Greatest Common Factor and Factoring by Grouping

Introduction to Factoring

2∙3 = 6

4∙2 =

8

3∙3∙3∙3 = 81

8∙3∙5 =

120

Introduction to FactoringFactoring is the opposite of multiplication. Previously you were taught to factor integers. Factoring an integer means to write the integer as a product of two or more integers.

6x² + 5x -4 = (3x +4)(2x – 1)

Factoring

multiplying

In the product 3∙5 = 15 , for example, 3 and 5 are factors of 15

In this chapter you will learn how to factor polynomials. To factor a polynomial means to express the polynomial as a product of two or more polynomials.

Greatest Common Factor

Let’s factor integers first:

30 1∙30, 3∙10, 2∙15, or 2∙3∙5

The product of 2∙3∙5 consists only of prime numbers and is called the prime

factorization.

In this first section we will be using the greatest common factor (GCF) . The greatest common factor of two or more integers is the greatest factor common to each integer. It is useful to express the numbers as a product of prime factors

Identifying the GCF of Two Integers

Find the greatest common factor of each pair of integers.

12 and 20

Step 1. Find the GCF of the numerical coefficients.Find the prime factorization of each number.

212

2 63

Factors of 12 = 2 ∙ 2 ∙ 32 20

210

5Factors of 20 = 2 ∙ 2 ∙ 5

Find the product of common factors

The numbers 12 and 20 share two factors of 2. Therefore the greatest common factor is 2 ∙ 2 = 4

Identifying the GCF of Two Variable Factors

Find the greatest common factor of each pair of variables.

ab³ and a²b³

Find the GCF of the variable factors.

Factors of ab³ = a ∙ b ∙ b· bFactors of a²b³ = a· a ∙ b ∙ b· b

Find the product of common factors

The variables ab³ and a²b³ share one factors of a and three factors b. Therefore the greatest

common factor is ab³

Find the GCF

Finding the GCF of a list of Monomials

Step 1: Find the GCF of the numerical coefficients.

Step 2: Find the GCF of the variable factors.

Step 3: The product of the factors found in Steps 1 and 2 is the GCF of the monomials.

Find the GCF of 8ab³, 20a²b³, and 28ab²

The GCF of the numerical coefficients 8, 20, and 28 is 4, the largest integer that is a factor of each integer. The GCF of the variable factors a, and a² is a, because a is the largest factor common to both powers of a. The variable b², and b³ is b², because b² is the largest factor common to both powers of b.

To see this in factored form:

8ab³ = 2·2·2 a b b b

20a²b³ = 2·2·5 a a b b b

28ab² = 2·2·7 a b b

GCF = 4ab²

Terms with Common Factors

Example 1Factor our a common factor:

2 2

2

2 2

6 18 3

3

6( 3

6 6

) 6 18:

6

Check

y y

y

y y

Solution:

26 18y

Noting that 6 is a common factor

Using the distributive law

Suppose in Example 1 That the common factor 2 were used.

2 2

2

6 18 3 9

3 9

2 2

2

y y

y

Solution:

Noting that 2 is a common factor Using the distributive law

If you look at 3y² - 9 it has a common factor, 3. This is the reason you always factor out the largest, or greatest, common factor, the polynomial is completely factored by factoring out a 3.

2 2

2 2

6 18 32 3 2 3

2 3 3 6( 3)

y y

y y

Remember to multiply the two common factors: 2· 3 = 6

Find the greatest common factor of

Solution:

The greatest common factor is

4 530 20x x

4 4

4

4 5 230 20 3 2

(

5

3 )0 2

2 5

1

x x

x

x x x

x

410x

Write an expression equivalent of by factoring out the greatest common factor

Solution:Step 1 Find the greatest positive common factor of the coefficients

Step 2 Find the greatest common factor of the powers of p

6 2 5 3 4 48 4 10p q p q p q

6 2 5 3 4 48 4 10p q p q p q

Greatest common factor =

8, -4, 10 Greatest common factor = 2

4 22 p q

4p

Step 3 Find the greatest common factor of the powers of q

2 3 4, ,q q q

6 5 4, ,p p p

Greatest common factor = 2q

is the greatest common factor of the given polynomial

26 2 5 3 4 4 4 2

2

2 4 2 4 2

4 22

8 4 2 2 2

2

10 4 2 5

4 2 5

p q p q p q

p q

p q p q p q p pq q

p pq q

Factored Completely

When polynomials are factored completely, the factors in the the resulting factorizations are said to be prime polynomials.When the leading coefficient is a negative number, we generally factor out a common factor with a negative coefficient.

There is a 5 in common in both terms that we could factor out. According to the directions though, we are supposed to factor out a negative real number so let's factor out –5.

5( 3 )x Remember you can always check to see if you've done this step correctly by re-distributing through. Let's check it.

1515 5x

Yes---it checks!

5 15x Trade the terms places for standard form.

Solving a binomial

15 5x

There is a 3y³ in common in both terms that we could factor out.

33 (5 4 )y yRemember you can always check to see if you've done this step correctly by re-distributing through. Let's check it.

315y 3 415 12y y Yes---it checks!

4 312 15y yTrade the terms places for standard form.

Solving a binomial

3 415 12y y

yxyyx 1042 2 There is a 2y in common in all three terms that we could factor out. According to the directions though, we are supposed to factor out a negative real number so let's factor out –2. There is also a y common in all three terms.

)52(2 2 xxy

Remember you can always check to see if you've done this step correctly by re-distributing through. Let's check it.

22x y xyx 42 2

Yes---it checks!

yxyyx 1042 2

Practice:Write an equivalent expression by factoring out a common factor with a negative coefficient.

3 2

) 4 24

) 2 6 2

a x

b x x x

Steps to Factoring by Grouping

1. Identify and factor out the GCF from all four terms.

2. Group the first pair of terms and the second pair of terms. Make sure you always connect the terms by addition. Factor out the GCF from the first pair of terms. Factor out the GCF from the second pair of terms. (Sometimes it is necessary to factor out the opposite of the GCF.)

3. If the two terms share a common binomial factor, factor out the binomial factor

Identify and factor out the GCF from all four terms.

When you see four terms, it is a clue to try factoring by grouping. Look at the first two terms first to see what is in common and then look at the second two terms.

3264 34 uuu

32 (2 3)u u

There is nothing in common in the second two terms but we want them to match what we have in parentheses in the first two terms so we'll factor out -1

321 3 uThese match so let's factor them out

1232 3 uu

Group the first pair of terms and the second pair of terms. Make sure you always connect the terms by addition

4 3(4 6 ) ( 2 3)u u u

When you see four terms, it is a clue to try factoring by grouping. Look at the first two terms first to see what is in common and then look at the second two terms.

4 3 216 40 12 30w w w w

22 [4 (2 5)w w w

There is a 3 in common in the second two terms but we want them to match what we have in parentheses in the first two terms so we'll factor out -3

3(2 5)]w These match so let's factor them out

22 [(2 5)(4 3)]w w w af

Identify and factor out the GCF from all four terms. In this case, the GCF is 2w

3 22 (8 20 6 15)w w w w 3 22 [(8 20 ) ( 6 15)]w w w w

Group the first pair of terms and the second pair of terms. Make sure you always connect the terms by addition

Factoring Trinomials: Grouping Method

Multiply

(2 3)( 2)x x

Multiply the binomials.

22 4 3 6x x x 22 7 6x x

Add the middle terms.

Factor

22 7 6x x

Rewrite the middle terms as a sum or difference of terms

22 4 3 6x x x

Factor by grouping

(2 3)( 2)x x

Grouping Method Factorax² + bx + c (a ≠ 0)

1. Multiply the coefficients of the first and last terms(ac).

2. Find two integers whose product is ac and whose sum is b. (if no pair of integers can be found, then the trinomial cannot be factored further and is a prime polynomial.)

3. Rewrite the middle terms bx as the sum of two terms whose coefficients are the integers found in Step 2.

4. Factor by grouping.

22 7 6x x Factor out the GCF from all the terms. In this case, the GCF is 1

Step 1. The trinomial is written in the form ax² +bx + c.Find the product ac = (2)(6) = 12

TIP: a = 2, b = 7, c = 3Step 2. List all the factors

of ac and search for the pair whose sum equals the value of b. That is, list the factors of 12 and find the pair whose sum equals 7.

12 12 1∙12 (-1)(-12)2∙6 (2)(6)3∙4 (3)(4)

The numbers 3 and 4 satisfy both conditions: 3 ∙ 4 = 12 3 + 4 = 7

22 7 6x x Step 3. Write the middle term of the trinomial as the sum of two terms whose coefficients are the selected pair of numbers: 3 and 4 2 32 64x xx

Step 4. Factor by grouping.

2(2 3 ) (4 6)x x x (2 3) 2(2 3)x x x

( 2)x (2 3)x

22 8 3x x First rewrite the polynomial in the form ax² + bx +c

Step 1. The trinomial is written in the form ax² +bx + c.Find the product ac = (8)(-3)

TIP: a = 8, b = -2, c = -3Step 2. List all the factors

of ac and search for the pair whose sum equals the value of b. That is, list the factors of -24 and find the pair whose sum equals -2.

-24 -24 -1∙24 (-8)(3)-2∙12 (-12)(2)-4 ∙6 (-6)(4)-3∙8 (-24)(1)

The numbers -6 and 4 satisfy both conditions: -6∙ 4 = -24 -6 + 4 = -2

The GCF is 1

28 2 3x x

(2 1)x

28 2 3x x Step 3. Write the middle term of the trinomial as the sum of two terms whose coefficients are the selected pair of numbers: -6 and 4 2 68 34x xx

Step 4. Factor by grouping.

2(8 6 ) (4 3)x x x 2 (4 3) 1(4 3)x x x

(4 3)x