Factoring (Math)
See how to factor a polynomial.
20 examples and their solutions.
How to Find the Factors of a Number
Example
Factors of 30
Solution 30 = 1⋅30- [1]
= 2⋅15- [2]
= 3⋅10
= 4⋅7.5- [3]
= 5⋅6
(factors) = 1, 2, 3, 5, 6, 10, 15, 30
= 2⋅15- [2]
= 3⋅10
= 4⋅7.5- [3]
= 5⋅6
(factors) = 1, 2, 3, 5, 6, 10, 15, 30
[1]
1 and 30 are natural numbers.
So 1 and 30 are the factors of 30.
So 1 and 30 are the factors of 30.
[2]
Likewise, 2 and 15 are the factors of 30.
[3]
7.5 is not a natural number.
So 4 and 7.5 are not the factors of 30.
So 4 and 7.5 are not the factors of 30.
Example
Factors of 16
Solution 16 = 1⋅16
= 2⋅8
= 3⋅163
= 4⋅4
(factors) = 1, 2, 4, 8, 16
= 2⋅8
= 3⋅163
= 4⋅4
(factors) = 1, 2, 4, 8, 16
Prime Number
2, 3, 5, 7, 11, 13, 17, ...
A prime number is a number
that has only two factors:
1 and itself.
2, 3, 5, 7, 11, 13, 17, ...
are prime numbers.
1 has only one factor: 1.
So 1 is not a prime.
Prime Factorization
Example
Prime factorization of 60
Solution 60 = 6⋅10
= 2⋅3⋅2⋅5
= 22⋅3⋅5
= 2⋅3⋅2⋅5
= 22⋅3⋅5
Change 60 to a product of two factors: 6⋅10.
Repeat this until only prime numbers remain.
Repeat this until only prime numbers remain.
Example
Prime factorization of 200
Solution 200 = 2⋅100
= 2⋅10⋅10
= 2⋅2⋅5⋅2⋅5
= 23⋅52
= 2⋅10⋅10
= 2⋅2⋅5⋅2⋅5
= 23⋅52
Greatest Common Factor
Example
GCF of 18 and 60
Solution 18 = 3⋅6
= 3⋅2⋅3
= 2⋅32
60 = 6⋅10
= 2⋅3⋅2⋅5
= 22⋅3⋅5
18 = 2⋅32
60 = 22⋅3⋅5
(GCF) = 2⋅3- [1]
= 6
= 3⋅2⋅3
= 2⋅32
60 = 6⋅10
= 2⋅3⋅2⋅5
= 22⋅3⋅5
18 = 2⋅32
60 = 22⋅3⋅5
(GCF) = 2⋅3- [1]
= 6
[1]
Compare the same base powers
and write the less exponent power
in the GCF.
and write the less exponent power
in the GCF.
Example
GCF of 6a3c and 2a2bc2
Solution 6a3c = 2⋅3⋅a3⋅c2
2a2bc2 = 2⋅a2⋅b⋅c2
(GCF) = 2a2c
2a2bc2 = 2⋅a2⋅b⋅c2
(GCF) = 2a2c
Least Common Multiple
Example
LCM of 18 and 60
Solution 18 = 3⋅6
= 3⋅2⋅3
= 2⋅32
60 = 6⋅10
= 2⋅3⋅2⋅5
= 22⋅3⋅5
18 = 2⋅32
60 = 22⋅3⋅5
(LCM) = 22⋅32⋅5- [1]
= 4⋅9⋅5
= 20⋅9
= 180
= 3⋅2⋅3
= 2⋅32
60 = 6⋅10
= 2⋅3⋅2⋅5
= 22⋅3⋅5
18 = 2⋅32
60 = 22⋅3⋅5
(LCM) = 22⋅32⋅5- [1]
= 4⋅9⋅5
= 20⋅9
= 180
[1]
Compare the same base powers
and write the greater exponent power
in the LCM.
and write the greater exponent power
in the LCM.
Example
LCM of 6a3c and 2a2bc2
Solution 6a3c = 2⋅3⋅a3⋅c2
2a2bc2 = 2⋅a2⋅b⋅c2
(LCM) = 2⋅3⋅a3⋅b⋅c2
= 6a3bc2
2a2bc2 = 2⋅a2⋅b⋅c2
(LCM) = 2⋅3⋅a3⋅b⋅c2
= 6a3bc2
Factoring: Common Monomial Factor
Example
x3 + 3x
Solution x3 + 3x
= x(x2 + 3)
= x(x2 + 3)
First write the GCF of x3 and +3x: x.
x3/x = x2
+3x/x = +3
x3/x = x2
+3x/x = +3
Example
a2 - 2ab + 4a
Solution a2 - 2ab + 4a
= a(a - 2b + 4)
= a(a - 2b + 4)
First write the GCF of a2, -2ab, +4a: a.
a2/a = a
-2ab/a = -2a
+4a/a = +4
a2/a = a
-2ab/a = -2a
+4a/a = +4
Factoring by Grouping
Example
a2 - 2a + 5a - 10
Solution a2 - 2a + 5a - 10
= a(a - 2) + 5(a - 2)- [1]
= (a + 5)(a - 2)
= a(a - 2) + 5(a - 2)- [1]
= (a + 5)(a - 2)
[1]
Split the polynomial into two groups,
a2 - 2a and 5a - 10,
and factor each group.
a2 - 2a and 5a - 10,
and factor each group.
x2 - 4xy - 3x + 12y
Solution x2 - 4xy - 3x + 12y
= x(x - 4y) - 3(x - 4y)
= (x - 3)(x - 4y)
= x(x - 4y) - 3(x - 4y)
= (x - 3)(x - 4y)
Factoring a Quadratic Trinomial
Example
x2 + 3x + 2
Solution x2 + 3x + 2
1⋅2 = +2
1 + 2 = +3- [1]
= (x + 1)(x + 2)- [2]
1⋅2 = +2
1 + 2 = +3- [1]
= (x + 1)(x + 2)- [2]
[1]
Pick a pair of numbers
whose product is +2
and whose sum is +3.
1, 2 satisfy these conditions.
whose product is +2
and whose sum is +3.
1, 2 satisfy these conditions.
[2]
Use +1 and +2 to write the answer.
x2 - 5x + 6
Solution x2 - 5x + 6
-2⋅(-3) = +6
-2 + (-3) = -5- [1]
= (x - 2)(x - 3)- [2]
-2⋅(-3) = +6
-2 + (-3) = -5- [1]
= (x - 2)(x - 3)- [2]
[1]
Pick a pair of numbers
whose product is +6
and whose sum is -5.
-2, -3 satisfy these conditions.
whose product is +6
and whose sum is -5.
-2, -3 satisfy these conditions.
[2]
Use -2 and -3 to write the answer.
x2 - x - 12
Solution x2 - x - 12
3⋅(-4) = -12
3 + (-4) = -1- [1]
= (x + 3)(x - 4)- [2]
3⋅(-4) = -12
3 + (-4) = -1- [1]
= (x + 3)(x - 4)- [2]
[1]
Pick a pair of numbers
whose product is -12
and whose sum is -1.
3, -4 satisfy these conditions.
whose product is -12
and whose sum is -1.
3, -4 satisfy these conditions.
[2]
Use +3 and -4 to write the answer.
x2 + 5xy - 24y2
Solution x2 + 5xy - 24y2
-3⋅8 = -24
-3 + 8 = +5- [1]
= (x - 3)(x + 8)- [2]
-3⋅8 = -24
-3 + 8 = +5- [1]
= (x - 3)(x + 8)- [2]
[1]
Pick a pair of numbers
whose product is -24
and whose sum is +5.
-3, 8 satisfy these conditions.
whose product is -24
and whose sum is +5.
-3, 8 satisfy these conditions.
[2]
Use -3 and +8 to write the answer.
2x2 + 7x + 6
Solution 2x2 + 7x + 6
3⋅2 = +6
3 + 2⋅2 = +7
= (2x + 3)(x + 2)
3⋅2 = +6
3 + 2⋅2 = +7
= (2x + 3)(x + 2)
[1]
Pick a pair of numbers
whose product is +6
and whose sum satisfies
[number 1] + 2⋅[number 2] = +7.
3, 2 satisfy these conditions.
whose product is +6
and whose sum satisfies
[number 1] + 2⋅[number 2] = +7.
3, 2 satisfy these conditions.
[2]
Use +3 and +2 to write the answer.
2 should be in the front part.
2 should be in the front part.
Factoring a2 ± 2ab + b2
Formula
a2 ± 2ab + b2
= (a ± b)2
= (a ± b)2
Example
x2 + 6x + 9
Solution x2 + 6x + 9
= x2 + 2⋅x⋅3 + 32
= (x + 3)2
= x2 + 2⋅x⋅3 + 32
= (x + 3)2
Example
x2 - 10x + 25
Solution x2 - 10x + 25
= x2 - 2⋅x⋅5 + 52
= (x - 5)2
= x2 - 2⋅x⋅5 + 52
= (x - 5)2
Factoring a2 - b2
Formula
a2 - b2
= (a + b)(a - b)
= (a + b)(a - b)
Example
x2 - 81
Solution x2 - 81
= x2 - 92
= (x + 9)(x - 9)
= x2 - 92
= (x + 9)(x - 9)
Example
16a2 - 49b2
Solution 16a2 - 49b2
= (4a)2 - (7b)2
= (4a + 7b)(4a - 7b)
= (4a)2 - (7b)2
= (4a + 7b)(4a - 7b)
Example
x4 - 1
Solution x4 - 1
= (x2)2 - 12
= (x2 + 1)(x2 - 1)
= (x2 + 1)(x2 - 12)
= (x2 + 1)(x + 1)(x - 1)
= (x2)2 - 12
= (x2 + 1)(x2 - 1)
= (x2 + 1)(x2 - 12)
= (x2 + 1)(x + 1)(x - 1)
Factoring a3 + b3
Formula
a3 + b3
= (a + b)(a2 - ab + b2)
= (a + b)(a2 - ab + b2)
Example
x3 + 8
Solution x3 + 8
= x3 + 23
= (x + 2)(x2 - x⋅2 + 22)
= (x + 2)(x2 - 2x + 4)
= x3 + 23
= (x + 2)(x2 - x⋅2 + 22)
= (x + 2)(x2 - 2x + 4)
Factoring a3 - b3
Formula
a3 - b3
= (a - b)(a2 + ab + b2)
= (a - b)(a2 + ab + b2)
Example
x3 - 125
Solution x3 - 125
= x3 - 53
= (x - 5)(x2 + x⋅5 + 52)
= (x - 5)(x2 + 5x + 25)
= x3 - 53
= (x - 5)(x2 + x⋅5 + 52)
= (x - 5)(x2 + 5x + 25)
Example
x6 - 1
Solution x6 - 1
= (x3)2 - 12
= (x3 + 1)(x3 - 1)
= (x3 + 13)(x3 - 13)
= (x + 1)(x2 - x⋅1 + 12)(x - 1)(x2 + x⋅1 + 12)
= (x + 1)(x2 - x + 1)(x - 1)(x2 + x + 1)
= (x3)2 - 12
= (x3 + 1)(x3 - 1)
= (x3 + 13)(x3 - 13)
= (x + 1)(x2 - x⋅1 + 12)(x - 1)(x2 + x⋅1 + 12)
= (x + 1)(x2 - x + 1)(x - 1)(x2 + x + 1)