Factoring by Grouping Lesson

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".

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

Let's look at a few examples.
Example 1: Factor each.
54x³ - 45x² + 60x - 50
Step 1) What is the GCF of all terms?
GCF(54x³, 45x², 60x, 50) = 1
Step 2) Arrange the terms into two groups of two terms each, such that each group has a common factor:
(54x³ - 45x²) + (60x - 50)
Step 3) Factor out the GCF or -GCF from each group:
9x²(6x - 5) + 10(6x - 5)
We will factor out the common binomial factor (6x - 5):
(9x² + 10)(6x - 5)
Example 2: Factor each.
30x³ - 105x² + 24x - 84
Step 1) What is the GCF of all terms?
GCF(30x³, 105x², 24x, 84) = 3
Since the GCF is not 1, factor out the GCF:
3[10x³ - 35x² + 8x - 28]
Step 2) Arrange the terms into two groups of two terms each, such that each group has a common factor:
3[(10x³ - 35x²) + (8x - 28)]
Step 3) Factor out the GCF or -GCF from each group:
3[5x²(2x - 7) + 4(2x - 7)]
We will factor out the common binomial factor (2x - 7):
3(5x² + 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)

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