剩余因子定理的介绍应用据/strong>:据/p>
考虑一般性的据span class="katex">
N据/span>TH.据/span>-degree多项式:据/p>
F据/span>(据/span>X据/span>)据/span>=据/span>一种据/span>N据/span>X据/span>N据/span>+据/span>一种据/span>N据/span>-据/span>1据/span>X据/span>N据/span>-据/span>1据/span>+据/span>一种据/span>N据/span>-据/span>2据/span>X据/span>N据/span>-据/span>2据/span>+据/span>⋯据/span>+据/span>一种据/span>1据/span>X据/span>+据/span>一种据/span>0.据/span>。据/span>
为了确定所有的据span class="katex">
(据/span>N据/span>+据/span>1据/span>)据/span>系数,据span class="katex">
一种据/span>0.据/span>那据/span>一种据/span>1据/span>那据/span>一种据/span>2据/span>那据/span>......据/span>那据/span>一种据/span>N据/span>,我们至少需要知道据span class="katex">
(据/span>N据/span>+据/span>1据/span>)据/span>点据span class="katex">
(据/span>X据/span>那据/span>F据/span>(据/span>X据/span>)据/span>)据/span>。它是最常见的,我们通过解决这个系数来确定这些系数据一种Href="//www.parkandroid.com/wiki/system-of-linear-equations/" class="wiki_link" title="线性方程系统GyD.F4y2Ba" target="_blank">线性方程系统据/a>同时。但是,如图所以据span class="katex">
N据/span>变大,手工解决它们变得越来越不切实际。那么有更好的选择吗?是的,有,它涉及直接应用据一种Href="//www.parkandroid.com/wiki/remainder-factor-theorem/" class="wiki_link" title="剩余因子定理GyD.F4y2Ba" target="_blank">剩余因子定理据/a>。为了说明这一点,让我们考虑以下多项式:据/p>
考虑二次多项式据span class="katex">
F据/span>(据/span>X据/span>)据/span>满足据/p>
F据/span>(据/span>1据/span>)据/span>=据/span>1据/span>3.据/span>那据/span>F据/span>(据/span>2据/span>)据/span>=据/span>2据/span>3.据/span>那据/span>F据/span>(据/span>3.据/span>)据/span>=据/span>3.据/span>3.据/span>那据/span>
我们有兴趣在二次多项式中找到未知数,据span class="katex">
F据/span>(据/span>X据/span>)据/span>=据/span>一种据/span>X据/span>2据/span>+据/span>B.据/span>X据/span>+据/span>C据/span>。回想一下,通过剩下的定理,我们有剩余的据span class="katex">
一种据/span>X据/span>2据/span>+据/span>B.据/span>X据/span>+据/span>C据/span>在划分时据span class="katex">
X据/span>-据/span>1据/span>那据/span>X据/span>-据/span>2据/span>那据/span>X据/span>-据/span>3.据/span>成为据span class="katex">
1据/span>3.据/span>那据/span>2据/span>3.据/span>那据/span>3.据/span>3.据/span>, 分别。这告诉我们我们需要同时解决以下等式:据/p>
一种据/span>(据/span>1据/span>)据/span>2据/span>+据/span>B.据/span>(据/span>1据/span>)据/span>+据/span>C据/span>一种据/span>(据/span>2据/span>)据/span>2据/span>+据/span>B.据/span>(据/span>2据/span>)据/span>+据/span>C据/span>一种据/span>(据/span>3.据/span>)据/span>2据/span>+据/span>B.据/span>(据/span>3.据/span>)据/span>+据/span>C据/span>=据/span>=据/span>=据/span>1据/span>3.据/span>2据/span>3.据/span>3.据/span>3.据/span>。据/span>
然而,同时解决它们可以繁琐和/或乏味。那么有替代方法吗?就在这里!实际上,我们可以同时解决这些方程,以及这样做的方式是应用剩余因子定理的第二部分:因子定理。据/p>
因子定理据/strong>
让据span class="katex">
F据/span>(据/span>X据/span>)据/span>是一个多项式据span class="katex">
F据/span>(据/span>C据/span>)据/span>=据/span>0.据/span>对于一些常数据span class="katex">
C据/span>。然后据span class="katex">
X据/span>-据/span>C据/span>是一个因素据span class="katex">
F据/span>(据/span>X据/span>)据/span>。相反,如果据span class="katex">
X据/span>-据/span>C据/span>是一个因素据span class="katex">
F据/span>(据/span>X据/span>)据/span>, 然后据span class="katex">
F据/span>(据/span>C据/span>)据/span>=据/span>0.据/span>。据/p>
但它与我们的多项式有关吗?好吧,如果我们考虑一个虚拟多项式据span class="katex">
G据/span>(据/span>X据/span>)据/span>这样它的根部为1,2和3.然后,通过因子定理,我们有据span class="katex">
G据/span>(据/span>X据/span>)据/span>=据/span>一种据/span>(据/span>X据/span>-据/span>1据/span>)据/span>(据/span>X据/span>-据/span>2据/span>)据/span>(据/span>X据/span>-据/span>3.据/span>)据/span>对于一些常数据span class="katex">
一种据/span>。同样,我们知道据span class="katex">
F据/span>(据/span>K.据/span>)据/span>=据/span>K.据/span>3.据/span>为了据span class="katex">
K.据/span>=据/span>1据/span>那据/span>2据/span>那据/span>3.据/span>。因此,喜欢据span class="katex">
G据/span>(据/span>X据/span>)据/span>,等式据span class="katex">
F据/span>(据/span>K.据/span>)据/span>-据/span>K.据/span>3.据/span>=据/span>0.据/span>还有1,2和3的根。因此,我们可以找到关系据span class="katex">
G据/span>(据/span>X据/span>)据/span>和据span class="katex">
F据/span>(据/span>X据/span>)据/span>要得到据span class="katex">
G据/span>(据/span>X据/span>)据/span>=据/span>F据/span>(据/span>X据/span>)据/span>-据/span>X据/span>3.据/span>=据/span>一种据/span>(据/span>X据/span>-据/span>1据/span>)据/span>(据/span>X据/span>-据/span>2据/span>)据/span>(据/span>X据/span>-据/span>3.据/span>)据/span>。我们知道,据/p>
F据/span>(据/span>X据/span>)据/span>-据/span>X据/span>3.据/span>=据/span>一种据/span>(据/span>X据/span>-据/span>1据/span>)据/span>(据/span>X据/span>-据/span>2据/span>)据/span>(据/span>X据/span>-据/span>3.据/span>)据/span>⇒据/span>F据/span>(据/span>X据/span>)据/span>=据/span>X据/span>3.据/span>+据/span>一种据/span>(据/span>X据/span>-据/span>1据/span>)据/span>(据/span>X据/span>-据/span>2据/span>)据/span>(据/span>X据/span>-据/span>3.据/span>)据/span>。据/span>
但自从我们知道据span class="katex">
F据/span>(据/span>X据/span>)据/span>是二次多项式,系数据span class="katex">
X据/span>3.据/span>在据span class="katex">
F据/span>(据/span>X据/span>)据/span>必须是0,所以据span class="katex">
一种据/span>被迫成为据span class="katex">
-据/span>1据/span>。然后我们有据span class="katex">
F据/span>(据/span>X据/span>)据/span>=据/span>X据/span>3.据/span>-据/span>(据/span>X据/span>-据/span>1据/span>)据/span>(据/span>X据/span>-据/span>2据/span>)据/span>(据/span>X据/span>-据/span>3.据/span>)据/span>=据/span>6.据/span>X据/span>2据/span>-据/span>1据/span>1据/span>X据/span>+据/span>6.据/span>。所以,我们已成功发现了未知的值据span class="katex">
一种据/span>那据/span>B.据/span>和据span class="katex">
C据/span>来自多项式的据span class="katex">
F据/span>(据/span>X据/span>)据/span>,我们不必同时解决以下任何方程式,这将浪费很多时间。据/p>
一种据/span>(据/span>1据/span>)据/span>2据/span>+据/span>B.据/span>(据/span>1据/span>)据/span>+据/span>C据/span>一种据/span>(据/span>2据/span>)据/span>2据/span>+据/span>B.据/span>(据/span>2据/span>)据/span>+据/span>C据/span>一种据/span>(据/span>3.据/span>)据/span>