Common Core Math

Evaluate the Difference of Two Squares | Difference of Two Squares Problems

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Are you looking for How to Evaluate the Difference of Two Squares Problems? We have given all Difference of Two Squares problems along with the evaluation of Difference of Two Squares with detailed explanation. Students can refer to all factorization problems on our website and begin their practice to score good marks in the exam.

Solved Problems to Evaluate the Difference of Two Squares

Use the formula of the difference of two squares to evaluate the following algebraic expressions:

(i) (202)2 – (123)2

Solution:
Given expression is (202)2 – (123)2
The above equation (202)2 – (123)2 is in the form of a2 – b2.
(202)2 – (123)2
Now, apply the formula of a2 – b2 = (a + b) (a – b), where a = 202 and b = 123
(202 + 123) (202 – 123)
(325) (79)
25675

The final answer is 25675.

(ii) (600)2 – (598)2

Solution:
Given expression is (600)2 – (598)2
The above equation (600)2 – (598)2 is in the form of a2 – b2.
(600)2 – (598)2
Now, apply the formula of a2 – b2 = (a + b) (a – b), where a = 600 and b = 598
(600 + 598) (600 – 598)
(1198) (2)
2396

The final answer is 2396.

(iii) (4.2)2 – (2.1)2

Solution:
Given expression is (4.2)2 – (2.1)2
The above equation (4.2)2 – (2.1)2 is in the form of a2 – b2.
(4.2)2 – (2.1)2
Now, apply the formula of a2 – b2 = (a + b) (a – b), where a = 4.2 and b = 2.1
(4.2 + 2.1) (4.2 – 2.1)
(6.3) (2.1)
13.23

The final answer is 13.23.

(iv) (97.8)2 – (0.4)2

Solution:
Given expression is (97.8)2 – (0.4)2
The above equation (97.8)2 – (0.4)2 is in the form of a2 – b2.
(97.8)2 – (0.4)2
Now, apply the formula of a2 – b2 = (a + b) (a – b), where a = 97.8 and b = 0.4
(97.8 + 0.4) (97.8 – 0.4)
(98.2) (97.4)
9564.68

The final answer is 9564.68.

(v) (8.4)2 – (1.8)2

Solution:
Given expression is (8.4)2 – (1.8)2
The above equation (8.4)2 – (1.8)2 is in the form of a2 – b2.
(8.4)2 – (1.8)2
Now, apply the formula of a2 – b2 = (a + b) (a – b), where a = 8.4 and b = 1.8
(8.4 + 1.8) (8.4 – 1.8)
(10.2) (6.6)
67.32

The final answer is 67.32.

Factorize the Difference of Two Squares | How to find Factors of Difference of Squares?

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Factoring a polynomial is the product of the two or more polynomials. Learn How to Factorize the Difference of Two Squares in this article. Break down all the huge algebraic expressions into small factors with the help of factorization. Solved Problems on Factoring the Difference of Two Squares are explained clearly along with the solutions. Visit all factorization problems and get complete knowledge of the factorization concept.

Solved Problems on How to Factorize the Difference of Two Squares

1. Factorize the following algebraic expressions

(i) m2 – 121

Solution:
Given expression is m2 – 121
Rewrite the above expression.
m2 – (11)2 
The above equation m2 – (11)2 is in the form of a2 – b2.
m2 – (11)2 
Now, apply the formula of a2 – b2 = (a + b) (a – b), where a = m and b = 11
(m + 11) (m – 11)

The final answer is (m + 11) (m – 11)

(ii) 49a2 – 16b2

Solution:
Given expression is 49a2 – 16b2
Rewrite the above expression.
(7a)2 – (4b)2 
The above equation (7a)2 – (4b)2  is in the form of a2 – b2.
(7a)2 – (4b)2 
Now, apply the formula of a2 – b2 = (a + b) (a – b), where a = 7a and b = 4b
(7a + 4b) (7a – 4b)

The final answer is (7a + 4b) (7a – 4b)

2. Factor the following

(i) 48m2 – 243n2

Solution:
Given expression is 48m2 – 243n2
Take 3 common
3{16m2 – 81n2}
Rewrite the above expression.
3{(4m)2 – (9n)2} 
The above equation {(4m)2 – (9n)2}   is in the form of a2 – b2.
{(4m)2 – (9n)2} 
Now, apply the formula of a2 – b2 = (a + b) (a – b), where a = 4m and b = 9n
(4m + 9n) (4m – 9n)
3{(4m + 9n) (4m – 9n)}

The final answer is 3{(4m + 9n) (4m – 9n)}

(ii) 3a3 – 48a

Solution:
Given expression is 3a3 – 48a
Take 3 common
3a{a2 – 16}
Rewrite the above expression.
3a{(a)2 – (4)2} 
The above equation {(a)2 – (4)2}    is in the form of a2 – b2.
{(a)2 – (4)2} 
Now, apply the formula of a2 – b2 = (a + b) (a – b), where a = a and b = 4
(a + 4) (a – 4)
3a{(a + 4) (a – 4)}

The final answer is 3a{(a + 4) (a – 4)}

3. Factor the expressions

(i) 25(a + 3b)2 – 16 (a – 3b)2

Solution:
Given expression is 25(a + 3b)2 – 16 (a – 3b)2
Rewrite the above expression.
{[5(a + 3b)]2 – [4 (a – 3b)]2} 
The above equation {[5(a + 3b)]2 – [4 (a – 3b)]2} is in the form of a2 – b2.
{[5(a + 3b)]2 – [4 (a – 3b)]2}
Now, apply the formula of a2 – b2 = (a + b) (a – b), where a = 5(a + 3b) and b = 4 (a – 3b)
(5(a + 3b) + 4 (a – 3b)) (5(a + 3b) – [4 (a – 3b)])
(5a + 15b + 4a – 12b) (5a + 15b – 4a + 12b)
(9a + 3b) (a + 27b)
3(3a + b) (a + 27b)

The final answer is 3(3a + b) (a + 27b)

(ii) 4x2 – 16/(25x2)

Solution:
Given expression is 4x2 – 16/(25x2)
Rewrite the above expression.
{[2x]2 – [4/5x]2}
The above equation {[2x]2 – [4/5x]2} is in the form of a2 – b2.
{[2x]2 – [4/5x]2}
Now, apply the formula of a2 – b2 = (a + b) (a – b), where a = 2x and b = 4/5x
(2x + 4/5x) (2x – 4/5x)

The final answer is (2x + 4/5x) (2x – 4/5x)

Factoring Differences of Squares | How do you find the Factors of the Difference of Two Squares?

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Join the list of successful students who learned all the Factorization problems using the best material provided on our website. Yes, it’s true. Students can easily score good marks in the exam using our information on Factoring Differences of Squares. Also, we are giving all Factorization-Related Solved Problems, Extra Questions, and also Multiple Choice Questions and Answers. Utilize this opportunity and start your practice now and test your knowledge on the concept.

Factoring the Differences of Two Squares Examples

1. Factorize the following algebraic expressions

(i) 64 – a2

Solution:
Given expression is 64 – a2
Rewrite the above expression.
82 – a2
The above equation 82 – a2 is in the form of a2 – b2.
[(8)2 – (a)2]
Now, apply the formula of a2 – b2 = (a + b) (a – b), where a = 8 and b = a
(8 + a) (8 – a)

The final answer is (8 + a) (8 – a)

(ii) 3m2 – 27n2

Solution:
Given expression is 3m2 – 27n2
Rewrite the above expression. Take 3 common.
3 (m2 – (3n)2) where 9n2 = (3n)2
The above equation (m2 – (3n)2)  is in the form of a2 – b2.
[(m)2 – (3n)2]
Now, apply the formula of a2 – b2 = (a + b) (a – b), where a = m and b = 3n
(m + 3n) (m – 3n)
3{(m + 3n) (m – 3n)}

The final answer is 3{(m + 3n) (m – 3n)}

(iii) a3 – 25a

Solution:
Given expression is a3 – 25a
Rewrite the above expression. Take a common.
a (a2 – 25)
a ((a)2 – (5)2)
The above equation ((a)2 – (5)2) is in the form of a2 – b2.
((a)2 – (5)2)
Now, apply the formula of a2 – b2 = (a + b) (a – b), where a = a and b = 5
(a + 5) (a – 5)
a {(a + 5) (a – 5)}

The final answer is a {(a + 5) (a – 5)}

2. Factor the expressions

(i) 81x2 – (y – z)2

Solution:
Given expression is 81x2 – (y – z)2
Rewrite the above expression.
(9x)2 – (y – z)2
The above equation ((9x)2 – (y – z)2) is in the form of a2 – b2.
((9x)2 – (y – z)2)
Now, apply the formula of a2 – b2 = (a + b) (a – b), where a = 9x and b = y – z
(9x + (y – z)) (9x – (y – z))
(9x + y – z) (9x – y + z)

The final answer is (9x + y – z) (9x – y + z)

(ii) 25(a + b)2 – 36(a – 2b)2.

Solution:
Given expression is 25(a + b)2 – 36(a – 2b)2
Rewrite the above expression.
{5(a + b)}2 – {6(a – 2b)}2
The above equation {5(a + b)}2 – {6(a – 2b)}2 is in the form of a2 – b2.
((5(a + b))2 – (6(a – 2b))2)
Now, apply the formula of a2 – b2 = (a + b) (a – b), where a = 5(a + b) and b = 6(a – 2b)
[5(a + b) + 6(a – 2b)] [5(a + b) – 6(a – 2b)]
[5a + 5b + 6a – 12b] [5a + 5b – 6a + 12b]
[11a – 7b] [17b – a]

The final answer is [11a – 7b] [17b – a]

(iii) (m – 2)2 – (m – 3)2

Solution:
Given expression is (m – 2)2 – (m – 3)2
The above equation (m – 2)2 – (m – 3)2 is in the form of a2 – b2.
(m – 2)2 – (m – 3)2
Now, apply the formula of a2 – b2 = (a + b) (a – b), where a = m – 2 and b = m – 3
[(m – 2) + (m – 3)] [(m – 2) – (m – 3)]
[m – 2 + m – 3] [m – 2 – m + 3]
[2m – 5] [1]
[2m – 5]

The final answer is [2m – 5]

Factorization of Perfect Square | How to Factor a Perfect Square?

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Factorization of Perfect Square is the process of finding factors for an equation which is in the form of a2 + 2ab + b2 or a2 – 2ab + b2. Get to know the step by step procedure involved for finding factors of a perfect square. Have a look at the different examples taken to illustrate the Factorization of Perfect Square Problems. By following this article, you will better understand the concept and solving process of perfect square factorization.

(i) a2 + 2ab + b2 = (a + b)2 = (a + b) (a + b)
(ii) a2 – 2ab + b2 = (a – b)2 = (a – b) (a – b)

Factorization of Perfect Square Solved Examples

1. Factorize the perfect square completely
(i) 16a2 + 25b2 + 40ab

Solution:
Given expression is 16a2 + 25b2 + 40ab
The given expression 16a2 + 25b2 + 40ab is in the form a2 + 2ab + b2.
So find the factors of given expression using a2 + 2ab + b2 = (a + b)2 = (a + b) (a + b) where a = 4a, b = 5b
Apply the formula and substitute the a and b values.
16a2 + 25b2 + 40ab
(4a)2 + 2 (4a) (5b) + (5b)2
(4a + 5b)2
(4a + 5b) (4a + 5b)

Factors of the 16a2 + 25b2 + 40ab are (4a + 5b) (4a + 5b)

(ii) 9x2 – 42xy + 49y2

Solution:
Given expression is 9x2 – 42xy + 49y2
The given expression 9x2 – 42xy + 49y2 is in the form a2 – 2ab + b2.
So find the factors of given expression using a2 – 2ab + b2 = (a – b)2 = (a – b) (a – b) where a = 3x, b = 7y
Apply the formula and substitute the a and b values.
9x2 – 42xy +49y2
(9x)2 – 2 (9x) (7y) + (7y)2
(9x – 7y)2
(9x – 7y) (9x – 7y)

Factors of the 9x2 – 42xy + 49y2 are (9x – 7y) (9x – 7y)

(iii) 25m2 + 80m + 64

Solution:
Given expression is 25m2 + 80m + 64
The given expression 25m2 + 80m + 64 is in the form a2 + 2ab + b2.
So find the factors of given expression using a2 + 2ab + b2 = (a + b)2 = (a + b) (a + b) where a = 5m, b = 8
Apply the formula and substitute the a and b values.
25m2 + 80m + 64
(5m)2 + 2 (5m) (8) + (8)2
(5m + 8)2
(5m + 8) (5m + 8)

Factors of the 25m2 + 80m + 64 are (5m + 8) (5m + 8)

(iv) a2 + 6a + 8

Solution:
Given expression is a2 + 6a + 8.
The Given expression is a2 + 6a + 8 is not a perfect square.
Add and subtract 1 to make the given expression a2 + 6a + 8 is not a perfect square.
a2 + 6a + 8 + 1 – 1
a2 + 6a + 9 – 1
The above expression a2 + 6a + 9 is in the form a2 + 2ab + b2.
So find the factors of the expression using a2 + 2ab + b2 = (a + b)2 = (a + b) (a + b) where a = a, b = 3
Apply the formula and substitute the a and b values.
a2 + 6a + 9
(a)2 + 2 (a) (3) + (3)2
(a + 3)2
(a + 3)2 – 1
(a + 3)2 – (1)2
(a + 3 + 1) (a + 3 – 1)
(a + 4) (a + 2)

Factors of the a2 + 6a + 8 are (a + 4) (a + 2)

2. Factor using the identity

(i) 4x4 + 1

Solution:
Given expression is 4x4 + 1.
The Given expression is 4x4 + 1 is not a perfect square.
Add and subtract 4x² to make the given expression 4x4 + 1 is not a perfect square.
4x4 + 1 + 4x² – 4x²
4x4 + 4x² + 1 – 4x²
The above expression 4x4 + 4x² + 1 is in the form a2 + 2ab + b2.
So find the factors of the expression using a2 + 2ab + b2 = (a + b)2 = (a + b) (a + b) where a = 2x², b = 1
Apply the formula and substitute the a and b values.
4x4 + 4x² + 1
(2x²)2 + 2 (2x²) (1) + (1)2
(2x² + 1)2
(2x² + 1)2 – 4x²
(2x² + 1)2 – (2x)2
(2x² + 1 + 2x) (2x² + 1 – 2x)
(2x² + 2x + 1) (2x² – 2x + 1)

Factors of the 4×4 + 1 are (2x² + 2x + 1) (2x² – 2x + 1)

(ii) (a + 2b)2 + 2(a + 2b) (3b – a) + (3b – a)2

Solution:
Given expression is (a + 2b)2 + 2(a + 2b) (3b – a) + (3b – a)2
The given expression (a + 2b)2 + 2(a + 2b) (3b – a) + (3b – a)2 is in the form a2 + 2ab + b2.
So find the factors of given expression using a2 + 2ab + b2 = (a + b)2 = (a + b) (a + b) where a = a + 2b, b = 3b – a
Apply the formula and substitute the a and b values.
(a + 2b)2 + 2(a + 2b) (3b – a) + (3b – a)2
(a + 2b)2 + 2 (a + 2b) (3b – a) + (3b – a)2
(a + 2b + 3b – a)2
(5b)2
25b2

Factors of the (a + 2b)2 + 2(a + 2b) (3b – a) + (3b – a)2 are 25b2

Factoring Terms by Regrouping | Factoring by Regrouping Steps, Examples

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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) p2r + pqr + pc + pqs + q2s + qc

Solution:
Given expression is p2r + pqr + pc + pqs + q2s + qc
Rearrange the terms
p2r + pqr + pqs + q2s + pc + qc
Group the first two terms, middle two terms, and last two terms.
The first two terms are p2r + pqr, middle terms are pqs + q2s, 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) s3k + s2(k – m) – s(m + n) – n

Solution:
Given expression is s3k + s2(k – m) – s(m + n) – n
Rearrange the terms
s3k + s2k – s2m – sm – sn – n
Group the first two terms, middle two terms, and last two terms.
The first two terms are s3k + s2k, the middle terms are – s2m – sm, and the second two terms are – sn – n
Take s2k common from the first two terms.
s2k (s + 1)
Take – sm common from the middle two terms.
– sm (s + 1)
Take -n common from the last two terms.
-n (s + 1)
s2k (s + 1) – sm (s + 1) – n (s + 1)
Then, take (s + 1) common from the above expression.
(s + 1) (s2k – sm – n)

The final answer is (s + 1) (s2k – 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) a3 – 2a2 + ma + a – 2m – 2

Solution:
Given expression is a3 – 2a2 + ma + a – 2m – 2
Rearrange the terms
a3 – 2a2 + ma – 2m+ a – 2
Group the first two terms, middle two terms, and last two terms.
The first two terms are a3 – 2a2, the middle terms are ma – 2m, and the last two terms are a – 2
Take a2 common from the first two terms.
a2 (a – 2)
Take m common from the middle two terms.
m (a – 2)
Take 1 common from the last two terms.
1 (a – 2)
a2 (a – 2) + m (a – 2) + 1 (a – 2)
Then, take (a – 2) common from the above expression.
(a – 2) (a2 + m + 1)

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

Factorize by Regrouping The Terms | How to Find Factors by Regrouping?

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Factorize by Regrouping The Terms to find factors of an algebraic expression. Rewrite the given expression to form different groups and take out the common factor. Finding factors is easy with the regrouping process. Follow all the problems given below and get complete knowledge on Factorization by Regrouping. Find the simplest method to find factors i.e. regrouping method.

Procedure to find Factors by Regrouping

Follow the below process and solve any difficult expression factors in minutes. They are as such

Step 1: Note down the given expression. From the given algebraic expression form the groups of the given expression in such a way that a common factor can be taken out from every group.
Step 2: Factorize each group.
Step 3: At last, take out the common factor of the groups formed.

Solved Examples on Factorization of Algebraic Expressions

1. Factoring the following expressions

(i) mn (a2 + b2) – ab (m2 + n2)

Solution:
Given expression is mn (a2 + b2) – ab (m2 + n2)
Rearrange the terms
mna2 – abm2 + mnb2 – abn2
Group the first two terms and last two terms.
The first two terms are mna2 – abm2  and the second two terms are mnb2 – abn2
Take ma common from the first two terms.
ma (na – bm)
Take -nb common from the second two terms.
-nb (na – bm)
ma (na – bm) -nb (na – bm)
Then, take (na – bm) common from the above expression.
(na – bm) (ma – nb)

The final answer is (na – bm) (ma – nb).

(ii) 2am – 4an – 3bm + 6nb

Solution:
Given expression is 2am – 4an – 3bm + 6nb
Rearrange the terms
2am – 3bm – 4an + 6nb
Group the first two terms and last two terms.
The first two terms are 2am – 3bm and the second two terms are – 4an + 6nb
Take m common from the first two terms.
m (2a – 3b)
Take -2n common from the second two terms.
-2n (2a – 3b)
m (2a – 3b) -2n (2a – 3b)
Then, take (2a – 3b) common from the above expression.
(2a – 3b) (m – 2n)

The final answer is (2a – 3b) (m – 2n).

(iii) – 6 – 12t + 18t2

Solution:
Given expression is – 6 – 12t + 18t2
Rearrange the terms
18t2 – 12t – 6
Then, take 6 as common from the above expression.
6 (3t2 – 2t – 1)

The final answer is 6 (3t2 – 2t – 1).

2. Factorize the expression

(i) mn – m – n + 1

Solution:
Given expression is mn – m – n + 1
Rearrange the terms
mn – n – m + 1
Group the first two terms and last two terms.
The first two terms are mn – n and the second two terms are – m + 1
Take n common from the first two terms.
n (m – 1)
Take -1 common from the second two terms.
-1(m – 1)
n (m – 1) – 1(m – 1)
Then, take (m – 1) common from the above expression.
(m – 1) (n – 1)

The final answer is (m – 1) (n – 1).

(ii) pm + pn – qm – qn

Solution:
Given expression is pm + pn – qm – qn
Rearrange the terms
pm – qm + pn – qn
Group the first two terms and last two terms.
The first two terms are pm – qm and the second two terms are pn – qn
Take m common from the first two terms.
m (p – q)
Take n common from the second two terms.
n (p – q)
m (p – q) + n (p – q)
Then, take (p – q) common from the above expression.
(p – q) (m + n)

The final answer is (p – q) (m + n).

Factoring Terms by Grouping | How to Factor Terms by Grouping?

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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) 12a2 + 6ab – 4ma – 2mb

Solution:
Given expression is 12a2 + 6ab – 4ma – 2mb.
Group the first two terms and last two terms.
The first two terms are 12a2 + 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) am2 – bm2 + an2 – bn2 + ar2 – br2

Solution:
Given expression is am2 – bm2 + an2 – bn2 + ar2 – br2.
Group the first two terms, middle two terms, and last two terms.
The first two terms are am2 – bm2, the middle two terms are + an2 – bn2, and the last two terms are + ar2 – br2.
Take m2 common from the first two terms.
m2 (a – b)
Take n2 common from the middle two terms.
n2 (a – b)
Take r2 common from the middle two terms.
r2 (a – b)
m2 (a – b) + n2 (a – b) + r2 (a – b)
Then, take (a – b) common from the above expression.
(a – b) (m2 + n2 + r2)

The final answer is (a – b) (m2 + n2 + r2).

(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) 3m3 + 5m2 + 3m + 5

Solution:
Given expression is 3m3 + 5m2 + 3m + 5.
Group the first two terms and last two terms.
The first two terms are 3m3 + 5m2 and the second two terms are 3m + 5.
Take m2 common from the first two terms.
m2 (3m + 5)
Take 1 common from the second two terms.
1 (3m + 5)
m2 (3m + 5) + 1 (3m + 5)
Then, take (3m + 5) common from the above expression.
(3m + 5) (m2 + 1)

The final answer is (3m + 5) (m2 + 1).

(iii) b3 + 3b2 + b + 3

Solution:
Given expression is b3 + 3b2 + b + 3.
Group the first two terms and last two terms.
The first two terms are b3 + 3b2 and the second two terms are b + 3.
Take b2 common from the first two terms.
b2 (b + 3)
Take 1 common from the second two terms.
1 (b + 3)
b2 (b + 3) + 1 (b + 3)
Then, take (b + 3) common from the above expression.
(b + 3) (b2 + 1)

The final answer is (b + 3) (b2 + 1).

(iv) 1 + s + s2t + s3t

Solution:
Given expression is 1 + s + s2t + s3t.
Group the first two terms and last two terms.
The first two terms are 1 + s and the second two terms are s2t + s3t.
Take 1 common from the first two terms.
1 (1 + s)
Take s2t common from the second two terms.
s2t (1 + s)
1 (1 + s) + s2t (1 + s)
Then, take (1 + s) common from the above expression.
(1 + s) (1 + s2t)

The final answer is (1 + s) (1 + s2t).

(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).

Factorize by Grouping the Terms | How to Factor by Grouping Terms?

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Factorize by grouping the terms consists of two or more product terms as resultant value. To divide the expression into product terms, we need to identify the greatest common factor which divides all the remaining terms in the expression. Step by step process on how to do Factorization by Grouping the Terms are clearly given in this article.

Procedure to Factorize by Grouping the Terms

Refer to the below-mentioned step by step process and learn the Factoring by Grouping Terms. They are along the lines

  1. Note down the given expression.
  2. Group the first two terms and last two terms.
  3. Factor out the greatest common factor from each group.
  4. Finally, you will get two or more product terms as result.

Solved Examples on Factorization by Grouping Terms

1. Factorize grouping the following expressions?

(i) 18x³y³ – 27x²y³ + 36x³y²

Solution:
The given expression is 18x³y³ – 27x²y³ + 36x³y²
Factor out the greatest common factor from the above expression.
That is, 9x²y²(2xy – 3y + 4x).

Therefore, the final solution for the expression 18x³y³ – 27x²y³ + 36x³y² is 9x²y²(2xy – 3y + 4x).

(ii) 12a²b³ – 21a³b²

Solution:
The given expression is 12a²b³ – 21a³b²
Factor out the greatest common factor from the above expression.
3a²b²(4b – 7a).

Finally, the solution for the expression 12a²b³ – 21a³b² is 3a²b²(4b – 7a).

(iii) a³ – a² + a – 1.

Solution:
The given expression is a³ – a² + a – 1
Group the first two terms and last two terms.
Here, first two terms are a³ – a² and the last two terms are a – 1.
So, (a³ – a²) + (a – 1).
Now, factor out the greatest common factor from each group.
That is, a²(a – 1) + (a – 1).
(a – 1) (a² + 1).

Therefore, solution for the expression a³ – a² + a – 1 is (a – 1) (a² + 1).

(iv) prs + qurs – pt – qut

Solution:
The given expression is prs + qurs – pt – qut.
Group the first two terms and last two terms.
Here, the first two terms are prs + qurs and last two terms are – pt – qut.
Then, (prs + qurs) – (pt + qut).
Now, factor out the greatest common factor from the above two groups.
That is, rs(p + qu) – t(p + qu).
(p + qu) (rs – t).

Therefore, the solution for the expression prs + qurs – pt – qut is (p + qu) (rs – t).

(v) a² – 3a – ab + 3b
Solution:

The given expression is
a² – 3a – ab + 3b.
Group the first two terms and last two terms.
Here, the first two terms are a²-3a and the last two terms are ab + 3b.
Then, (a² – 3a) – (ab – 3b).
Now, factor out the greatest common factor from the above two groups.
That is a(a – 3) – b(a – 3).
(a – 3) (a – b).

Therefore, solution for the expression is a² – 3a – ab + 3b is (a – 3) (a – b).

2. How to factorize by grouping the following expressions?
(i) 2q^4 – q³ + 4q – 2
Solution:
The given expression is
2q^4 – q³ + 4q – 2.
Group the first two terms and last two terms.
Here, the first two terms are 2q^4 – q³ and the last two terms are 4q – 2.
Then, (2q^4 – q^3) + (4q-2).
Now, factor out the greatest common factor from the above two groups.
That is, q^3(2q – 1) +2(2q – 1).
(2q – 1) (q^3 + 2).

Therefore, solution for the expression is 2q4 – q3 + 4q – 2 is (2q-1) (q^3+2).

(ii) ac + bc – ad – bd.

Solution:
The given expression is ac + bc – ad – bd.
Group the first two terms and last two terms.
Here, the first two terms are ac+bc and the last two terms are – ad-bd.
Then, (ac + bc) – (ad + bd).
Now, factor out the greatest common factor from the above two groups.
That is, c(a+b) -d(a+b).
(a+b) (c-d).

Therefore, solution for the expression is ac + bc – ad – bd is (a+b) (c-d).

(iii) pa – pb – qa –qb.

Solution:
The given expression is pa – pb – qa–qb.
Group the first two terms and last two terms.
Here, the first two terms are pa – pb, and the last two terms are qa – qb.
Then, (pa – pb) – (qa -qb).
Now, factor out the greatest common factor from the above two groups.
That is, p(a – b) – q(a – b).
(p – q) (a – b).

Therefore, the solution for the expression is pa – pb – qa – qb is (p – q) (a – b).

3. How to factorize by grouping the algebraic expressions?
(i)x^2z^2 + xzw + xyz + yw

Solution:
The given expression is
x^2z^2 + xzw + xyz + yw.
Group the first two terms and last two terms.
Here, the first two terms are x^2z^2 + xzw and the last two terms are xyz + yw.
Then, (x^2z^2 + xzw) + (xyz + yw).
Now, factor out the greatest common factor from the above two groups.
That is, xz(xz + w) + y(xz + w).
(xz + y) (xz + w).

Therefore, solution for the expression is x^2z^2 + xzw + xyz + yw is (xz + y) (xz + w).

(ii) 5x + xy + 5y + y^2

Solution:
The given expression is 5x + xy + 5y + y^2
Group the first two terms and last two terms.
Here, the first two terms are 5x + xy, and the last two terms are 5y + y^2.
Then, (5x + xy) + (5y + y^2).
Now, factor out the greatest common factor from the above two groups.
That is, x(5 + y) + y(5 + y).
(5 + y) (x + y).

Therefore, solution for the expression is 5x + xy + 5y + y^2 is (5 + y) (x + y).

(iii) xy – yz – xz + z^2

Solution:
The given expression is xy – yz – xz + z^2
Group the first two terms and last two terms.
Here, the first two terms are xy – yz and the last two terms are – xz + z^2.
Then, (xy – yz) – (xz – z^2).
Now, factor out the greatest common factor from the above two groups.
That is, y(x – z) – z(x – z).
(x – z) (y – z).

Therefore, solution for the expression xy – yz – xz + z^2 is (x – z) (y – z).

4. Factorize the expressions
(i) a^4 + a^3 + 2a + 2

Solution:
The given expression is a^4 + a^3 + 2a + 2.
Group the first two terms and last two terms.
Here, the first two terms are a^4 + a^3 and the last two terms are 2a + 2.
Then, (a^4 + a^3) + (2a + 2).
Now, factor out the greatest common factor from the above two groups.
That is, a^3(a + 1) + 2(a + 1).
(a + 1) (a^3 + 2).

Therefore, solution for the expression a^4 + a^3 + 2a + 2 is (a + 1) (a^3 + 2).

(ii) a²b² + d²b² – cd² – ca²

Solution:
The given expression is a²b² + d²b² – cd² – ca²
Group the first two terms and last two terms.
Here, the first two terms are a²b² + d²b² and the last two terms are – cd² – ca².
Then, (a^2b^2 + d^2b^2) – (cd^2 + ca^2).
Now, factor out the greatest common factor from the above two groups.
That is, b^2(a^2 + d^2) – c(d^2 + a^2).
(a^2 + d^2) (b^2 – c).

Therefore, solution for the expression a2b2 + d2b2 – cd2 – ca2 is(a^2 + d^2) (b^2 – c).

5. Factorize by grouping the terms
(x^2 + 3x)^2 – 2(x^2 + 3x) – y(x^2 + 3x) + 2y

Solution:
Thegiven expression is (x^2 + 3x)^2 – 2(x^2 + 3x) – y(x^2 + 3x) + 2y.
Group the first two terms and last two terms.
here, first two terms are (x^2 + 3x)^2 – 2(x^2 + 3x) and the last two terms are –y(x^2 + 3x) + 2y.
Then, [(x^2 + 3x)^2 – 2(x^2 + 3x)] –[y(x^2 + 3x) – 2y].
That is, (x^2 + 3x)(x^2 + 3x – 2) – y(x^2 + 3x – 2).
(x^2 + 3x – 2)(x^2 + 3x – y).

Therefore, solution for the expression (x2 + 3x)2 – 2(x2 + 3x) – y(x2 + 3x) + 2y is (x^2 + 3x – 2)(x^2 + 3x – y).

Factorization by Grouping | What is Factorization by Grouping?

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Are you looking for the different problems on Factorization by Grouping? Then, you are in the right place. We have given all types of factorization problems on our website. Students can learn to Factorize by Grouping the Terms in these articles. While solving Factorization by grouping problems, students need to group the terms with common factors before factoring.

How to Factor by Grouping?

Have a look at the Factoring by Grouping Steps and learn how to solve related problems easily. Follow the guidelines provided and perform factorization by grouping method. They are as under

  • Note down the given expression, group the first two terms and last two terms.
  • Find out the greatest common factor(GCF) from the first term and second term.
  • Now, find the common factor from the above two groups.
  • Finally factor out the terms in terms of product.

Factorization by Grouping Examples

1. Factor grouping the expressions?
1 + x + xy + x²y.

Solution: Given Expression is 1 + x + xy + x²y.
Group the first two terms and last two terms.
First two terms are 1 + x and the last two terms are xy + x²y.
(1+ x) + (xy + x²y).
Find out the common factor from the above two groups.
(1+x) + xy(1+x).
Factor out the terms in terms of product.
(1+x) (1+xy).

By factor grouping the expression 1 + x + xy + x²y., we will get (1+x) (1+xy).

2. How to factor by grouping the following algebraic expressions?

(i) x² – xy + xz – zy.

Solution:
Given Expression is x² – xy + xz – zy
Group the first two terms and last two terms.
First two terms are x² – xy and the last two terms are xz – zy.
(x² – xy) + (xz – zy)
Find out the common factor from the above two groups.
x(x – y) + z(x – y)
Factor out the terms in terms of product.
(x + z) (x – y)

Factor by grouping the expression x² – xy + xz – zy, we will get the result as (x+z) (x-y).

(ii) x² + 3x + xy + 3y.

Solution:
Given Expression is x² + 3x + xy + 3y.
Group the first two terms and last two terms.
First two terms are x² + 3x and the last two terms are xy + 3y.
(x² + 3x) + (xy + 3y)
Find out the common factor from the above two groups.
x(x + 3) + y(x + 3)
Factor out the terms in terms of product.
(x+y) (x+3).

Factor by grouping the expression x² + 3x + xy + 3y, we will get solution as (x+y) (x+3).

3. Factorize the algebraic expressions.

(i) 2a + ba + 2b + b²

Solution:
Given Expression is 2a + ba + 2b + b²
Group the first two terms and last two terms.
First two terms are 2a + ba and the last two terms are 2b + b².
(2a + ba)+ (2b + b²).
Find out the common factor from the above two groups.
a(2 + b) + b(2 + b).
Factor out the terms in terms of product.
(a + b) (2 + b).

By factorizing the expression 2a + ba + 2b + b², we will get (a + b) (2 + b).

(ii) b² – yb + 5b– 5y.

Solution:
Given Expression is b² – yb + 5b– 5y.
Group the first two terms and last two terms.
First, two terms are b² – yb and the last two terms are 5b – 5y.
(b² – yb)+ (5b – 5y).
Find out the common factor from the above two groups.
b(b – y) + 5(b – y).
Factor out the terms in terms of product.
(5 + b) (b – y).

By factorizing the expression b² – yb + 5b– 5y, we will get (5 + b) (b – y).

(iii) pq – rq – ps + rs.

Solution:
Given Expression is pq – rq – ps + rs.
Group the first two terms and last two terms.
First two terms are pq – rq and the last two terms are – ps + rs.
(pq – rq) – (ps – rs).
Find out the common factor from the above two groups.
q(p – r) -s(p – r).
Factor out the terms in terms of product.
(p – r) (q – s).

By factorizing the expression pq – rq – ps + rs, we will get (p – r) (q – s).

(iv) ab – 2ac – db + 2dc.

Solution:
The given expression is ab – 2ac – db + 2dc.
Group the first two terms and last two terms.
Fist two terms are ab – 2ac and the last two terms are – db + 2dc.
(ab – 2ac) – (db – 2dc).
Find out the common factor from the above two groups.
a(b – 2c) -d(b – 2c).
Factor out the terms in terms of product.
(b – 2c) (a – d).

By factorizing the expression ab – 2ac – db + 2dc., we will get (b – 2c) (a – d).

(v) pq² – 3rqs – pqs + 3rs²

Solution:
The given expression is pq² – 3rqs – pqs + 3rs²
Group the first two terms and last two terms.
First two terms are pq^2 – 3rqs and the last two terms are –pqs + 3rs².
(pq² – 3rqs) – (pqs – 3rs²).
Find out the common factor from the above two groups.
q(pq – 3rs) -s(pq – 3rs).
Factor out the terms in terms of product.
(pq – 3rs) (q – s).

By factorizing the expression pq² – 3rqs – pqs + 3rs², we will get (pq – 3rs) (q – s).

4. Factor each of the following expressions by grouping

(i) a² – 3a – ab + 3b.

Solution:
Given expression is a² – 3a – ab + 3b
Group the first two terms and last two terms.
First two terms are a² – 3a and the last two terms are – ab + 3b.
(a² – 3a) – (ab – 3b).
Find out the common factor from the above two groups.
a(a – 3) – b(a – 3).
Factor out the terms in terms of product.
(a – 3) (a – b).

By factorizing the expression a² – 3a – ab + 3b, we will get (a – 3) (a – b).

(ii) pq² + rq² + 2p + 2r.

Solution:
Given expression is pq² + rq² + 2p + 2r
Group the first two terms and last two terms.
First two terms are pq² + rq² and the last two terms are 2p + 2r.
(pq² + rq²) + (2p + 2r).
Find out the common factor from the above two groups.
q²(p +r) + 2(p+r).
Factor out the terms in terms of product.
(q²+ 2) (p + r).

By factorizing the expression pq² + rq² + 2p + 2r, we will get (q²+ 2) (p + r).

(iii) 2pq² + 3pqr – 2sqr – 3sr²

Solution:
Given expression is 2pq² + 3pqr – 2sqr – 3sr²
Group the first two terms and last two terms.
First two terms are 2pq² + 3pqr and the last two terms are – 2sqr – 3sr².
(2pq² + 3pqr) – ( 2sqr + 3sr²).
Find out the common factor from the above two groups.
pq(2q + 3r) – sr(2q + 3r).
Factor out the terms in terms of product.
(2q + 3r) (pq – sr).

By factorizing the expression 2pq² + 3pqr – 2sqr – 3sr², we will get (2q + 3r) (pq – sr).

(iv) par² + qars – pnrs – qns²

Solution:
The given expression is par² + qars – pnrs – qns²
Group the first two terms and last two terms.
First two terms are par² + qars and the last two terms are – pnrs – qns².
(par² + qars) – (pnrs – qns²).
Find out the common factor from the above two groups.
ar(pr + qs) – ns(pr + qs).
Factor out the terms in terms of product.
(pr + qs) (ar – ns).

By factorizing the expression par² + qars – pnrs – qns², we will get (pr + qs) (ar – ns).

5. Factorize
(i) (a + b) (2a + 5) – (a + b) (a + 3).

Solution:
The given expression is (a + b) (2a + 5) – (a + b) (a + 3).
Find out the common factor from the above expression.
(a + b) [(2a + 5) – (a + 3)].
Expand the terms.
(a + b)(2a + 5 – a – 3).
Simplify the second term.
(a + b) (a + 2).

By factorizing the expression (a + b) (2a + 5) – (a + b) (a + 3), we will get (a + b)(a + 2).

(ii) 6xy – y² + 12xz – 2yz.

Solution:
The given expression is 6xy – y² + 12xz – 2yz.
Group the first two terms and last two terms.
First two terms are 6xy – y² and the last two terms are 12xz – 2yz.
(6xy – y²) + (12xz – 2yz).
Find out the common factor from the above two groups.
y(6x – y)+ 2z(6x – y).
Factor out the terms in terms of product.
(6x – y) (y + 2z).

By factorizing the expression 6xy – y² + 12xz – 2yz, we will get (6x – y) (y + 2z).

Factorization when Binomial is Common | What is a Common Binomial?

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Students can find the factors for an algebraic expression when one of its factors is binomial. Find given algebraic expression Factorization when Binomial is Common Factor. All related problems are included in this article along with a clear explanation. Therefore, students can practice every problem available in this article and improve their knowledge. The process of solving the factorization problem is very simple if you follow the procedure we explained below. Go through the complete article and learn different methods to solve factorization problems.

Factorization of Algebraic Expressions when Binomial is Common

Follow the simple and easy guidelines on Factorization of Algebraic Expressions When Binomial is Common. They are as follows

Step 1: In the first step, find the common binomial factor.
Step 2: Note down the given expression as the product of this binomial and the quotient obtained on dividing the given expression by this binomial.

Solved Examples on Factorization When Binomial is a Common

1. Factorize the algebraic expressions.

(i) 7b(3x – 4y) + 3a(3x – 4y)

Solution:
Given expression is 7b(3x – 4y) + 3a(3x – 4y)
In the given expression, the binomial factor is (3x – 4y) as it is common in both terms.
Take the (3x – 4y) common and multiply it with the remained terms.
(3x – 4y) (7b + 3a)

The final answer is (3x – 4y) (7b + 3a)

(ii) 12(9a + 6b)² – 4(9a + 6b)

Solution:
Given expression is 12(9a + 6b)² – 4(9a + 6b)
12 (9a + 6b) (9a + 6b) -4(9a + 6b)
In the given expression, the binomial factor is (9a + 6b) as it is common in both terms.
Take the (9a + 6b) common and multiply it with the remained terms.
(9a + 6b)(12(9a + 6b) – 4)
(9a + 6b)(108a + 72b – 48)
The final answer is (9a + 6b)(108a + 72b – 48)

2. Factorize the expression 10r(m – 2n) – 8m + 16n

Solution:
Given expression is 10r(m – 2n) – 8m + 16n
Lets take -8m  + 16n from the above equation.
Take -8 common from -8m + 16n
-8(m – 2n)
Place -8(m – 2n) in 10r(2m – 4n) – 8m + 16n equation.
10r(m – 2n) -8(m – 2n)
In the above expression, the binomial factor is (m – 2n) as it is common in both terms.
Take the (m – 2n) common and multiply it with the remained terms.
(m – 2n)(10r – 8)

The final answer is (m – 2n)(10r – 8)

3. Factorize (a – 4b)^2 – 7a + 28b

Solution:
(a – 4b)(a – 4b) – 7a + 28b
Given expression is (a – 4b)(a – 4b) – 7a + 28b
Lets take – 7a + 28b from the above equation.
Take -7 common from – 7a + 28b
-7(a – 4b)
Place -7(a – 4b) in (a – 4b)(a – 4b) – 7a + 28b equation.
(a – 4b)(a – 4b) – 7(a – 4b)
In the above expression, the binomial factor is (a – 4b) as it is common in both terms.
Take the (a – 4b) common and multiply it with the remained terms.
(a – 4b)(a – 4b – 7)

The final answer is (a – 4b)(a – 4b – 7)

Binomial is a Common Factor | Factorisation when a Binomial is Common

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All solved examples of Factorization problems when Binomial is a common factor are included in this article. Check how to find the binomial factor and practice all problems given here. Compare the terms in the given expression and factor out the greatest common factor term from the expression. Based on the greatest common factor, we will get the solution for the expression that is in terms of the product of two or more terms.

How to Factorize taking out the Binomial Common Factor?

Go through the below step by step procedure and do Factorization when a Binomial is a Common Factor. They are along the lines

  1. Note down the given expression
  2. Note down the first term and second term from the expression
  3. Compare the two terms and observe the greatest common factor
  4. Now, factor out the greatest common factor from the expression
  5. Finally, we will get the result in the form of the product of two or more terms.

Solved Binomial Factor Examples

1. Factorize the expression (6x + 1)² – 5(6x + 1)?

Solution:
The given expression is (6x + 1)² – 5(6x + 1)
Here, the first term is (6x + 1)² and second term is – 5(6x + 1)
By comparing the above two terms, we can observe the greatest common factor and that is (6x + 1)
Now, factor out the greatest common factor from the expression
That is, (6x + 1) [(6x + 1) – 5]
(6x + 1)(6x – 4)

Therefore, the resultant value for the expression (6x + 1)² – 5(6x + 1) is (6x + 1)(6x – 4)

2. Factorize the algebraic expression 4x(y – z) + 3(y – z)?

Solution:
The given expression is 4x(y – z) + 3(y – z)
Here, the first term is 4x(y – z) and the second term is 3(y – z)
By comparing the above two terms, we can observe the greatest common factor and that is (y – z)
Now, factor out the greatest common factor from the expression
That is, (y – z)(4x + 3)

Therefore, the resultant value for the expression 4x(y – z) + 3(y – z) is (y – z)(4x + 3)

3. Factorize the expression (3x – 4y) (p – q) + (3x – 2y) (p – q)?

Solution:
The given expression is (3x – 4y) (p – q) + (3x – 2y) (p – q)
Here, the first term is (3x – 4y) (p – q) and the second term is (3x – 2y) (p – q)
By comparing the above two terms, we can observe the greatest common factor and that is (p – q)
Now, factor out the greatest common factor from the expression
That is, (p – q)[(3x – 4y) + (3x – 2y)]
= (p – q)(6x – 6y)

Therefore, the resultant value for the expression (3x – 4y) (p – q) + (3x – 2y) (p – q) is (p – q)(6x – 6y)

4. Factorize the expression 6(2x + y)² +8(2x + y)?

Solution:
The given expression is 6(2x + y)² +8(2x + y)
Here, the first term is 6(2x + y)² and the second term is 8(2x + y)
By comparing the above two terms, we can observe the greatest common factor and that is (2x + y)
Now, factor out the greatest common factor from the expression
That is, (2x + y)[6(2x + y) + 8]
(2x + y)(12x + 6y + 8)

Therefore, the resultant value for the expression 6(2x + y)² +8(2x + y) is (2x + y)(12x + 6y + 8)