# 多项式

## 多项式的定义

### 多项式及非多项式

${\displaystyle x^{10}+60x^{8}-33x^{5}+42x^{2}-3}$ ${\displaystyle ({\sqrt {2}}-{\sqrt[{3}]{5}})x^{2}-{\sqrt[{4}]{6}}}$ ${\displaystyle \pi ^{e}x^{10^{10}}-6x+1}$ ${\displaystyle (3+i)x^{5}-2ix^{2}+1}$

${\displaystyle 2x^{3}+10x^{-1}}$ ${\displaystyle 5x^{\sqrt {2}}+x^{3\pi }+2x-1}$ ${\displaystyle x^{3}-5x^{\frac {2}{3}}+4{\sqrt {x}}-3{\sqrt[{3}]{x^{5}}}}$ ${\displaystyle {\sqrt {1+x^{2}}}}$

## 多项式的计算

 當中間有些項的係數為零時，最好將這些項也給寫出來，以減少錯誤，而寫答案時，係數為零的項可省略不寫


### 加法

${\displaystyle f(x)+g(x)=(2x^{4}+3x^{3}+2x^{2}+5x+5)+(5x^{4}+10x^{3}+x^{2}+3x+2)=7x^{4}+13x^{3}+3x^{2}+8x+7}$

### 减法

${\displaystyle f(x)-g(x)=(36{\color {Orange}-5})x^{5}+(7{\color {Orange}+73})x^{4}+(66{\color {Orange}+11})x^{3}+(36{\color {Orange}+11})x^{2}+(66{\color {Orange}-5})x+(6{\color {Orange}-3})=31x^{5}+80x^{4}+77x^{3}+47x^{2}+61x+3}$

### 除法

#### 多项式长除法

 例子：用多项式长除法求解${\displaystyle x^{2}+36x+155}$  除以 ${\displaystyle x+5}$ 如下写上被除式与除式 ${\displaystyle {\begin{array}{rl}\\x+5\!\!\!\!&{\big )}\!\!\!{\begin{array}{lll}\hline \,x^{2}+36x+155\end{array}}\end{array}}}$  首先要消掉${\displaystyle x^{2}}$ 这一项，${\displaystyle x}$ 乘上${\displaystyle x+5}$ 是二次多项式，把${\displaystyle x}$ 记到商式部份 ${\displaystyle {\begin{array}{rl}&~~\,x\\x+5\!\!\!\!&{\big )}\!\!\!{\begin{array}{lll}\hline \,x^{2}+36x+155\end{array}}\\\end{array}}}$  把${\displaystyle x(x+5)=x^{2}+5x}$ 写到被除式的下方 ${\displaystyle {\begin{array}{rl}&~~\,x\\x+5\!\!\!\!&{\big )}\!\!\!{\begin{array}{lll}\hline \,x^{2}+36x+155\end{array}}\\&\!\!\!\!-{\underline {(x^{2}+5x)~~~}}\\\end{array}}}$  与被除式相减 ${\displaystyle {\begin{array}{rl}&~~\,x\\x+5\!\!\!\!&{\big )}\!\!\!{\begin{array}{lll}\hline \,x^{2}+36x+155\end{array}}\\&\!\!\!\!-{\underline {(x^{2}+5x)~~~~~~~~~~~}}\\&\!\!\!\!~~~~~(36{\color {Orange}-5})x+155~~~\\\end{array}}}$  把相减得到的结果看成另外一个被除式，重复以上步骤 ${\displaystyle {\begin{array}{rl}&~~\,x+31\\x+5\!\!\!\!&{\big )}\!\!\!{\begin{array}{lll}\hline \,x^{2}+36x+155\end{array}}\\&\!\!\!\!-{\underline {(x^{2}+5x)~~~~~~~~~~~}}\\&\!\!\!\!~~~~~~~~~~~~~31x+155~~~\\&\!\!\!\!~~~~~~~-{\underline {(31x+155)~~~}}\\&\!\!\!\!~~~~~~~~~~~~~~~~~~~~~~~~~~~~~0~~~\\\end{array}}}$  在这个例子中商式为${\displaystyle x+31}$ ，馀式为0

#### 综合除法

${\displaystyle {\begin{array}{cc}{\begin{array}{r}~\\k\\~\end{array}}&{\begin{array}{|rrrr}f(x)&~\\~&~\\\hline q(x)&r\\\end{array}}\end{array}}}$

 例子：用综合除法求解${\displaystyle x^{3}-12x^{2}-42}$ 除以${\displaystyle x-3}$ 在顶部写上被除式的各项系数，左边写上k ${\displaystyle {\begin{array}{cc}{\begin{array}{r}\\3\\\end{array}}&{\begin{array}{|rrrr}1&-12&0&-42\\&&&\\\hline \end{array}}\end{array}}}$  把被除式最高次项的系数写下来 ${\displaystyle {\begin{array}{cc}{\begin{array}{r}\\3\\\\\end{array}}&{\begin{array}{|rrrr}1&-12&0&-42\\&&&\\\hline 1&&&\\\end{array}}\end{array}}}$  乘上左边的常数再放上第二行 ${\displaystyle {\begin{array}{cc}{\begin{array}{r}\\3\\\\\end{array}}&{\begin{array}{|rrrr}1&-12&0&-42\\&3&&\\\hline 1&&&\\\end{array}}\end{array}}}$  与上面的系数相加，写到下面 ${\displaystyle {\begin{array}{cc}{\begin{array}{c}\\3\\\\\end{array}}&{\begin{array}{|rrrr}1&-12&0&-42\\&3&&\\\hline 1&-9&&\\\end{array}}\end{array}}}$  重复乘法加法运算，直到除法结束 ${\displaystyle {\begin{array}{cc}{\begin{array}{c}\\3\\\\\end{array}}&{\begin{array}{|rrrr}1&-12&0&-42\\&3&-27&-81\\\hline 1&-9&-27&-123\end{array}}\end{array}}}$  结果得出商式为${\displaystyle x^{2}-9x-27}$ ，馀式为${\displaystyle -123}$

#### 馀式定理

 证明馀式定理由多项式除法得到： ${\displaystyle f(x)=(x-a)q(x)+r}$  因为除式为一次多项式，所以馀式一定是常数，代入${\displaystyle x=a}$ ，得到${\displaystyle f(a)=r}$

${\displaystyle f(x)=x^{2}+36x+155}$  除以 ${\displaystyle x+5}$ 的馀式为${\displaystyle f(-5)=(-5)^{2}+36(-5)+155=25-180+155=0}$
${\displaystyle f(x)=x^{3}-12x^{2}-42}$ 除以${\displaystyle x-3}$ 的馀式为${\displaystyle f(3)=(3)^{3}-12(3)^{2}-42=27-108-42=-123}$
${\displaystyle f(x)=x^{3}-7x-55}$ 除以${\displaystyle x-4}$ 的馀式为${\displaystyle f(4)=(4)^{3}-7(4)-55=64-28-55=36-55=-19}$

### 习题

1. ${\displaystyle f(x)=2x^{4}+3x^{3}+2x^{2}+5x+5}$  除以${\displaystyle x+2}$ 的馀式
2. 用长除法求${\displaystyle f(x)=x^{3}+36x^{2}+36x+155}$  除以${\displaystyle x^{2}+5x+5}$ 的馀式

## 因式分解

• ${\displaystyle x^{2}+31x-180}$

## 方程式求解

### 一元二次方程式

 例子：找出${\displaystyle 4x^{2}+7x-2}$ 的所有根即求${\displaystyle 4x^{2}+7x-2=0}$ 的所有解。 代入${\displaystyle a=4,b=7,c=-2}$ 到求根公式，得到： ${\displaystyle x={\frac {-7\pm {\sqrt {7^{2}-4(4)(-2)}}}{2(4)}}={\frac {-7\pm {\sqrt {49+32}}}{8}}={\frac {-7\pm {\sqrt {81}}}{8}}={\frac {-7\pm 9}{8}}}$  ${\displaystyle x={\frac {2}{8}},x={\frac {-16}{8}}}$  ${\displaystyle x={\frac {1}{4}},x=-2}$

 例子：因式分解${\displaystyle 4x^{2}+7x-2}$ 从上例中我们得到多项式的根为${\displaystyle x={\frac {1}{4}}}$  and ${\displaystyle x=-2}$ ，于是多项式可分解成： ${\displaystyle 4x^{2}+7x-2=C(x+2)(x-{\frac {1}{4}})=C(x^{2}+{\frac {7}{4}}x-{\frac {1}{2}})}$  可见 ${\displaystyle C=4}$ 时等式成立，于是多项式可因式分解为： ${\displaystyle 4x^{2}+7x-2=4(x+2)(x-{\frac {1}{4}})=(x+2)(4x-1)}$

#### 配方法

1. ${\displaystyle ax^{2}+bx+c=0\Rightarrow x^{2}+{\frac {b}{a}}x+{\frac {c}{a}}=0}$
1. ${\displaystyle x^{2}+{\frac {b}{a}}x+{\frac {c}{a}}=0\Rightarrow x^{2}+2\left({\frac {b}{2a}}\right)x+{\frac {c}{a}}=0}$
1. ${\displaystyle x^{2}+2\left({\frac {b}{2a}}\right)x+{\frac {c}{a}}=0\Rightarrow x^{2}+2\left({\frac {b}{2a}}\right)x+\left({\frac {b}{2a}}\right)^{2}+{\frac {c}{a}}=\left({\frac {b}{2a}}\right)^{2}}$
1. ${\displaystyle x^{2}+2\left({\frac {b}{2a}}\right)x+\left({\frac {b}{2a}}\right)^{2}+{\frac {c}{a}}=\left({\frac {b}{2a}}\right)^{2}\Rightarrow x^{2}+2\left({\frac {b}{2a}}\right)x+\left({\frac {b}{2a}}\right)^{2}=\left({\frac {b}{2a}}\right)^{2}-{\frac {c}{a}}}$
1. ${\displaystyle x^{2}+2\left({\frac {b}{2a}}\right)x+\left({\frac {b}{2a}}\right)^{2}=\left({\frac {b}{2a}}\right)^{2}-{\frac {c}{a}}\Rightarrow (x+{\frac {b}{2a}})^{2}=\left({\frac {b}{2a}}\right)^{2}-{\frac {4ac}{4a^{2}}}={\frac {b^{2}-4ac}{4a^{2}}}}$
1. ${\displaystyle (x+{\frac {b}{2a}})^{2}={\frac {b^{2}-4ac}{4a^{2}}}\Rightarrow x+{\frac {b}{2a}}=\pm {\frac {\sqrt {b^{2}-4ac}}{2a}}}$
1. ${\displaystyle x+{\frac {b}{2a}}=\pm {\frac {\sqrt {b^{2}-4ac}}{2a}}\Rightarrow x={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}}$
##### 根与系数的关系

• ${\displaystyle w+z=-{\frac {b}{a}}}$
• ${\displaystyle wz={\frac {c}{a}}}$