考研高等代数真题分类汇编04

感冒的梵高 • 2023-06-09 16:54 • 杂文

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
${ f(x)=/frac{1}{3} g(x)=/frac{2}{3}(x-1)^{2}/left(x+/frac{1}{2}/right)(x-/sqrt{3} /mathrm{i})(x+/sqrt{3} /mathrm{i}) . }$