Factoring Terms by Regrouping concept and examples are given in this article. Students who are searching for the best way to solve problems of finding factors can follow this article. All the tricks and tips to learn **Factorization** problems are given in this article. All the students need to do is solve all the problems and test their knowledge. Score good marks in the exam by solving all the problems given in this article.

## Factoring Terms by Regrouping Solved Examples

1. Factorize the expression

(i) p^{2}r + pqr + pc + pqs + q^{2}s + qc

Solution:

Given expression is p^{2}r + pqr + pc + pqs + q^{2}s + qc

Rearrange the terms

p^{2}r + pqr + pqs + q^{2}s + pc + qc

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

The first two terms are p^{2}r + pqr, middle terms are pqs + q^{2}s, and the last two terms are pc + qc

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

pr (p + q)

Take qs common from the second two terms.

qs (p + q)

Take c common from the last two terms.

c (p + q)

pr (p + q) + qs (p + q) + c (p + q)

Then, take (p + q) common from the above expression.

(p + q) (pr + qs + c)

The final answer is (p + q) (pr + qs + c).

(ii) s^{3}k + s^{2}(k – m) – s(m + n) – n

Solution:

Given expression is s^{3}k + s^{2}(k – m) – s(m + n) – n

Rearrange the terms

s^{3}k + s^{2}k – s^{2}m – sm – sn – n

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

The first two terms are s^{3}k + s^{2}k, the middle terms are – s^{2}m – sm, and the second two terms are – sn – n

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

s^{2}k (s + 1)

Take – sm common from the middle two terms.

– sm (s + 1)

Take -n common from the last two terms.

-n (s + 1)

s^{2}k (s + 1) – sm (s + 1) – n (s + 1)

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

(s + 1) (s^{2}k – sm – n)

The final answer is (s + 1) (s^{2}k – sm – n).

2. How to factorize by grouping the following expressions?

(i) px – qx + qy + ry – rx – py

Solution:

Given expression is px – qx + qy + ry – rx – py

Rearrange the terms

px – qx – rx + qy + ry – py

Group the first three terms, and last three terms.

The first three terms are px – qx – rx, and the last three terms are qy + ry – py

Take x^{ }common from the first three terms.

x (p – q – r)

Take y common from the last three terms.

-y (p – q – r)

x (p – q – r) – y (p – q – r)

Then, take (p – q – r) common from the above expression.

(p – q – r) (x – y)

The final answer is (p – q – r) (x – y).

(ii) a^{3} – 2a^{2} + ma + a – 2m – 2

Solution:

Given expression is a^{3} – 2a^{2} + ma + a – 2m – 2

Rearrange the terms

a^{3} – 2a^{2} + ma – 2m+ a – 2

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

The first two terms are a^{3} – 2a^{2}, the middle terms are ma – 2m, and the last two terms are a – 2

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

a^{2} (a – 2)

Take m common from the middle two terms.

m (a – 2)

Take 1 common from the last two terms.

1 (a – 2)

a^{2} (a – 2) + m (a – 2) + 1 (a – 2)

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

(a – 2) (a^{2} + m + 1)

The final answer is (a – 2) (a2 + m + 1).