# 因式分解

## 因式分解定理

${\displaystyle f(x)=p_{1}(x)p_{2}(x)p_{3}(x)\cdots p_{s}(x)=q_{1}(x)q_{2}(x)\cdots q_{t}(x)}$

${\displaystyle p_{i}(x)=c_{i}q_{i}(x)(i=1,2,\cdots ,s)}$ ,其中${\displaystyle c_{i}(i=1,2,\cdots ,s)}$ 是一些非零常数

## 分解方法

### 公因式分解(抽)

1、分解必須要彻底（即分解後之因式均不能再做分解）

2、結果最後只留下小括號

3、結果的多項式首項為正。 在一個公式內把其公因子抽出，例子：

• ${\displaystyle 7a+98ab}$
• 其中，${\displaystyle 7a}$ 是公因子。因此，因式分解後得到的答案是：${\displaystyle 7a(1+14b)}$
• ${\displaystyle 51a^{4}b^{7}+24a^{3}b^{2}+75a^{5}b^{5}}$
• 其中，${\displaystyle 3a^{3}b^{2}}$ 是公因子。因此，因式分解後得到的答案是：${\displaystyle 3a^{3}b^{2}(17ab^{5}+25a^{2}b^{3}+8)}$

### 公式重組(拼)

• ${\displaystyle 3a^{2}b-5y+12a^{3}b^{2}-20aby}$
${\displaystyle =(3a^{2}b+12a^{3}b^{2})-(5y+20aby)}$
${\displaystyle =3a^{2}b(1+4ab)-5y(1+4ab)}$
${\displaystyle =(1+4ab)(3a^{2}b-5y)}$
• ${\displaystyle 15n^{2}+2m-3n-10mn}$
${\displaystyle =(15n^{2}-3n)+(2m-10mn)}$
${\displaystyle =3n(5n-1)+2m(1-5n)}$
${\displaystyle =3n(5n-1)-2m(5n-1)}$
${\displaystyle =(5n-1)(3n-2m)}$

### 添項法(增)

• ${\displaystyle x^{4}+x^{2}+1}$
${\displaystyle =x^{4}+x^{2}+x^{2}-x^{2}+1}$
${\displaystyle =x^{4}+2x^{2}-x^{2}+1}$
${\displaystyle =x^{4}+2x^{2}+1-x^{2}}$
${\displaystyle =(x^{2})^{2}+(2)(x^{2})(1)+(1)^{2}-x^{2}}$
${\displaystyle =(x^{2}+1)^{2}-x^{2}}$
${\displaystyle =(x^{2}+1-x)(x^{2}+1+x)}$
${\displaystyle =(x^{2}-x+1)(x^{2}+x+1)}$

### 分項法(拆)

${\displaystyle x^{3}-7x+6}$

• 其中，${\displaystyle -7x}$ 可以被拆成${\displaystyle -x}$ ${\displaystyle -6x}$ 。所以，${\displaystyle x^{3}-7x+6}$ 可以被寫成${\displaystyle x^{3}-x-6x+6}$ 。因此，
${\displaystyle x^{3}-7x+6}$
${\displaystyle =x^{3}-x-6x+6}$
${\displaystyle =(x^{3}-x)-(6x-6)}$
${\displaystyle =x(x^{2}-1)-6(x-1)}$
${\displaystyle =x(x+1)(x-1)-6(x-1)}$
${\displaystyle =(x(x+1)-6)(x-1)}$
${\displaystyle =(x^{2}+x-6)(x-1)}$

${\displaystyle (x^{2}+x-6)(x-1)}$
${\displaystyle =(x^{2}+3x-2x-6)(x-1)}$
${\displaystyle =((x^{2}+3x)-(2x+6))(x-1)}$
${\displaystyle =(x(x+3)-2(x+3))(x-1)}$
${\displaystyle =(x-2)(x+3)(x-1)}$
${\displaystyle =(x-1)(x-2)(x+3)}$

### 兩個n次方數之和與差

${\displaystyle a^{3}+b^{3}\,\!}$ 可分解為${\displaystyle (a+b)(a^{2}-ab+b^{2})\,\!}$

${\displaystyle a^{3}-b^{3}\,\!}$ 可分解為${\displaystyle (a-b)(a^{2}+ab+b^{2})\,\!}$

${\displaystyle a^{n}-b^{n}=(a-b)(a^{n-1}+a^{n-2}b+......+b^{n-1})}$

${\displaystyle a^{n}+b^{n}=(a+b)(a^{n-1}-a^{n-2}b+......+b^{n-1})}$

## 一次因式檢驗法

• ${\displaystyle p|a_{n}}$
• ${\displaystyle q|a_{0}}$

• ${\displaystyle p-q|f(1)}$
• ${\displaystyle p+q|f(-1)}$

## 注释

1. ^ 也有polynomial factorisationfactoring的用法
2. ^ 因式即多項式。

## 延伸閱讀

• Burnside, William Snow; Panton, Arthur William (1960) [1912], The Theory of Equations with an introduction to the theory of binary algebraic forms (Volume one), Dover
• Dickson, Leonard Eugene (1922), First Course in the Theory of Equations, New York: John Wiley & Sons
• Fite, William Benjamin (1921), College Algebra (Revised), Boston: D. C. Heath & Co.
• Klein, Felix (1925), Elementary Mathematics from an Advanced Standpoint; Arithmetic, Algebra, Analysis, Dover
• Selby, Samuel M., CRC Standard Mathematical Tables (18th ed.), The Chemical Rubber Co