This article shows how to calculate algebraic Greatest Common Factor by means of simple examples.
In Case 1 the method produces a unique GCF (if there is one).
In Case 2 the method produces several possibilities, any one of which could be the GCF depending on parameters supplied.
In Case 3 the method solves for a given variable if there are two equations containing two variables.
Case 1: the unique GCF
edit
3
x
3
+
23
x
2
−
19
x
−
231
{\displaystyle 3x^{3}+23x^{2}-19x-231}
6
x
3
+
31
x
2
−
162
x
−
715
{\displaystyle 6x^{3}+31x^{2}-162x-715}
15
x
2
+
124
x
+
253
{\displaystyle 15x^{2}+124x+253}
69
x
2
+
844
x
+
2167
{\displaystyle 69x^{2}+844x+2167}
3
x
+
11
{\displaystyle 3x+11}
3
x
+
11
{\displaystyle 3x+11}
(
3
x
+
11
)
{\displaystyle (3x+11)}
Case 2: solving the cubic
edit
Equal roots occur where the function and the slope of the function both equal 0.
a
x
3
+
b
x
2
+
c
x
+
d
.
.
.
.
(
f
0
)
3
a
x
2
+
2
b
x
+
c
.
.
.
.
(
f
1
)
{\displaystyle {\begin{aligned}&ax^{3}+bx^{2}+cx+d\,\,\,\,....\,\,\,\,(f0)\\&3ax^{2}+2bx+c\,\,\,\,....\,\,\,\,(f1)\end{aligned}}}
(f0) contains maximum power of x equal to 3. (f1) contains maximum power of x equal to 2.
Combine (f0) and (f1) to produce a second function containing maximum power of x equal to 2.
c
(
f
0
)
−
d
(
f
1
)
,
(
+
a
c
)
x
x
+
(
−
3
a
d
+
b
c
)
x
+
(
−
2
b
d
+
c
c
)
.
.
.
.
(
f
2
)
{\displaystyle c(f0)-d(f1),\,\,\,\,(+ac)xx+(-3ad+bc)x+(-2bd+cc)\,\,\,\,....\,\,\,\,(f2)}
c
(
f
2
)
−
(
−
2
b
d
+
c
c
)
(
f
1
)
,
(
+
6
a
b
d
−
2
a
c
c
)
x
+
(
−
3
a
c
d
+
4
b
b
d
−
b
c
c
)
.
.
.
.
(
f
3
)
3
a
(
f
2
)
−
a
c
(
f
1
)
,
(
−
9
a
d
+
b
c
)
x
+
(
−
6
b
d
+
2
c
c
)
.
.
.
.
(
f
4
)
{\displaystyle {\begin{aligned}c(f2)-(-2bd+cc)(f1),\,\,\,\,&(+6abd-2acc)x+(-3acd+4bbd-bcc)\,\,\,\,....\,\,\,\,(f3)\\3a(f2)-ac(f1),\,\,\,\,&(-9ad+bc)x+(-6bd+2cc)\,\,\,\,....\,\,\,\,(f4)\end{aligned}}}
(
+
6
a
b
d
−
2
a
c
c
)
(
f
4
)
−
(
−
9
a
d
+
b
c
)
(
f
3
)
,
−
27
a
a
d
d
+
18
a
b
c
d
−
4
a
c
c
c
−
4
b
b
b
d
+
b
b
c
c
.
.
.
.
(
f
5
)
(
−
6
b
d
+
2
c
c
)
(
f
3
)
−
(
−
3
a
c
d
+
4
b
b
d
−
b
c
c
)
(
f
4
)
,
−
27
a
a
d
d
+
18
a
b
c
d
−
4
a
c
c
c
−
4
b
b
b
d
+
b
b
c
c
.
.
.
.
(
f
6
)
{\displaystyle {\begin{aligned}(+6abd-2acc)(f4)-(-9ad+bc)(f3),\,\,\,\,&-27aadd+18abcd-4accc-4bbbd+bbcc\,\,\,\,....\,\,\,\,(f5)\\(-6bd+2cc)(f3)-(-3acd+4bbd-bcc)(f4),\,\,\,\,&-27aadd+18abcd-4accc-4bbbd+bbcc\,\,\,\,....\,\,\,\,(f6)\end{aligned}}}
Look at (f3) and (f4) again. From (f3) and (f4):
x
=
−
(
−
3
a
c
d
+
4
b
b
d
−
b
c
c
)
(
+
6
a
b
d
−
2
a
c
c
)
=
−
(
−
6
b
d
+
2
c
c
)
(
−
9
a
d
+
b
c
)
−
(
−
6
b
d
+
2
c
c
)
(
+
6
a
b
d
−
2
a
c
c
)
=
−
(
−
3
a
c
d
+
4
b
b
d
−
b
c
c
)
(
−
9
a
d
+
b
c
)
(
−
6
b
d
+
2
c
c
)
(
+
6
a
b
d
−
2
a
c
c
)
−
(
−
3
a
c
d
+
4
b
b
d
−
b
c
c
)
(
−
9
a
d
+
b
c
)
=
0
(
−
6
b
d
+
2
c
c
)
(
+
6
a
b
d
−
2
a
c
c
)
−
(
−
3
a
c
d
+
4
b
b
d
−
b
c
c
)
(
−
9
a
d
+
b
c
)
=
−
27
a
a
c
d
d
+
18
a
b
c
c
d
−
4
a
c
c
c
c
−
4
b
b
b
c
d
+
b
b
c
c
c
=
c
(
−
27
a
a
d
d
+
18
a
b
c
d
−
4
a
c
c
c
−
4
b
b
b
d
+
b
b
c
c
)
{\displaystyle {\begin{aligned}&x={\frac {-(-3acd+4bbd-bcc)}{(+6abd-2acc)}}={\frac {-(-6bd+2cc)}{(-9ad+bc)}}\\\\&-(-6bd+2cc)(+6abd-2acc)=-(-3acd+4bbd-bcc)(-9ad+bc)\\\\&(-6bd+2cc)(+6abd-2acc)-(-3acd+4bbd-bcc)(-9ad+bc)=0\\\\&(-6bd+2cc)(+6abd-2acc)-(-3acd+4bbd-bcc)(-9ad+bc)\\\\&=-27aacdd+18abccd-4acccc-4bbbcd+bbccc\\\\&=c(-27aadd+18abcd-4accc-4bbbd+bbcc)\\\end{aligned}}}
(
−
27
a
a
d
d
+
18
a
b
c
d
−
4
a
c
c
c
−
4
b
b
b
d
+
b
b
c
c
)
{\displaystyle (-27aadd+18abcd-4accc-4bbbd+bbcc)}
= the discriminant of the solution of the cubic equation.
Example 1
(
+
2
)
x
3
+
(
−
3
)
x
2
+
(
−
44
)
x
+
(
+
105
)
.
.
.
.
(
e
1
)
{\displaystyle (+2)x^{3}+(-3)x^{2}+(-44)x+(+105)\ ....\ (e1)}
+
9327
−
2881
x
.
.
.
.
(
f
3
)
+
2881
−
879
x
.
.
.
.
(
f
4
)
{\displaystyle {\begin{aligned}+9327-2881x\ ....\ (f3)\\\\+2881-879x\ ....\ (f4)\end{aligned}}}
Example 2
(
+
4
)
x
3
+
(
−
40
)
x
2
+
(
+
133
)
x
+
(
−
147
)
.
.
.
.
(
e
2
)
{\displaystyle (+4)x^{3}+(-40)x^{2}+(+133)x+(-147)\ ....\ (e2)}
+
7
−
2
x
.
.
.
.
(
f
3
)
+
7
−
2
x
.
.
.
.
(
f
4
)
{\displaystyle {\begin{aligned}+7-2x\ ....\ (f3)\\\\+7-2x\ ....\ (f4)\end{aligned}}}
Example 3
(
+
27
)
x
3
+
(
+
297
)
x
2
+
(
+
1089
)
x
+
(
+
1331
)
.
.
.
.
(
e
3
)
{\displaystyle (+27)x^{3}+(+297)x^{2}+(+1089)x+(+1331)\ ....\ (e3)}
0 .... (f4)
+
121
+
66
x
+
9
x
x
.
.
.
.
(
f
1
)
+
121
+
66
x
+
9
x
x
.
.
.
.
(
f
2
)
{\displaystyle {\begin{aligned}+121+66x+9xx\ ....\ (f1)\\\\+121+66x+9xx\ ....\ (f2)\end{aligned}}}
(
f
1
)
,
x
=
−
66
/
(
2
∗
9
)
=
−
11
/
3
,
{\displaystyle (f1),\ \ x=-66/(2*9)=-11/3,}
Case 3: Solving for a given variable in two equations with two unknowns.
edit
Let two functions in
p
{\displaystyle p}
Q
{\displaystyle Q}
a
p
3
+
3
a
p
Q
+
b
p
2
+
b
Q
+
c
p
+
d
.
.
.
.
(
f
0
)
3
a
p
2
+
a
Q
+
2
b
p
+
c
.
.
.
.
(
f
1
)
{\displaystyle {\begin{aligned}ap^{3}+3apQ+bp^{2}+bQ+cp+d\ ....\ (f0)\\3ap^{2}+aQ+2bp+c\ ....\ (f1)\end{aligned}}}
a
,
b
,
c
,
d
{\displaystyle a,b,c,d}
p
{\displaystyle p}
Q
{\displaystyle Q}
p
{\displaystyle p}
(
+
a
)
p
p
p
+
(
+
b
)
p
p
+
(
+
3
Q
a
+
c
)
p
+
(
+
Q
b
+
d
)
.
.
.
.
(
f
2
)
(
+
3
a
)
p
p
+
(
+
2
b
)
p
+
(
+
Q
a
+
c
)
.
.
.
.
(
f
3
)
{\displaystyle {\begin{aligned}(+a)ppp+(+b)pp+(+3Qa+c)p+(+Qb+d)\ ....\ (f2)\\(+3a)pp+(+2b)p+(+Qa+c)\ ....\ (f3)\end{aligned}}}
(f2) contains maximum power of
p
{\displaystyle p}
p
{\displaystyle p}
Combine (f2) and (f3) to produce a second function containing maximum power of
p
{\displaystyle p}
(
+
Q
a
+
c
)
(
f
2
)
−
(
+
Q
b
+
d
)
(
f
3
)
,
(
+
Q
a
a
+
a
c
)
p
p
+
(
−
2
Q
a
b
−
3
a
d
+
b
c
)
p
+
(
+
3
Q
Q
a
a
+
4
Q
a
c
−
2
Q
b
b
−
2
b
d
+
c
c
)
.
.
.
.
(
f
4
)
{\displaystyle {\begin{aligned}(+Qa+c)(f2)-(+Qb+d)(f3),\\&\\(+Qaa+ac)&pp+\\(-2Qab-3ad+bc)&p+\\(+3QQaa+4Qac-2Qbb-2bd+cc)&\ ....\ (f4)\end{aligned}}}
p
{\displaystyle p}
p
{\displaystyle p}
(
3
a
)
(
f
4
)
−
(
+
Q
a
a
+
a
c
)
(
f
3
)
,
(
−
8
Q
a
b
−
9
a
d
+
b
c
)
p
+
(
+
8
Q
Q
a
a
+
10
Q
a
c
−
6
Q
b
b
−
6
b
d
+
2
c
c
)
.
.
.
.
(
f
5
)
(
+
Q
a
+
c
)
(
f
4
)
−
(
+
3
Q
Q
a
a
+
4
Q
a
c
−
2
Q
b
b
−
2
b
d
+
c
c
)
(
f
3
)
,
(
−
8
Q
Q
a
a
a
−
10
Q
a
a
c
+
6
Q
a
b
b
+
6
a
b
d
−
2
a
c
c
)
p
+
(
−
8
Q
Q
a
a
b
−
3
Q
a
a
d
−
9
Q
a
b
c
+
4
Q
b
b
b
−
3
a
c
d
+
4
b
b
d
−
b
c
c
)
.
.
.
.
(
f
6
)
{\displaystyle {\begin{aligned}(3a)(f4)-(+Qaa+ac)(f3),\\&\\(-8Qab-9ad+bc)&p+\\(+8QQaa+10Qac-6Qbb-6bd+2cc)&\ ....\ (f5)\\\\(+Qa+c)(f4)-(+3QQaa+4Qac-2Qbb-2bd+cc)(f3),\\&\\(-8QQaaa-10Qaac+6Qabb+6abd-2acc)&p+\\(-8QQaab-3Qaad-9Qabc+4Qbbb-3acd+4bbd-bcc)&\ ....\ (f6)\end{aligned}}}
(f5) and (f6) both have maximum power of
p
{\displaystyle p}
p
{\displaystyle p}
c
15
=
−
8
Q
a
b
−
9
a
d
+
b
c
,
{\displaystyle c15=-8Qab-9ad+bc,}
and let
c
05
=
+
8
Q
Q
a
a
+
10
Q
a
c
−
6
Q
b
b
−
6
b
d
+
2
c
c
.
{\displaystyle c05=+8QQaa+10Qac-6Qbb-6bd+2cc.}
c
16
=
−
8
Q
Q
a
a
a
−
10
Q
a
a
c
+
6
Q
a
b
b
+
6
a
b
d
−
2
a
c
c
,
{\displaystyle c16=-8QQaaa-10Qaac+6Qabb+6abd-2acc,}
and let
c
06
=
−
8
Q
Q
a
a
b
−
3
Q
a
a
d
−
9
Q
a
b
c
+
4
Q
b
b
b
−
3
a
c
d
+
4
b
b
d
−
b
c
c
.
{\displaystyle c06=-8QQaab-3Qaad-9Qabc+4Qbbb-3acd+4bbd-bcc.}
(
c
15
)
(
f
6
)
−
(
c
16
)
(
f
5
)
,
+
64
Q
Q
Q
Q
a
a
a
a
a
+
160
Q
Q
Q
a
a
a
a
c
−
32
Q
Q
Q
a
a
a
b
b
+
132
Q
Q
a
a
a
c
c
−
56
Q
Q
a
a
b
b
c
+
4
Q
Q
a
b
b
b
b
+
27
Q
a
a
a
d
d
−
18
Q
a
a
b
c
d
+
40
Q
a
a
c
c
c
+
4
Q
a
b
b
b
d
−
25
Q
a
b
b
c
c
+
4
Q
b
b
b
b
c
+
27
a
a
c
d
d
−
18
a
b
c
c
d
+
4
a
c
c
c
c
+
4
b
b
b
c
d
−
b
b
c
c
c
…
(
f
7
)
(
c
06
)
(
f
5
)
−
(
c
05
)
(
f
6
)
,
+
64
Q
Q
Q
Q
a
a
a
a
a
+
160
Q
Q
Q
a
a
a
a
c
−
32
Q
Q
Q
a
a
a
b
b
+
132
Q
Q
a
a
a
c
c
−
56
Q
Q
a
a
b
b
c
+
4
Q
Q
a
b
b
b
b
+
27
Q
a
a
a
d
d
−
18
Q
a
a
b
c
d
+
40
Q
a
a
c
c
c
+
4
Q
a
b
b
b
d
−
25
Q
a
b
b
c
c
+
4
Q
b
b
b
b
c
+
27
a
a
c
d
d
−
18
a
b
c
c
d
+
4
a
c
c
c
c
+
4
b
b
b
c
d
−
b
b
c
c
c
…
(
f
8
)
{\displaystyle {\begin{aligned}&(c15)(f6)-(c16)(f5),\\\\&+64QQQQaaaaa+160QQQaaaac-32QQQaaabb+132QQaaacc-56QQaabbc+4QQabbbb+27Qaaadd\\&-18Qaabcd+40Qaaccc+4Qabbbd-25Qabbcc+4Qbbbbc+27aacdd-18abccd+4acccc+4bbbcd-bbccc\dots (f7)\\\\&(c06)(f5)-(c05)(f6),\\\\&+64QQQQaaaaa+160QQQaaaac-32QQQaaabb+132QQaaacc-56QQaabbc+4QQabbbb+27Qaaadd\\&-18Qaabcd+40Qaaccc+4Qabbbd-25Qabbcc+4Qbbbbc+27aacdd-18abccd+4acccc+4bbbcd-bbccc\dots (f8)\end{aligned}}}
p
{\displaystyle p}
(
+
64
a
a
a
a
a
)
Q
4
+
(
+
160
a
a
a
a
c
−
32
a
a
a
b
b
)
Q
3
+
(
+
132
a
a
a
c
c
−
56
a
a
b
b
c
+
4
a
b
b
b
b
)
Q
2
+
(
+
27
a
a
a
d
d
−
18
a
a
b
c
d
+
40
a
a
c
c
c
+
4
a
b
b
b
d
−
25
a
b
b
c
c
+
4
b
b
b
b
c
)
Q
+
(
+
27
a
a
c
d
d
−
18
a
b
c
c
d
+
4
a
c
c
c
c
+
4
b
b
b
c
d
−
b
b
c
c
c
)
…
(
f
9
)
{\displaystyle {\begin{aligned}(+64aaaaa)&Q^{4}+\\(+160aaaac-32aaabb)&Q^{3}+\\(+132aaacc-56aabbc+4abbbb)&Q^{2}+\\(+27aaadd-18aabcd+40aaccc+4abbbd-25abbcc+4bbbbc)&Q+\\(+27aacdd-18abccd+4acccc+4bbbcd-bbccc)&\ \dots \ (f9)\end{aligned}}}
The constant term of (f9)
=
+
27
a
a
c
d
d
−
18
a
b
c
c
d
+
4
a
c
c
c
c
+
4
b
b
b
c
d
−
b
b
c
c
c
=
c
(
+
27
a
a
d
d
−
18
a
b
c
d
+
4
a
c
c
c
+
4
b
b
b
d
−
b
b
c
c
)
{\displaystyle {\begin{aligned}&=+27aacdd-18abccd+4acccc+4bbbcd-bbccc\\&=c(+27aadd-18abcd+4accc+4bbbd-bbcc)\end{aligned}}}
(
+
27
a
a
d
d
−
18
a
b
c
d
+
4
a
c
c
c
+
4
b
b
b
d
−
b
b
c
c
)
{\displaystyle (+27aadd-18abcd+4accc+4bbbd-bbcc)}
p
{\displaystyle p}
