Lesson Objectives

- Demonstrate an understanding of how to factor out the GCF from a polynomial
- Learn how to factor by grouping

## How to Factor by Grouping

In the last lesson, we learned how to factor out the GCF from a polynomial. In this lesson, we will learn how to factor a four-term polynomial using a process called "factoring by grouping".

Example 1: Factor each.

54x

Step 1) What is the GCF of all terms?

GCF(54x

Step 2) Arrange the terms into two groups of two terms each, such that each group has a common factor:

(54x

Step 3) Factor out the GCF or -GCF from each group:

9x

We will factor out the common binomial factor (6x - 5):

(9x

Example 2: Factor each.

30x

Step 1) What is the GCF of all terms?

GCF(30x

Since the GCF is not 1, factor out the GCF:

3[10x

Step 2) Arrange the terms into two groups of two terms each, such that each group has a common factor:

3[(10x

Step 3) Factor out the GCF or -GCF from each group:

3[5x

We will factor out the common binomial factor (2x - 7):

3(5x

Example 3: Factor each.

20xy + 40 + 16x + 50y

Step 1) What is the GCF of all terms?

GCF(20xy, 40, 16x, 50y) = 2

Since the GCF is not 1, factor out the GCF:

2[10xy + 20 + 8x + 25y]

Step 2) Arrange the terms into two groups of two terms each, such that each group has a common factor:

2[10xy + 20 + 8x + 25y]

Since 8x and 25y don't have a common factor other than 1, let's rearrange terms:

2[(10xy + 25y) + (8x + 20)]

Step 3) Factor out the GCF or -GCF from each group:

2[5y(2x + 5) + 4(2x + 5)]

We will factor out the common binomial factor (2x + 5):

2[(5y + 4)(2x + 5)]

Example 4: Factor each.

30xy + 6 - 10x - 18y

Step 1) What is the GCF of all terms?

GCF(30xy, 6, 10x, 18y) = 2

Since the GCF is not 1, factor out the GCF:

2[15xy + 3 - 5x - 9y]

Step 2) Arrange the terms into two groups of two terms each, such that each group has a common factor:

2[15xy + 3 - 5x - 9y]

Since 5x and 9y don't have a common factor other than 1, let's rearrange terms:

2[(15xy - 5x) + (3 - 9y)]

Step 3) Factor out the GCF or -GCF from each group:

2[5x(3y - 1) + 3(1 - 3y)]

Notice how we have opposites:

(3y - 1) and (1 - 3y) are opposites.

If we factor out a -1 from either, we will have a common binomial factor (3y - 1):

2[5x(3y - 1) + (-3)(-1 + 3y)]

2[5x(3y - 1) - 3(3y - 1)]

We will factor out the common binomial factor (3y - 1):

2(5x - 3)(3y - 1)

Example 5: Factor each.

5xy + 12 + 15x + 4y

Step 1) What is the GCF of all terms?

GCF(5xy, 12, 15x, 4y) = 1

Step 2) Arrange the terms into two groups of two terms each, such that each group has a common factor:

5xy + 12 + 15x + 4y

Since 5xy and 12 don't have a common factor other than 1, let's rearrange terms:

(5xy + 15x) + (12 + 4y)

Step 3) Factor out the GCF or -GCF from each group:

5x(y + 3) + 4(3 + y)

5x(y + 3) + 4(y + 3)

We will factor out the common binomial factor (y + 3):

(5x + 4)(y + 3)

### Factoring a Four-Term Polynomial by Grouping

- Look for the GCF of all terms. When the GCF is not 1, factor out the GCF
- Arrange the terms into two groups of two terms each, such that each group has a common factor
- In some cases, the common factor will be 1 or -1

- When we factor the GCF or -GCF out from each group, we should be left with a common binomial factor
- When we succeed and obtain a common binomial factor, we factor out the common binomial factor
- When a common binomial factor is not produced, we need to try a different grouping

Example 1: Factor each.

54x

^{3}- 45x^{2}+ 60x - 50Step 1) What is the GCF of all terms?

GCF(54x

^{3}, 45x^{2}, 60x, 50) = 1Step 2) Arrange the terms into two groups of two terms each, such that each group has a common factor:

(54x

^{3}- 45x^{2}) + (60x - 50)Step 3) Factor out the GCF or -GCF from each group:

9x

^{2}(6x - 5) + 10(6x - 5)We will factor out the common binomial factor (6x - 5):

(9x

^{2}+ 10)(6x - 5)Example 2: Factor each.

30x

^{3}- 105x^{2}+ 24x - 84Step 1) What is the GCF of all terms?

GCF(30x

^{3}, 105x^{2}, 24x, 84) = 3Since the GCF is not 1, factor out the GCF:

3[10x

^{3}- 35x^{2}+ 8x - 28]Step 2) Arrange the terms into two groups of two terms each, such that each group has a common factor:

3[(10x

^{3}- 35x^{2}) + (8x - 28)]Step 3) Factor out the GCF or -GCF from each group:

3[5x

^{2}(2x - 7) + 4(2x - 7)]We will factor out the common binomial factor (2x - 7):

3(5x

^{2}+ 4)(2x - 7)Example 3: Factor each.

20xy + 40 + 16x + 50y

Step 1) What is the GCF of all terms?

GCF(20xy, 40, 16x, 50y) = 2

Since the GCF is not 1, factor out the GCF:

2[10xy + 20 + 8x + 25y]

Step 2) Arrange the terms into two groups of two terms each, such that each group has a common factor:

2[10xy + 20 + 8x + 25y]

Since 8x and 25y don't have a common factor other than 1, let's rearrange terms:

2[(10xy + 25y) + (8x + 20)]

Step 3) Factor out the GCF or -GCF from each group:

2[5y(2x + 5) + 4(2x + 5)]

We will factor out the common binomial factor (2x + 5):

2[(5y + 4)(2x + 5)]

Example 4: Factor each.

30xy + 6 - 10x - 18y

Step 1) What is the GCF of all terms?

GCF(30xy, 6, 10x, 18y) = 2

Since the GCF is not 1, factor out the GCF:

2[15xy + 3 - 5x - 9y]

Step 2) Arrange the terms into two groups of two terms each, such that each group has a common factor:

2[15xy + 3 - 5x - 9y]

Since 5x and 9y don't have a common factor other than 1, let's rearrange terms:

2[(15xy - 5x) + (3 - 9y)]

Step 3) Factor out the GCF or -GCF from each group:

2[5x(3y - 1) + 3(1 - 3y)]

Notice how we have opposites:

(3y - 1) and (1 - 3y) are opposites.

If we factor out a -1 from either, we will have a common binomial factor (3y - 1):

2[5x(3y - 1) + (-3)(-1 + 3y)]

2[5x(3y - 1) - 3(3y - 1)]

We will factor out the common binomial factor (3y - 1):

2(5x - 3)(3y - 1)

Example 5: Factor each.

5xy + 12 + 15x + 4y

Step 1) What is the GCF of all terms?

GCF(5xy, 12, 15x, 4y) = 1

Step 2) Arrange the terms into two groups of two terms each, such that each group has a common factor:

5xy + 12 + 15x + 4y

Since 5xy and 12 don't have a common factor other than 1, let's rearrange terms:

(5xy + 15x) + (12 + 4y)

Step 3) Factor out the GCF or -GCF from each group:

5x(y + 3) + 4(3 + y)

5x(y + 3) + 4(y + 3)

We will factor out the common binomial factor (y + 3):

(5x + 4)(y + 3)

#### Skills Check:

Example #1

Factor each completely. $$42xy + 5 - 7x - 30y$$

Please choose the best answer.

A

$$(7x - 5)(6y - 1)$$

B

$$(7x - 1)(6y + 5)$$

C

$$(7x + 5)(7x + 1)$$

D

$$(6x - 1)(3x + 2)$$

E

$$(7x + 5)(6y + 1)$$

Example #2

Factor each completely. $$3az - 80xc - 24ac + 10xz$$

Please choose the best answer.

A

$$(z + 10x)(3a + 8c)$$

B

$$(z + 10x)(3a - 8c)$$

C

$$(3a + 10x)(z - 8c)$$

D

$$(3a - 10x)(z + 10x)$$

E

$$2(x - 10z)(2a - c)$$

Example #3

Factor each completely. $$30uv - 18 - 6u + 90v$$

Please choose the best answer.

A

$$6(u - 1)(5v - 3)$$

B

$$6(u - 1)(5v + 3)$$

C

$$(u - 1)(5v - 3)$$

D

$$2(u + 2)(3v - 1)$$

E

$$6(u + 3)(5v - 1)$$

Congrats, Your Score is 100%

Better Luck Next Time, Your Score is %

Try again?

Ready for more?

Watch the Step by Step Video Lesson Take the Practice Test