📝 题目
例 20 设 $P\left( x\right) = {x}^{6} - 2{x}^{2} - x + 3$ .
(1)分别把 $P\left( x\right)$ 表示成(x - 1)幂与 $\left( {x + 1}\right)$ 幂的多项式；
(2)求证: $P\left( x\right) = 0$ 在 $\left| x\right| \geq 1$ 上无实根；
(3) 求证: $P\left( x\right) = 0$ 无实根.
💡 答案与解析
解(1)对 $P\left( x\right)$ 求各阶导数得
$$ {P}^{\prime }\left( x\right) = 6{x}^{5} - {4x} - 1,\;{P}^{\prime \prime }\left( x\right) = {30}{x}^{4} - 4, $$
$$ {P}^{\prime \prime \prime }\left( x\right) = {120}{x}^{3},\;{P}^{\left( 4\right) }\left( x\right) = {360}{x}^{2}, $$
$$ {P}^{\left( 5\right) }\left( x\right) = {720x},\;{P}^{\left( 6\right) }\left( x\right) = {720}. $$
由此推出
$$ P\left( x\right) = \mathop{\sum }\limits_{{k = 0}}^{6}\frac{{P}^{\left( k\right) }\left( 1\right) }{k!}{\left( x - 1\right) }^{k} $$
$$ = 1 + \left( {x - 1}\right) + {13}{\left( x - 1\right) }^{2} + {20}{\left( x - 1\right) }^{3} $$
$$ + {15}{\left( x - 1\right) }^{4} + 6{\left( x - 1\right) }^{5} + {\left( x - 1\right) }^{6}\text{ ; } \tag{6.22} $$
$$ P\left( x\right) = \mathop{\sum }\limits_{{k = 0}}^{6}\frac{{P}^{\left( k\right) }\left( {-1}\right) }{k!}{\left( x + 1\right) }^{k} $$
$$ = 3 - 3\left( {x + 1}\right) + {13}{\left( x + 1\right) }^{2} - {20}{\left( x + 1\right) }^{3} $$
$$ + {15}{\left( x + 1\right) }^{4} - 6{\left( x + 1\right) }^{5} + {\left( x + 1\right) }^{6}. \tag{6.23} $$
(2)利用第 (1) 小题的结果, 有
$\left. \begin{array}{l} \left( {6.22}\right) 式 \Rightarrow P\left( x\right) = \mathop{\sum }\limits_{{k = 0}}^{6}\frac{{P}^{\left( k\right) }\left( 1\right) }{k!}{\left( x - 1\right) }^{k} > 0\left( {\forall x \geq 1}\right) \\ \left( {6.23}\right) 式 \Rightarrow P\left( x\right) = \mathop{\sum }\limits_{{k = 0}}^{6}\frac{{P}^{\left( k\right) }\left( {-1}\right) }{k!}{\left( x + 1\right) }^{k} > 0\left( {\forall x \leq - 1}\right) \end{array}\right\}$
$$ \Rightarrow P\left( x\right) > 0\;\left( {\left| x\right| \geq 1}\right) , $$
所以在 $\left| x\right| \geq 1$ 上 $P\left( x\right) = 0$ 无实根.
(3)利用第 (2) 小题结果,只需再证 $P\left( x\right)$ 在(-1,1)上无实根. 将 $P\left( x\right) = 0$ 改写成
$$ {x}^{6} = 2{x}^{2} + x - 3 \tag{6.24} $$
而 $2{x}^{2} + x - 3 = \left( {x - 1}\right) \left( {{2x} + 3}\right)$ ,其图形如图 2.14 所示. 在(-1,1) 上,(6.24)式左边 $\geq 0$ ,右边 $< 0$ ,所以方程 (6.24) 在(-1,1)上无实根,即 $P\left( x\right) = 0$ 在(-1,1)上无实根.
\begin{center} \includegraphics[max width=0.2\textwidth]{images/021.jpg} \end{center} \hspace*{3em}
图 2.14