \pi}{5}, k=0,1,2,3,4 . }$
容易发现 ${\omega_{0}, \omega_{1}, \omega_{2}, \omega_{3}, \omega_{4}}$ 两两不等, 从而它们是 ${t^{5}+1}$ 的全部复数根, 即有
${ t^{5}+1=\left(t-\omega_{0}\right)\left(t-\omega_{1}\right)\left(t-\omega_{2}\right)\left(t-\omega_{3}\right)\left(t-\omega_{4}\right) }$ 另外, 还容易发现
${ \left\{\begin{array}{l} \omega_{4}=\cos \frac{9 \pi}{5}+\mathrm{i} \sin \frac{9 \pi}{5}=\cos \frac{\pi}{5}-\mathrm{i} \sin \frac{\pi}{5}=\bar{\omega}_{0} \\ \omega_{3}=\cos \frac{7 \pi}{5}+\mathrm{i} \sin \frac{7 \pi}{5}=\cos \frac{3 \pi}{5}-\mathrm{i} \sin \frac{3 \pi}{5}=\bar{\omega}_{1} \\ \omega_{2}=\cos \frac{5 \pi}{5}+\mathrm{i} \sin \frac{5 \pi}{5}=-1 \end{array}\right. }$
于是
${\begin{aligned} t^{5}+1 & =\left(t-\omega_{0}\right)\left(t-\omega_{1}\right)\left(t-\omega_{2}\right)\left(t-\omega_{3}\right)\left(t-\omega_{4}\right) \\ & =(t+1)\left[\left(t-\omega_{0}\right)\left(t-\bar{\omega}_{0}\right)\right]\left[\left(t-\omega_{1}\right)\left(t-\bar{\omega}_{1}\right)\right] \\ & =(t+1)\left(t^{2}-2 t \cos \frac{\pi}{5}+1\right)\left(t^{2}-2 t \sin \frac{3 \pi}{5}+1\right) \end{aligned} }$
从而结合 ${t=\frac{x}{2}}$ 便有
${\begin{aligned} f(x) & =32\left(t^{5}+1\right)=32(t+1)\left(t^{2}-2 t \cos \frac{\pi}{5}+1\right)\left(t^{2}-2 t \cos \frac{3 \pi}{5}+1\right) \\ & =(x+2)\left(x^{2}-4 x \cos \frac{\pi}{5}+4\right)\left(x^{2}-4 x \cos \frac{3 \pi}{5}+4\right) \end{aligned} }$

将 ${f(x)=x^{5}+x^{4}+1}$ 分解为有理数域上不可约多项式的乘积.

解答:首先由于 ${f(\pm 1) \neq 0}$ , 所以 ${f(x)}$ 在有理数域上不存在一次因式, 进而 ${f(x)}$ 只可能分解为二次与三次整系数多项式的乘积, 再结合 ${f(x)}$ 首一可设
${ f(x)=x^{5}+x^{4}+1=\left(x^{2}+a x+b\right)\left(x^{3}+c x^{2}+d x+e\right) . }$
其中 ${a, b, c, d, e}$ 均为整数. 由对应系数相等可知
$a+c=1$
$d+a c+b=0$
$e+a d+b c=0$
$a e+b d=0$
$b e=1$
由\ref{eq1.6}可知 ${b=e=\pm 1}$ , 结合\ref{eq1.5}可知 ${a+d=0}$ , 即 ${d=-a}$ , 而由\ref{eq1.2}可知 ${c=1-a}$ . 下面分情况讨论:
当 ${b=e=-1}$ 时, 由\ref{eq1.4}可知 ${-1-a^{2}-(1-a)=0}$ , 即 ${a^{2}-a+2=0}$ , 显然无解.

当 ${b=e=1}$ 时, 由\ref{eq1.3},\ref{eq1.4}可知
${ \begin{array}{c} -a+a(1-a)+1=0 \\ 1-a^{2}+(1-a)=0 . \end{array} }$
解得 ${a=1}$ , 进而 ${b=1, c=0, d=-1, e=1}$ , 即有
${ f(x)=\left(x^{2}+x+1\right)\left(x^{3}-x+1\right) }$
而根据 ${f(x)}$ 无有理根可知 ${x^{2}+x+1}$ 与 ${x^{3}-x+1}$ 均无有理根, 从而它们在有理数域上不可约.

求多项式 ${f(x)=\dfrac{2}{3} x^{5}-x^{4}+2 x^{3}-\frac{8}{3} x^{2}+1}$ 在复数域上的标准分解式.

解答:为了方便, 记
${ g(x)=3 f(x)=2 x^{5}-3 x^{4}+6 x^{3}-8 x^{2}+3 . }$
容易 ${g(x)}$ 存在有理根 1 , 由此可知
${\begin{aligned} g(x) & =2 x^{4}(x-1)-x^{3}(x-1)+5 x^{2}(x-1)-3 x(x-1)-3(x-1) \\ & =(x-1)\left(2 x^{4}-x^{3}+5 x^{2}-3 x-3\right) \end{aligned} }$
而明显 ${2 x^{4}-x^{3}+5 x^{2}-3 x-3}$ 依旧以 1 为根, 进而
${\begin{aligned} 2 x^{4}-x^{3}+5 x^{2}-3 x-3 & =2 x^{3}(x-1)+x^{2}(x-1)+6 x(x-1)+3(x-1) \\ & =(x-1)\left(2 x^{3}+x^{2}+6 x+3\right) . \end{aligned} }$
而此时容易发现 ${2 x^{3}+x^{2}+6 x+3}$ 以 ${-\dfrac{1}{2}}$ 为根, 于是
${\begin{aligned} 2 x^{3}+x^{2}+6 x+3 & =x^{2}(2 x+1)+3(2 x+1)=(2 x+1)\left(x^{2}+3\right) \\ & =2\left(x+\frac{1}{2}\right)(x-\sqrt{3} \mathrm{i})(x+\sqrt{3} \mathrm{i}) . \end{aligned} }$
综上可知
${ f(x)=\frac{1}{3} g(x)=\frac{2}{3}(x-1)^{2}\left(x+\frac{1}{2}\right)(x-\sqrt{3} \mathrm{i})(x+\sqrt{3} \mathrm{i}) . }$