數體
P
{\displaystyle P}
f
(
x
)
{\displaystyle f(x)}
p
i
(
x
)
{\displaystyle p_{i}(x)}
在複數域上，每個不可約多項式都是一次的，因此高於一次的複系數多項式，都可以唯一地分解為多個一次式之積。
在實數體上，不可約的多項式都是一次或二次的，因此高於一次的實系數多項式，都可以唯一地分解為一次、二次多項式之積。
在有理數體上，不可約多項式可以有任何次。例如，在有理數範圍內，當
n
{\displaystyle n}
x
{\displaystyle x}
x
n
+
2
{\displaystyle x^{n}+2}
[ 1]
數體F上每個次數
≥
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)}
原則：
分解必須 要徹底（即分解後之因式均不能再做分解）
結果最後只留下小括號
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)}
兩個立方數之和
a
3
+
b
3
=
(
a
+
b
)
(
a
2
−
a
b
+
b
2
)
{\displaystyle a^{3}+b^{3}=(a+b)(a^{2}-ab+b^{2})}
兩個立方數之差
a
3
−
b
3
=
(
a
−
b
)
(
a
2
+
a
b
+
b
2
)
{\displaystyle a^{3}-b^{3}=(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})}
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
=
(
3
a
2
b
+
12
a
3
b
2
)
−
(
5
y
+
20
a
b
y
)
=
3
a
2
b
(
1
+
4
a
b
)
−
5
y
(
1
+
4
a
b
)
=
(
1
+
4
a
b
)
(
3
a
2
b
−
5
y
)
{\displaystyle {\begin{aligned}&3a^{2}b-5y+12a^{3}b^{2}-20aby\\=&(3a^{2}b+12a^{3}b^{2})-(5y+20aby)\\=&3a^{2}b(1+4ab)-5y(1+4ab)\\=&(1+4ab)(3a^{2}b-5y)\end{aligned}}}
15
n
2
+
2
m
−
3
n
−
10
m
n
=
(
15
n
2
−
3
n
)
+
(
2
m
−
10
m
n
)
=
3
n
(
5
n
−
1
)
+
2
m
(
1
−
5
n
)
=
3
n
(
5
n
−
1
)
−
2
m
(
5
n
−
1
)
=
(
5
n
−
1
)
(
3
n
−
2
m
)
{\displaystyle {\begin{aligned}&15n^{2}+2m-3n-10mn\\=&(15n^{2}-3n)+(2m-10mn)\\=&3n(5n-1)+2m(1-5n)\\=&3n(5n-1)-2m(5n-1)=&(5n-1)(3n-2m)\end{aligned}}}
透過添項然後減掉，然後再抽出公因數，例子：
x
4
+
x
2
+
1
=
x
4
+
x
2
+
x
2
−
x
2
+
1
=
x
4
+
2
x
2
−
x
2
+
1
=
x
4
+
2
x
2
+
1
−
x
2
=
(
x
2
+
1
)
2
−
x
2
=
(
x
2
+
1
−
x
)
(
x
2
+
1
+
x
)
=
(
x
2
−
x
+
1
)
(
x
2
+
x
+
1
)
{\displaystyle {\begin{aligned}&x^{4}+x^{2}+1\\=&x^{4}+x^{2}+x^{2}-x^{2}+1\\=&x^{4}+2x^{2}-x^{2}+1\\=&x^{4}+2x^{2}+1-x^{2}\\=&\left(x^{2}+1\right)^{2}-x^{2}\\=&\left(x^{2}+1-x\right)\left(x^{2}+1+x\right)\\=&\left(x^{2}-x+1\right)\left(x^{2}+x+1\right)\end{aligned}}}
或者透過分裂某項，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
=
x
3
−
x
−
6
x
+
6
=
(
x
3
−
x
)
−
(
6
x
−
6
)
=
x
(
x
2
−
1
)
−
6
(
x
−
1
)
=
x
(
x
+
1
)
(
x
−
1
)
−
6
(
x
−
1
)
=
[
x
(
x
+
1
)
−
6
]
(
x
−
1
)
=
(
x
2
+
x
−
6
)
(
x
−
1
)
{\displaystyle {\begin{aligned}&x^{3}-7x+6\\=&x^{3}-x-6x+6\\=&\left(x^{3}-x\right)-\left(6x-6\right)\\=&x\left(x^{2}-1\right)-6\left(x-1\right)\\=&x\left(x+1\right)\left(x-1\right)-6\left(x-1\right)\\=&\left[x\left(x+1\right)-6\right]\left(x-1\right)\\=&\left(x^{2}+x-6\right)\left(x-1\right)\end{aligned}}}
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
)
=
(
x
2
+
3
x
−
2
x
−
6
)
(
x
−
1
)
=
[
(
x
2
+
3
x
)
−
(
2
x
+
6
)
]
(
x
−
1
)
=
(
x
(
x
+
3
)
−
2
(
x
+
3
)
)
(
x
−
1
)
=
(
x
−
2
)
(
x
+
3
)
(
x
−
1
)
=
(
x
−
1
)
(
x
−
2
)
(
x
+
3
)
{\displaystyle {\begin{aligned}&\left(x^{2}+x-6\right)\left(x-1\right)\\=&\left(x^{2}+3x-2x-6\right)\left(x-1\right)\\=&\left[\left(x^{2}+3x\right)-\left(2x+6\right)\right]\left(x-1\right)\\=&\left(x\left(x+3\right)-2\left(x+3\right)\right)\left(x-1\right)\\=&\left(x-2\right)\left(x+3\right)\left(x-1\right)\\=&\left(x-1\right)\left(x-2\right)\left(x+3\right)\end{aligned}}}
十字交乘法（cross method），也叫做十字相乘法。它實際上是拆項法 的一個變形，只不過用十字形矩陣來表示。
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
^ 多 项 式 (PDF) . 清華大學出版社. [2025-01-04 ] .
![{\displaystyle {\begin{aligned}&x^{3}-7x+6\\=&x^{3}-x-6x+6\\=&\left(x^{3}-x\right)-\left(6x-6\right)\\=&x\left(x^{2}-1\right)-6\left(x-1\right)\\=&x\left(x+1\right)\left(x-1\right)-6\left(x-1\right)\\=&\left[x\left(x+1\right)-6\right]\left(x-1\right)\\=&\left(x^{2}+x-6\right)\left(x-1\right)\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e173081941a261a4a517cd7652052ef66b5d5f4f)
![{\displaystyle {\begin{aligned}&\left(x^{2}+x-6\right)\left(x-1\right)\\=&\left(x^{2}+3x-2x-6\right)\left(x-1\right)\\=&\left[\left(x^{2}+3x\right)-\left(2x+6\right)\right]\left(x-1\right)\\=&\left(x\left(x+3\right)-2\left(x+3\right)\right)\left(x-1\right)\\=&\left(x-2\right)\left(x+3\right)\left(x-1\right)\\=&\left(x-1\right)\left(x-2\right)\left(x+3\right)\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/eb483e17739ac90192cc6ff1f789567260990a3f)
