# Factoring

See how to factor a polynomial.

20 examples and their solutions.

- N = a⋅b

→ a, b: factors of N - Prime Factorization:

12 = 2^{2}⋅3 - a
^{2}bc = a^{2}⋅b⋅c

ac^{2}= a⋅c^{2}

(GCF) = a⋅c - a
^{2}bc = a^{2}⋅b⋅c

ac^{2}= a⋅c^{2}

(LCM) = a^{2}⋅b⋅c^{2} - ax + ay + az

= a(x + y + z) - ax + ay + bx + by

= a(x + y) + b(x + y)

= (a + b)(x + y) - x
^{2}+ bx + c

m⋅n = +c

m + n = +b

= (x + m)(x + n) - a
^{2}± 2ab + b^{2}

= (a ± b)^{2} - a
^{2}- b^{2}

= (a + b)(a - b) - a
^{3}+ b^{3}

= (a + b)(a^{2}- ab + b^{2}) - a
^{3}- b^{3}

= (a - b)(a^{2}+ ab + b^{2})

## 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

[1] 1 and 30 are natural numbers.

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.

= 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.

[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.

### 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

= 2

Change 60 to a product of two factors: 6⋅10.

Repeat this until only prime numbers remain.

= 2⋅3⋅2⋅5

= 2

^{2}⋅3⋅5Change 60 to a product of two factors: 6⋅10.

Repeat this until only prime numbers remain.

### Example

Prime factorization of 200

Solution 200 = 2⋅100

= 2⋅10⋅10

= 2⋅2⋅5⋅2⋅5

= 2

= 2⋅10⋅10

= 2⋅2⋅5⋅2⋅5

= 2

^{3}⋅5^{2}## Greatest Common Factor

### Example

GCF of 18 and 60

Solution 18 = 3⋅6

= 3⋅2⋅3

= 2⋅3

60 = 6⋅10

= 2⋅3⋅2⋅5

= 2

18 = 2⋅3

60 = 2

(GCF) = 2⋅3- [1]

= 6

[1] Compare the same base powers

and write the less exponent power

in the GCF.

= 3⋅2⋅3

= 2⋅3

^{2}60 = 6⋅10

= 2⋅3⋅2⋅5

= 2

^{2}⋅3⋅518 = 2⋅3

^{2}60 = 2

^{2}⋅3⋅5(GCF) = 2⋅3- [1]

= 6

[1] Compare the same base powers

and write the less exponent power

in the GCF.

### Example

GCF of 6a

Solution ^{3}c and 2a^{2}bc^{2} 6a

2a

(GCF) = 2a

^{3}c = 2⋅3⋅a^{3}⋅c^{2}2a

^{2}bc^{2}= 2⋅a^{2}⋅b⋅c^{2}(GCF) = 2a

^{2}c## Least Common Multiple

### Example

LCM of 18 and 60

Solution 18 = 3⋅6

= 3⋅2⋅3

= 2⋅3

60 = 6⋅10

= 2⋅3⋅2⋅5

= 2

18 = 2⋅3

60 = 2

(LCM) = 2

= 4⋅9⋅5

= 20⋅9

= 180

[1] Compare the same base powers

and write the greater exponent power

in the LCM.

= 3⋅2⋅3

= 2⋅3

^{2}60 = 6⋅10

= 2⋅3⋅2⋅5

= 2

^{2}⋅3⋅518 = 2⋅3

^{2}60 = 2

^{2}⋅3⋅5(LCM) = 2

^{2}⋅3^{2}⋅5- [1]= 4⋅9⋅5

= 20⋅9

= 180

[1] Compare the same base powers

and write the greater exponent power

in the LCM.

### Example

LCM of 6a

Solution ^{3}c and 2a^{2}bc^{2} 6a

2a

(LCM) = 2⋅3⋅a

= 6a

^{3}c = 2⋅3⋅a^{3}⋅c^{2}2a

^{2}bc^{2}= 2⋅a^{2}⋅b⋅c^{2}(LCM) = 2⋅3⋅a

^{3}⋅b⋅c^{2}= 6a

^{3}bc^{2}## Factoring: Common Monomial Factor

### Example

x

Solution ^{3}+ 3x x

= x(x

First write the GCF of x

x

+3x/x = +3

^{3}+ 3x= x(x

^{2}+ 3)First write the GCF of x

^{3}and +3x: x.x

^{3}/x = x^{2}+3x/x = +3

### Example

a

Solution ^{2}- 2ab + 4a a

= a(a - 2b + 4)

First write the GCF of a

a

-2ab/a = -2a

+4a/a = +4

^{2}- 2ab + 4a= a(a - 2b + 4)

First write the GCF of a

^{2}, -2ab, +4a: a.a

^{2}/a = a-2ab/a = -2a

+4a/a = +4

## Factoring by Grouping

### Example

a

Solution ^{2}- 2a + 5a - 10 a

= a(a - 2) + 5(a - 2)- [1]

= (a + 5)(a - 2)

[1] Split the polynomial into two groups,

a

and factor each group.

^{2}- 2a + 5a - 10= a(a - 2) + 5(a - 2)- [1]

= (a + 5)(a - 2)

[1] Split the polynomial into two groups,

a

^{2}- 2a and 5a - 10,and factor each group.

x

Solution ^{2}- 4xy - 3x + 12y x

= x(x - 4y) - 3(x - 4y)

= (x - 3)(x - 4y)

^{2}- 4xy - 3x + 12y= x(x - 4y) - 3(x - 4y)

= (x - 3)(x - 4y)

## Factoring a Quadratic Trinomial

### Example

x

Solution ^{2}+ 3x + 2 x

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.

[2] Use +1 and +2 to write the answer.

^{2}+ 3x + 21⋅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.

[2] Use +1 and +2 to write the answer.

x

Solution ^{2}- 5x + 6 x

-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.

[2] Use -2 and -3 to write the answer.

^{2}- 5x + 6-2⋅(-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.

[2] Use -2 and -3 to write the answer.

x

Solution ^{2}- x - 12 x

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.

[2] Use +3 and -4 to write the answer.

^{2}- x - 123⋅(-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.

[2] Use +3 and -4 to write the answer.

x

Solution ^{2}+ 5xy - 24y^{2} x

-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.

[2] Use -3 and +8 to write the answer.

^{2}+ 5xy - 24y^{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.

[2] Use -3 and +8 to write the answer.

2x

Solution ^{2}+ 7x + 6 2x

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.

[2] Use +3 and +2 to write the answer.

2 should be in the front part.

^{2}+ 7x + 63⋅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.

[2] Use +3 and +2 to write the answer.

2 should be in the front part.

## Factoring a^{2} ± 2ab + b^{2}

### Formula

a

= (a ± b)

^{2}± 2ab + b^{2}= (a ± b)

^{2}### Example

x

Solution ^{2}+ 6x + 9 x

= x

= (x + 3)

^{2}+ 6x + 9= x

^{2}+ 2⋅x⋅3 + 3^{2}= (x + 3)

^{2}### Example

x

Solution ^{2}- 10x + 25 x

= x

= (x - 5)

^{2}- 10x + 25= x

^{2}- 2⋅x⋅5 + 5^{2}= (x - 5)

^{2}## Factoring a^{2} - b^{2}

### Formula

a

= (a + b)(a - b)

^{2}- b^{2}= (a + b)(a - b)

### Example

x

Solution ^{2}- 81 x

= x

= (x + 9)(x - 9)

^{2}- 81= x

^{2}- 9^{2}= (x + 9)(x - 9)

### Example

16a

Solution ^{2}- 49b^{2} 16a

= (4a)

= (4a + 7b)(4a - 7b)

^{2}- 49b^{2}= (4a)

^{2}- (7b)^{2}= (4a + 7b)(4a - 7b)

### Example

x

Solution ^{4}- 1 x

= (x

= (x

= (x

= (x

^{4}- 1= (x

^{2})^{2}- 1^{2}= (x

^{2}+ 1)(x^{2}- 1)= (x

^{2}+ 1)(x^{2}- 1^{2})= (x

^{2}+ 1)(x + 1)(x - 1)## Factoring a^{3} + b^{3}

### Formula

a

= (a + b)(a

^{3}+ b^{3}= (a + b)(a

^{2}- ab + b^{2})### Example

x

Solution ^{3}+ 8 x

= x

= (x + 2)(x

= (x + 2)(x

^{3}+ 8= x

^{3}+ 2^{3}= (x + 2)(x

^{2}- x⋅2 + 2^{2})= (x + 2)(x

^{2}- 2x + 4)## Factoring a^{3} - b^{3}

### Formula

a

= (a - b)(a

^{3}- b^{3}= (a - b)(a

^{2}+ ab + b^{2})### Example

x

Solution ^{3}- 125 x

= x

= (x - 5)(x

= (x - 5)(x

^{3}- 125= x

^{3}- 5^{3}= (x - 5)(x

^{2}+ x⋅5 + 5^{2})= (x - 5)(x

^{2}+ 5x + 25)### Example

x

Solution ^{6}- 1 x

= (x

= (x

= (x

= (x + 1)(x

= (x + 1)(x

^{6}- 1= (x

^{3})^{2}- 1^{2}= (x

^{3}+ 1)(x^{3}- 1)= (x

^{3}+ 1^{3})(x^{3}- 1^{3})= (x + 1)(x

^{2}- x⋅1 + 1^{2})(x - 1)(x^{2}+ x⋅1 + 1^{2})= (x + 1)(x

^{2}- x + 1)(x - 1)(x^{2}+ x + 1)