# 多項式

## 多項式的定義

### 多項式及非多項式

${\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}}}}$

## 多項式的計算

 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

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

### 減法

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

### 除法

#### 多項式長除法

 例子：用多項式長除法求解${\displaystyle x^{2}+36x+155}$ 除以 ${\displaystyle x+5}$如下寫上被除式與除式 ${\displaystyle {\begin{array}{rl}\\x+5\!\!\!\!&{\big )}\!\!\!{\begin{array}{lll}\hline \,x^{2}+36x+155\end{array}}\end{array}}}$ 首先要消掉${\displaystyle x^{2}}$這一項，${\displaystyle x}$乘上${\displaystyle x+5}$是二次多項式，把${\displaystyle x}$記到商式部份 ${\displaystyle {\begin{array}{rl}&~~\,x\\x+5\!\!\!\!&{\big )}\!\!\!{\begin{array}{lll}\hline \,x^{2}+36x+155\end{array}}\\\end{array}}}$ 把${\displaystyle x(x+5)=x^{2}+5x}$寫到被除式的下方 ${\displaystyle {\begin{array}{rl}&~~\,x\\x+5\!\!\!\!&{\big )}\!\!\!{\begin{array}{lll}\hline \,x^{2}+36x+155\end{array}}\\&\!\!\!\!-{\underline {(x^{2}+5x)~~~}}\\\end{array}}}$ 與被除式相減 ${\displaystyle {\begin{array}{rl}&~~\,x\\x+5\!\!\!\!&{\big )}\!\!\!{\begin{array}{lll}\hline \,x^{2}+36x+155\end{array}}\\&\!\!\!\!-{\underline {(x^{2}+5x)~~~~~~~~~~~}}\\&\!\!\!\!~~~~~(36{\color {Orange}-5})x+155~~~\\\end{array}}}$ 把相減得到的結果看成另外一個被除式，重復以上步驟 ${\displaystyle {\begin{array}{rl}&~~\,x+31\\x+5\!\!\!\!&{\big )}\!\!\!{\begin{array}{lll}\hline \,x^{2}+36x+155\end{array}}\\&\!\!\!\!-{\underline {(x^{2}+5x)~~~~~~~~~~~}}\\&\!\!\!\!~~~~~~~~~~~~~31x+155~~~\\&\!\!\!\!~~~~~~~-{\underline {(31x+155)~~~}}\\&\!\!\!\!~~~~~~~~~~~~~~~~~~~~~~~~~~~~~0~~~\\\end{array}}}$ 在這個例子中商式為${\displaystyle x+31}$，餘式為0

#### 綜合除法

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

 例子：用綜合除法求解${\displaystyle x^{3}-12x^{2}-42}$除以${\displaystyle x-3}$在頂部寫上被除式的各項系數，左邊寫上k ${\displaystyle {\begin{array}{cc}{\begin{array}{r}\\3\\\end{array}}&{\begin{array}{|rrrr}1&-12&0&-42\\&&&\\\hline \end{array}}\end{array}}}$ 把被除式最高次項的系數寫下來 ${\displaystyle {\begin{array}{cc}{\begin{array}{r}\\3\\\\\end{array}}&{\begin{array}{|rrrr}1&-12&0&-42\\&&&\\\hline 1&&&\\\end{array}}\end{array}}}$ 乘上左邊的常數再放上第二行 ${\displaystyle {\begin{array}{cc}{\begin{array}{r}\\3\\\\\end{array}}&{\begin{array}{|rrrr}1&-12&0&-42\\&3&&\\\hline 1&&&\\\end{array}}\end{array}}}$ 與上面的系數相加，寫到下面 ${\displaystyle {\begin{array}{cc}{\begin{array}{c}\\3\\\\\end{array}}&{\begin{array}{|rrrr}1&-12&0&-42\\&3&&\\\hline 1&-9&&\\\end{array}}\end{array}}}$ 重複乘法加法運算，直到除法結束 ${\displaystyle {\begin{array}{cc}{\begin{array}{c}\\3\\\\\end{array}}&{\begin{array}{|rrrr}1&-12&0&-42\\&3&-27&-81\\\hline 1&-9&-27&-123\end{array}}\end{array}}}$ 結果得出商式為${\displaystyle x^{2}-9x-27}$，餘式為${\displaystyle -123}$

#### 餘式定理

 證明餘式定理由多項式除法得到： ${\displaystyle f(x)=(x-a)q(x)+r}$ 因為除式為一次多項式，所以餘式一定是常數，代入${\displaystyle x=a}$，得到${\displaystyle f(a)=r}$

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

### 習題

1. ${\displaystyle f(x)=2x^{4}+3x^{3}+2x^{2}+5x+5}$ 除以${\displaystyle x+2}$的餘式
2. 用長除法求${\displaystyle f(x)=x^{3}+36x^{2}+36x+155}$ 除以${\displaystyle x^{2}+5x+5}$的餘式

## 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${\displaystyle 4x^{2}+7x-2}$從上例中我們得到多項式的根為${\displaystyle x={\frac {1}{4}}}$ and ${\displaystyle x=-2}$，於是多項式可分解成： ${\displaystyle 4x^{2}+7x-2=C(x+2)(x-{\frac {1}{4}})=C(x^{2}+{\frac {7}{4}}x-{\frac {1}{2}})}$ 可見 ${\displaystyle C=4}$時等式成立，於是多項式可因式分解為： ${\displaystyle 4x^{2}+7x-2=4(x+2)(x-{\frac {1}{4}})=(x+2)(4x-1)}$

#### 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. ${\displaystyle x+{\frac {b}{2a}}=\pm {\frac {\sqrt {b^{2}-4ac}}{2a}}\Rightarrow x={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}}$
##### 根與系數的關係

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