Finding factors for an algebraic expression is simple when it consists of only a few terms. But when it comes to an expression that has more than two or three terms, students feel difficult to solve those problems. Students need a better process to solve algebraic expression factorization. Every student who is looking for the best method to solve algebraic expression **factorization** can follow the grouping method.

Factoring Terms by Grouping is the easy and best method to solve different expressions easily. Also, the process of Factoring by Grouping The Terms is very simple compared to other methods.

## Procedure for Factoring Algebraic Expressions by Grouping

Follow the below steps to find the factorization of a given expression using the below steps.

(i) Take out a factor from each group from the groups of the given expression.

(ii) Factorize each group

(iii) Lastly, take out the common factor.

### Factoring Terms by Grouping Examples

1. Factoring of algebraic expression

(i) 2ma + mb + 2na + nb

Solution:

Given expression is 2ma + mb + 2na + nb.

Group the first two terms and last two terms.

The first two terms are 2ma + mb and the second two terms are 2na + nb.

Take m common from the first two terms.

m (2a + b)

Take n common from the second two terms.

n (2a + b)

m (2a + b) + n (2a + b)

Then, take (2a + b) common from the above expression.

(2a + b) (m + n)

The final answer is (2a + b) (m + n).

(ii) 3xm – ym – 3xn + yn

Solution:

Given expression is 3xm – ym – 3xn + yn.

Group the first two terms and last two terms.

The first two terms are 3xm – ym and the second two terms are – 3xn + yn.

Take m common from the first two terms.

m (3x – y)

Take -n common from the second two terms.

-n (3x – y)

m (3x – y) – n (3x – y)

Then, take (3x – y) common from the above expression.

(3x – y) (m – n)

The final answer is (3x – y) (m – n).

(iii) 12a^{2} + 6ab – 4ma – 2mb

Solution:

Given expression is 12a^{2} + 6ab – 4ma – 2mb.

Group the first two terms and last two terms.

The first two terms are 12a^{2} + 6ab and the second two terms are – 4ma – 2mb.

Take 6a common from the first two terms.

6a (2a + b)

Take -2m common from the second two terms.

-2m (2a + b)

6a (2a + b) – 2m (2a + b)

Then, take (2a + b) common from the above expression.

(2a + b) (6a – 2m)

The final answer is (2a + b) (6a – 2m).

(iv) am^{2} – bm^{2} + an^{2} – bn^{2} + ar^{2} – br^{2}

Solution:

Given expression is am^{2} – bm^{2} + an^{2} – bn^{2} + ar^{2} – br^{2}.

Group the first two terms, middle two terms, and last two terms.

The first two terms are am^{2} – bm^{2}, the middle two terms are + an^{2} – bn^{2}, and the last two terms are + ar^{2} – br^{2}.

Take m^{2} common from the first two terms.

m^{2} (a – b)

Take n^{2} common from the middle two terms.

n^{2} (a – b)

Take r^{2} common from the middle two terms.

r^{2} (a – b)

m^{2} (a – b) + n^{2} (a – b) + r^{2} (a – b)

Then, take (a – b) common from the above expression.

(a – b) (m^{2} + n^{2} + r^{2})

The final answer is (a – b) (m^{2} + n^{2} + r^{2}).

(v) ax – ay + bx – by

Solution:

Given expression is ax – ay + bx – by.

Group the first two terms and last two terms.

The first two terms are ax – ay and the second two terms are + bx – by.

Take a common from the first two terms.

a (x – y)

Take b common from the second two terms.

b (x – y)

a (x – y) + b (x – y)

Then, take (x – y) common from the above expression.

(x – y) (a + b)

The final answer is (x – y) (a + b).

2. Factoring the following algebraic expression

(i) 4a + 2ab + b + 2

Solution:

Given expression is 4a + 2ab + b + 2.

Group the first two terms and last two terms.

The first two terms are 4a + 2ab and the second two terms are b + 2.

Take 2a common from the first two terms.

2a (2 + b)

Take 1 common from the second two terms.

1 (b + 2)

2a (2 + b) + 1 (2 + b)

Then, take (2 + b) common from the above expression.

(2 + b) (2a + 1)

The final answer is (2 + b) (2a + 1).

(ii) 3m^{3} + 5m^{2} + 3m + 5

Solution:

Given expression is 3m^{3} + 5m^{2} + 3m + 5.

Group the first two terms and last two terms.

The first two terms are 3m^{3} + 5m^{2} and the second two terms are 3m + 5.

Take m^{2 }common from the first two terms.

m^{2} (3m + 5)

Take 1 common from the second two terms.

1 (3m + 5)

m^{2} (3m + 5) + 1 (3m + 5)

Then, take (3m + 5) common from the above expression.

(3m + 5) (m^{2} + 1)

The final answer is (3m + 5) (m^{2} + 1).

(iii) b^{3} + 3b^{2} + b + 3

Solution:

Given expression is b^{3} + 3b^{2} + b + 3.

Group the first two terms and last two terms.

The first two terms are b^{3} + 3b^{2} and the second two terms are b + 3.

Take b^{2}^{ }common from the first two terms.

b^{2} (b + 3)

Take 1 common from the second two terms.

1 (b + 3)

b^{2} (b + 3) + 1 (b + 3)

Then, take (b + 3) common from the above expression.

(b + 3) (b^{2} + 1)

The final answer is (b + 3) (b^{2} + 1).

(iv) 1 + s + s^{2}t + s^{3}t

Solution:

Given expression is 1 + s + s^{2}t + s^{3}t.

Group the first two terms and last two terms.

The first two terms are 1 + s and the second two terms are s^{2}t + s^{3}t.

Take 1^{ }common from the first two terms.

1 (1 + s)

Take s^{2}t common from the second two terms.

s^{2}t (1 + s)

1 (1 + s) + s^{2}t (1 + s)

Then, take (1 + s) common from the above expression.

(1 + s) (1 + s^{2}t)

The final answer is (1 + s) (1 + s^{2}t).

(v) m – 1 – (m – 1)^{2} + bm – b

Solution:

Given expression is m – 1 – (m – 1)^{2} + bm – b.

Group the first two terms, middle and last two terms.

The first two terms are m – 1, the middle term is – (m – 1)^{2}, and the last two terms is bm – b.

Take 1^{ }common from the first two terms.

1 (m – 1)

Take 1 common from the middle term.

– (m – 1)^{2
}Take b common from the last two terms.

b (m – 1)

1 (m – 1) – (m – 1)^{2}+ b (m – 1)

Then, take (m – 1) common from the above expression.

(m – 1) (1 – m + 1 + b)

(m – 1) (2 + b – m)

The final answer is (m – 1) (2 + b – m).