# Factor By Grouping - Explained in Easy Steps

Are you up for some great examples of factoring by grouping? Take a look at this easy tutorial which will show you how to solve certain mathematical problems using this method. So, let's put on our mathematics caps, shall we?

ScienceStruck Staff

Last Updated: Jun 12, 2018

Factoring, in mathematics, refers to simplifying or reducing a number, a polynomial, etc. to a result that cannot be reduced or simplified any further. Also known as factorization, this method is mostly used in simplifying polynomials.

For example, factoring the number 36 would give us a simplified result 2 x 2 x 3 x 3 (each of which is a prime number). Another simple example may include a polynomial 2x² - 2. Factoring it, will give you an irreducible polynomial 2x (x - 1).

Factoring by grouping is usually used in simplifying polynomials. Here, terms with common factors are grouped before they are factored. To help you understand the concept better, we have outlined some examples of factoring by grouping.

Factoring Polynomials by Grouping

Example #1

Factorize the polynomial: 5x³ - 10x² + 3x - 6

Always look for a GCF (Greatest Common Factor) in the polynomial.

In the given expression, 5x³ - 10x² + 3x - 6, there is none.

Group the first two terms and last two terms together [this method is called factor by grouping two and two], which will give you:

= (5x³ - 10x²) + (3x - 6)

**Step 1**Always look for a GCF (Greatest Common Factor) in the polynomial.

In the given expression, 5x³ - 10x² + 3x - 6, there is none.

**Step 2**Group the first two terms and last two terms together [this method is called factor by grouping two and two], which will give you:

= (5x³ - 10x²) + (3x - 6)

**Step 3**

Factor out the GCF from each of the two binomials. You will get:

= 5x² (x - 2) + 3 (x - 2)

**Step 4**

Factor out the common or shared binomial in this expression. This will give you the final factors:

= (x - 2) (5x² + 3)

Example #2

Factorize the polynomial: 6x³ - 18x² + 3x - 9

This expression has no GCF.

Grouping the first two terms and the last two terms gives:

= (6x³ - 18x²) + (3x - 9)

**Step 1**This expression has no GCF.

**Step 2**Grouping the first two terms and the last two terms gives:

= (6x³ - 18x²) + (3x - 9)

**Step 3**

Factoring out the GCF from each of the binomials gives:

= 6x² (x - 3) + 3 (x - 3)

**Step 4**

Factoring out the common term in the expression gives the final factors:

= (x - 3) (6x² + 3)

Example #3

Factorize the polynomial: p³ - 2p² - p + 2

This expression has no GCF.

Group the first two and the last two terms.

= (p³ - 2p²) - (p - 2)

**Step 1**This expression has no GCF.

**Step 2**Group the first two and the last two terms.

= (p³ - 2p²) - (p - 2)

[

**Note the change**: p + 2 becomes p - 2, because the latter is introduced inside the brackets. This to ensure that we do not change the original sign of 2 which is '+'. In other words, when you open the brackets, '-' '-' will give you '+'.]**Step 3**

Factor out the GCF from each of the binomials.

= p² (p - 2) - 1 (p - 2)

**Step 4**

What's the common factor in the expression? It is p - 2. So, you get:

= (p - 2) (p² - 1)

**Step 5**

Can you simplify the expressions any further? Yes, you can!

= (p - 2) (p + 1) (p - 1) [a² - b² = (a + b) (a - b)

Example #4

Factorize the polynomial: 16x² + 24xy + 9y² - 36z²

No GCF in the expression.

Using the two and two grouping method, we get:

= (16x² + 24xy) + (9y² - 36z²)

Factoring out the GCF from each of the term, we get:

= 4x (4x + 6y) + 3(3y² - 12z²)

**Step 1**No GCF in the expression.

**Step 2**Using the two and two grouping method, we get:

= (16x² + 24xy) + (9y² - 36z²)

**Step 3**Factoring out the GCF from each of the term, we get:

= 4x (4x + 6y) + 3(3y² - 12z²)

But, we cannot go any further than this. So, let's go back to

Group the first three terms together, leaving out the last.

= (16x² + 24xy + 9y²) - 36z²

**Step 2**, and let's try grouping the terms in a different manner.**Step 2**Group the first three terms together, leaving out the last.

= (16x² + 24xy + 9y²) - 36z²

**Step 3**

First, let's factorize the first term. Use the formula a² + 2ab + b².

= {(4x)² + 24xy + (3y)²} - 36z²

= {(4x)² + 2 (4x) (3y) + (3y)²} - 36z² [product of 2, 4x and 3y gives 24xy ]

= (4x + 3y)² - 36z²

= (4x + 3y)² - (6z)²

**Step 4**

Using the formula a² - b² = (a + b) (a - b), we get the final factors as:

= (4x + 3y + 6z) ( 4x + 3y - 6z)

Example #5

Factorize the polynomial: 4x² + 13x + 10

The given trinomial does not have a GCF.

**Note:**The given term is a polynomial of the form ax² + bx + c ['c' being a constant]. Here, we will use the method of grouping, but at a later stage of the factorization.**Step 1**The given trinomial does not have a GCF.

**Step 2**

Here, we will simplify the expression using the split method. Here's how it goes:

• First, find the product of a and c (a . c).

• Find the factors of (a . c) which when added, should give the center term 'b'.

• Put the factors with their appropriate signs (+ or -) in place of the center term.

• Finally, follow the method of grouping of common factors.

So, using the split method on 4x² + 13x + 10, we derive:

= 4x² + 8x + 5x + 10 [The sum of 8 and 5 is 13 (which is 'b'), and product is 40 (which is a.c)]

Now, group the first two and last two terms together. You will get:

= (4x² + 8x) + (5x + 10)

= 4x² + 8x + 5x + 10 [The sum of 8 and 5 is 13 (which is 'b'), and product is 40 (which is a.c)]

**Step 3**Now, group the first two and last two terms together. You will get:

= (4x² + 8x) + (5x + 10)

**Step 4**

Factor out the GCF from each of the binomials.

= 4x (x + 2) +5 (x + 2)

**Step 5**

Factor out the common term in the expression to get the final factors.

= (x + 2) (4x + 5)

Example #6

Factorize the polynomial: 4x²y + 20xy + 16y

This trinomial has a GCF, which is 4y. So, factoring out the GCF, we get:

= 4y (x² + 5x + 4)

Keep (4y) aside for now, and simplify the expression (x² + 5x + 4), using the split method, which will give you:

= x² + 4x + 1x + 4 [The sum of 4 and 1 is 5, and their product is 4]

**Step 1**This trinomial has a GCF, which is 4y. So, factoring out the GCF, we get:

= 4y (x² + 5x + 4)

**Step 2**Keep (4y) aside for now, and simplify the expression (x² + 5x + 4), using the split method, which will give you:

= x² + 4x + 1x + 4 [The sum of 4 and 1 is 5, and their product is 4]

**Step 3**

Grouping the first two and last two terms, we get:

= (x² + 4x) + (1x + 4)

**Step 4**

After factoring out the GCF from each of the terms, we get:

= x (x + 4) +1 (x + 4)

**Step 5**

Factoring out the common term, we get:

= (x + 4) (x + 1)

Therefore, the final factors are 4y (x + 4) (x + 1)

Some More Examples

**#1**. Factorize 4a² + ab - 16a - 4b

**Solution:**

= 4a² + ab - 16a - 4b

= (4a² + ab) - (16a + 4b) [Note the change in the sign of 4b]

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

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

**#2**. Factorize x³ - 3x² + 6x - 18

**Solution:**

= x³ - 3x² + 6x - 18

= (x³ - 3x²) + (6x - 18)

= x² (x - 3) + 6 (x - 3)

= (x - 3) (x² + 6)

**#3**. Factorize 16x²y + 32xy - 20y

**Solution:**

= 16x²y + 32xy - 20y

= 4y (4x² + 8x - 5)

= 4y (4x² + 10x - 2x - 5)

= 4y {(4x² + 10x) - (2x + 5)}

= 4y { 2x (2x + 5) - 1 (2x + 5)}

= 4y (2x + 5) (2x - 1)

**#4**. Factorize 16x² - 2x - 3

**Solution:**

= 16x² - 2x - 3

= 16x² - 8x + 6x - 3

= (16x² - 8x) + (6x - 3)

= 8x (2x - 1) + 3 (2x - 1)

= (8x + 3) (2x - 1)

**#5**. Factorize 2x³ - 8² - 9x + 36

**Solution:**

= 2x³ - 8x² - 9x + 36

= (2x³ - 8x²) - (9x - 36)

= 2x² (x - 4) - 9 (x - 4)

= (x - 4) (2x² - 9)

**#6**. Factorize x³ - 3x² + 10x - 30

**Solution:**

= x³ - 3x² + 10x - 30

= (x³ - 3x²) + (10x - 30)

= x² (x - 3) + 10 (x - 3)

= (x - 3) (x² + 10)

**#7**. Factorize 2a²b + 30ab + 112b

**Solution:**

= 2a²b + 30ab + 112b

= 2b (a² + 15a + 56)

= 2b (a² + 8a + 7a + 56)

= 2b {(a² + 8a) + (7a + 56)}

= 2b {a (a + 8) + 7 (a + 8)}

= 2b (a + 8) (a + 7)

***To check whether you have the right answer, multiply the final factors, and you should get the original polynomial.**Books are the best source of information on subjects such as factor by grouping. Regular practice with different polynomials will enhance your skills, and help you master this mathematical art of simplification. Always remember, in a subject like mathematics, "loads" of practice works like a charm!