一多項式 x 2 + cx + d 可因式分解成(x + a )(x + b )。其中：ab = d ，a + b = c 因式分解 ，在这里是指多項式因式分解 （英語：Polynomial Factorization [註 1] 多項式 分解為兩個或多個的因式 [註 2]
x
2
−
1
2
{\displaystyle x^{2}-1^{2}}
(
x
+
1
)
(
x
−
1
)
{\displaystyle \left(x+1\right)\left(x-1\right)}
x
2
−
y
2
{\displaystyle x^{2}-y^{2}}
(
x
+
y
)
(
x
−
y
)
{\displaystyle \left(x+y\right)\left(x-y\right)}
複整數 ，那么
x
2
+
1
2
{\displaystyle x^{2}+1^{2}}
(
x
+
i
)
(
x
−
i
)
{\displaystyle \left(x+i\right)\left(x-i\right)}
不可约多项式 （irreducible ）。也就是不能再分解了。
因式分解定理 数域F上每个次数
≥
1
{\displaystyle \geq 1}
f
(
x
)
{\displaystyle f(x)}
f
(
x
)
=
p
1
(
x
)
p
2
(
x
)
p
3
(
x
)
⋯
p
s
(
x
)
=
q
1
(
x
)
q
2
(
x
)
⋯
q
t
(
x
)
{\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)}
其中
p
i
(
x
)
(
i
=
1
,
2
,
⋯
,
s
)
{\displaystyle p_{i}(x)(i=1,2,\cdots ,s)}
q
j
(
x
)
(
j
=
1
,
2
,
⋯
,
t
)
{\displaystyle q_{j}(x)(j=1,2,\cdots ,t)}
s
=
t
{\displaystyle s=t}
p
i
(
x
)
=
c
i
q
i
(
x
)
(
i
=
1
,
2
,
⋯
,
s
)
{\displaystyle p_{i}(x)=c_{i}q_{i}(x)(i=1,2,\cdots ,s)}
c
i
(
i
=
1
,
2
,
⋯
,
s
)
{\displaystyle c_{i}(i=1,2,\cdots ,s)}
分解方法 公因式分解(抽) 原则：
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：
7
a
+
98
a
b
{\displaystyle 7a+98ab}
其中，
7
a
{\displaystyle 7a}
7
a
(
1
+
14
b
)
{\displaystyle 7a(1+14b)}
51
a
4
b
7
+
24
a
3
b
2
+
75
a
5
b
5
{\displaystyle 51a^{4}b^{7}+24a^{3}b^{2}+75a^{5}b^{5}}
其中，
3
a
3
b
2
{\displaystyle 3a^{3}b^{2}}
3
a
3
b
2
(
17
a
b
5
+
25
a
2
b
3
+
8
)
{\displaystyle 3a^{3}b^{2}(17ab^{5}+25a^{2}b^{3}+8)}
公式重組(拼) 透過公式重組，然後再抽出公因數，例子：
3
a
2
b
−
5
y
+
12
a
3
b
2
−
20
a
b
y
{\displaystyle 3a^{2}b-5y+12a^{3}b^{2}-20aby}
=
(
3
a
2
b
+
12
a
3
b
2
)
−
(
5
y
+
20
a
b
y
)
{\displaystyle =(3a^{2}b+12a^{3}b^{2})-(5y+20aby)}
=
3
a
2
b
(
1
+
4
a
b
)
−
5
y
(
1
+
4
a
b
)
{\displaystyle =3a^{2}b(1+4ab)-5y(1+4ab)}
=
(
1
+
4
a
b
)
(
3
a
2
b
−
5
y
)
{\displaystyle =(1+4ab)(3a^{2}b-5y)}
15
n
2
+
2
m
−
3
n
−
10
m
n
{\displaystyle 15n^{2}+2m-3n-10mn}
=
(
15
n
2
−
3
n
)
+
(
2
m
−
10
m
n
)
{\displaystyle =(15n^{2}-3n)+(2m-10mn)}
=
3
n
(
5
n
−
1
)
+
2
m
(
1
−
5
n
)
{\displaystyle =3n(5n-1)+2m(1-5n)}
=
3
n
(
5
n
−
1
)
−
2
m
(
5
n
−
1
)
{\displaystyle =3n(5n-1)-2m(5n-1)}
=
(
5
n
−
1
)
(
3
n
−
2
m
)
{\displaystyle =(5n-1)(3n-2m)}
添項法(增) 透過添項然後減掉，然後再抽出公因數，例子：
x
4
+
x
2
+
1
{\displaystyle x^{4}+x^{2}+1}
=
x
4
+
x
2
+
x
2
−
x
2
+
1
{\displaystyle =x^{4}+x^{2}+x^{2}-x^{2}+1}
=
x
4
+
2
x
2
−
x
2
+
1
{\displaystyle =x^{4}+2x^{2}-x^{2}+1}
=
x
4
+
2
x
2
+
1
−
x
2
{\displaystyle =x^{4}+2x^{2}+1-x^{2}}
=
(
x
2
)
2
+
(
2
)
(
x
2
)
(
1
)
+
(
1
)
2
−
x
2
{\displaystyle =(x^{2})^{2}+(2)(x^{2})(1)+(1)^{2}-x^{2}}
=
(
x
2
+
1
)
2
−
x
2
{\displaystyle =(x^{2}+1)^{2}-x^{2}}
=
(
x
2
+
1
−
x
)
(
x
2
+
1
+
x
)
{\displaystyle =(x^{2}+1-x)(x^{2}+1+x)}
=
(
x
2
−
x
+
1
)
(
x
2
+
x
+
1
)
{\displaystyle =(x^{2}-x+1)(x^{2}+x+1)}
分項法(拆) 透過分裂某項，然後再抽出公因數，例子：
x
3
−
7
x
+
6
{\displaystyle x^{3}-7x+6}
其中，
−
7
x
{\displaystyle -7x}
−
x
{\displaystyle -x}
−
6
x
{\displaystyle -6x}
x
3
−
7
x
+
6
{\displaystyle x^{3}-7x+6}
x
3
−
x
−
6
x
+
6
{\displaystyle x^{3}-x-6x+6}
x
3
−
7
x
+
6
{\displaystyle x^{3}-7x+6}
=
x
3
−
x
−
6
x
+
6
{\displaystyle =x^{3}-x-6x+6}
=
(
x
3
−
x
)
−
(
6
x
−
6
)
{\displaystyle =(x^{3}-x)-(6x-6)}
=
x
(
x
2
−
1
)
−
6
(
x
−
1
)
{\displaystyle =x(x^{2}-1)-6(x-1)}
=
x
(
x
+
1
)
(
x
−
1
)
−
6
(
x
−
1
)
{\displaystyle =x(x+1)(x-1)-6(x-1)}
=
(
x
(
x
+
1
)
−
6
)
(
x
−
1
)
{\displaystyle =(x(x+1)-6)(x-1)}
=
(
x
2
+
x
−
6
)
(
x
−
1
)
{\displaystyle =(x^{2}+x-6)(x-1)}
其中，
+
x
{\displaystyle +x}
+
3
x
{\displaystyle +3x}
−
2
x
{\displaystyle -2x}
x
2
+
x
−
6
{\displaystyle x^{2}+x-6}
x
2
+
3
x
−
2
x
−
6
{\displaystyle x^{2}+3x-2x-6}
(
x
2
+
x
−
6
)
(
x
−
1
)
{\displaystyle (x^{2}+x-6)(x-1)}
=
(
x
2
+
3
x
−
2
x
−
6
)
(
x
−
1
)
{\displaystyle =(x^{2}+3x-2x-6)(x-1)}
=
(
(
x
2
+
3
x
)
−
(
2
x
+
6
)
)
(
x
−
1
)
{\displaystyle =((x^{2}+3x)-(2x+6))(x-1)}
=
(
x
(
x
+
3
)
−
2
(
x
+
3
)
)
(
x
−
1
)
{\displaystyle =(x(x+3)-2(x+3))(x-1)}
=
(
x
−
2
)
(
x
+
3
)
(
x
−
1
)
{\displaystyle =(x-2)(x+3)(x-1)}
=
(
x
−
1
)
(
x
−
2
)
(
x
+
3
)
{\displaystyle =(x-1)(x-2)(x+3)}
十字交乘法 十字交乘法（cross method），也叫做十字相乘法。它实際上是拆項法 的一個變形，只不過用十字形矩陣來表示。
兩個n次方數之和與差 兩個立方數之和
a
3
+
b
3
{\displaystyle a^{3}+b^{3}\,\!}
(
a
+
b
)
(
a
2
−
a
b
+
b
2
)
{\displaystyle (a+b)(a^{2}-ab+b^{2})\,\!}
兩個立方數之差
a
3
−
b
3
{\displaystyle a^{3}-b^{3}\,\!}
(
a
−
b
)
(
a
2
+
a
b
+
b
2
)
{\displaystyle (a-b)(a^{2}+ab+b^{2})\,\!}
兩個n次方數之差
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})}
兩個奇數次方數之和
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 一個整係數的一元多項式
a
n
x
n
+
a
n
−
1
x
n
−
1
+
.
.
.
.
.
.
a
1
x
+
a
0
{\displaystyle a_{n}x^{n}+a_{n-1}x^{n-1}+......a_{1}x+a_{0}}
p
x
+
q
{\displaystyle px+q}
且p,q互質 ，則以下兩條必成立：（逆敘述並不真）
p
|
a
n
{\displaystyle p|a_{n}}
q
|
a
0
{\displaystyle q|a_{0}}
不過反過來說，即使當
p
|
a
n
{\displaystyle p|a_{n}}
q
|
a
0
{\displaystyle q|a_{0}}
p
x
+
q
{\displaystyle px+q}
a
n
x
n
+
a
n
−
1
x
n
−
1
+
.
.
.
.
.
.
a
1
x
+
a
0
{\displaystyle a_{n}x^{n}+a_{n-1}x^{n-1}+......a_{1}x+a_{0}}
另外一個看法是：
一個整係數的ｎ次多項式
a
n
x
n
+
a
n
−
1
x
n
−
1
+
.
.
.
.
.
.
a
1
x
+
a
0
{\displaystyle a_{n}x^{n}+a_{n-1}x^{n-1}+......a_{1}x+a_{0}}
p
x
−
q
{\displaystyle px-q}
且p,q互質 ，則：（逆敘述並不真）
p
−
q
|
f
(
1
)
{\displaystyle p-q|f(1)}
p
+
q
|
f
(
−
1
)
{\displaystyle p+q|f(-1)}
参见 注释 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 
