Sachin

Learn the process to Factorize the Trinomial ax Square Plus bx Plus c. One of the basic expressions for trinomial is ax² + bx + c. To find the ax² + bx + c factors, firstly, we need to find the two numbers and that is p and q. Here, the second term ‘b’ is the sum of the two numbers that is p + q = b. The product of the first and last terms is equal to the product of two numbers that is p * q = ac. Based on these two instructions, we need to find the values of p and q.

Steps to Factorize the Trinomial of Form ax² + bx + c?

1. Note down the given expression and compare it with the basic expression ax² + bx + c.
2. Note down the product and sum terms and find the two numbers.
3. Depends on the values of two numbers, expand the given expression.
4. Factor out the common terms.
5. Finally, we will get the product of two terms which is equal to the trinomial expression.

Solved Examples on Factoring Trinomials of the Form ax² + bx + c

1. Resolve into factors.

(i) 2s² + 9s + 10.

Solution:
The Given expression is 2s² + 9s + 10.
By comparing the given expression 2s² + 9s + 10 with the basic expression ax² + bx + c.
Here, a = 2, b = 9, and c = 10.
The sum of two numbers is p + q = b = 9 = 5 + 4.
The product of two number is p * q = a * c = 2 * 10 = 20 = 5 * 4.
From the above two instructions, we can write the values of two numbers p and q as 5 and 4.
Then, 2s² + 9s + 10 = 2s² + 5s + 4s + 20.
= 2s (s + 5) + 4 (s + 5).
Factor out the common terms.

Then, 2s² + 9s + 10 = (2s + 4) (s + 5).

(ii) 6s² + 7s –3

Solution:
The Given expression is 6s² + 7s – 3.
By comparing the given expression 6s² + 7s – 3 with the basic expression ax² + bx + c.
Here, a = 6, b = 7, and c = 3.
The sum of two numbers is p + q = b = 7 = 9 – 2.
The product of two number is p * q = a * c = 6 * 3 = 18 = 9 * 2.
From the above two instructions, we can write the values of two numbers p and q as 9 and 2.
Then, 6s² + 7s -3 = 6s² + 9s – 2s – 3.
= 6s² – 2s + 9s – 3.
= 2s (3s – 1) + 3(3s – 1).
Factor out the common terms.

Then, 6s² + 7s – 3 = (3s – 1) (2s + 3).

2. Factorize the trinomial.

(i) 2x² + 7x + 3.

Solution:
The Given expression is 2x² + 7x + 3.
By comparing the given expression 2x² + 7x + 3 with the basic expression ax² + bx + c.
Here, a = 2, b = 7, and c = 3.
The sum of two numbers is p + q = b = 7 = 6 + 1.
The product of two number is p * q = a * c = 2 * 3 = 6 = 6 * 1.
From the above two instructions, we can write the values of two numbers p and q as 6 and 1.
Then,2x² + 7x + 3 = 2x² + 6x + x + 3.
= 2x (x + 3) + (x + 3).
Factor out the common terms.

Then, 2x² + 7x + 3 = (x + 3) (2x + 1).

(ii) 3s² – 4s – 4.

Solution:
The Given expression is 3s² – 4s – 4.
By comparing the given expression 3s² – 4s – 4 with the basic expression ax² + bx + c.
Here, a = 3, b = – 4, and c = – 4.
The sum of two numbers is p + q = b = – 4 = – 6 + 2.
The product of two number is p * q = a * c = 3 * (- 4) = – 12 = (- 6) * 2.
From the above two instructions, we can write the values of two numbers p and q as – 6 and 2.
Then, 3s² – 4s – 4 = 3s² – 6s + 2s – 4.
= 3s (s – 2) + 2(s – 2).
Factor out the common terms.

Then, 3s² – 4s – 4 = (s – 2) (3s + 2).

Learn How to Factorize the Trinomial x² + px +q? A Trinomial is a three-term algebraic expression. By Factoring the trinomial expression, we will get the product of two binomial terms. Here, the trinomial expression contains three terms which are combined with the operations like addition or subtraction. Here, we need to find the coefficient values, and based on the coefficient values, we can find out the binomial terms as products of trinomial expression.

How to Factorize the Trinomial x² + px + q?

To find x^2 + px + q, we have to find the two terms (m + n) = p and mn = q.
Substitute (m + n) = p and mn = q in x^2 + px + q.
x^2 + px + q = x^2 + (m + n)x + mn.
By expanding the above expression, we will get
x^2 + px + q = x^2 + mx + nx + mn.
separate the common terms from the above expression.
that is, x(x + m) + n(x + m).
factor out the common term.
that is, (x + m) (x + n).
So, x^2 + px + q = (x + m)(x + n).

Factorization of Trinomial Steps

• Note down the given trinomial expression and compare the expression with the basic expression.
• Find out the product and sum of co-efficient values that is (m + n) and mn.
• Based on the above step, find out the two co-efficient values m and n.
• Finally, we will get the product of two terms which are equal to the trinomial expression.

Examples on Factoring Trinomials of Form x² + px + q

1. Resolve into factors

(i) a² + 3a -28

Solution:
Given Expression is a² + 3a -28.
Compare the a² + 3a -28 with the x^2 + px +q
Here, p = m + n = 3 and q = mn = -28
q is the product of two co-efficient. That is, 7 *(- 4) = -28
p is the sum of two co-efficient. That is 7 + ( – 4) = 3.
So, a² + 3a -28 = a² + [7 + (-4)]a – 28.
= a² + 7a – 4a – 28.
=a (a + 7) – 4(a + 7)
Factor out the common term.
That is, (a + 7) (a – 4).

Finally, the expression a² + 3a -28 = (a + 7) (a – 4).

(ii) a² + 8a + 15

Solution:
Given Expression is a² + 8a + 15.
Compare the a² + 8a + 15 with the x^2 + px +q.
Here, p = m + n = 8 and q = mn = 15.
q is the product of two co-efficient. That is, 5 * 3 = 15.
p is the sum of two co-efficient. That is 5 + 3 = 8.
So, a² + 8a + 15 = a² + (5 + 3)a + 15.
= a² + 5a + 3a + 15.
=a (a + 5) + 3(a + 5).
Factor out the common term.
That is, (a + 5) (a + 3).

Finally, the expression a² + 8a + 15= (a + 5) (a + 3).

2. Factorize the Trinomial

(i) a² + 15a + 56

Solution:
Given Expression is a² + 15a + 56.
Compare the a² + 15a + 56 with the x^2 + px +q.
Here, p = m + n = 15 and q = mn = 56.
q is the product of two co-efficient. That is, 7 * 8 = 56.
p is the sum of two co-efficient. That is 7 + 8 = 15.
So, a² + 15a + 56= a² + (7 + 8) a + 56.
= a² + 7a + 8a + 56.
=a (a + 7) + 8(a + 7).
Factor out the common term.
That is, (a + 7) (a + 8).

Finally, the expression a² + 15a + 56= (a + 7) (a + 8).

(ii) a² + a – 56

Solution:
Given Expression is a² + a – 56.
Compare the a² + a – 56 with the x^2 + px +q.
Here, p = m + n = 1 and q = mn = – 56.
q is the product of two co-efficient. That is, – 7 * 8 = – 56.
p is the sum of two co-efficient. That is – 7 + 8 = 1.
So, a² + a – 56 = a² + ( – 7 + 8)a – 56.
= a² – 7a + 8a – 56.
=a (a – 7) + 8(a – 7).
Factor out the common term.
That is, (a – 7) (a + 8).

Finally, the expression a^2 + a – 56 = (a – 7) (a + 8).

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.

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)

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]

If you are searching for help on Factorization of Perfect Square Trinomials you are the correct place. Check out all the problems explained in this article. We have given a detailed explanation for every individual problem along with correct answers. Also, we are providing all factorization concepts, worksheets on our website for free of cost. Therefore, learn every concept and improve your knowledge easily.

Learn to solve the given algebraic expressions using the below formulas.
(i) a² + 2ab + b² = (a + b)² = (a + b) (a + b)
(ii) a² – 2ab + b² = (a – b)² = (a – b) (a – b)

Factoring Perfect Square Trinomials Examples

1. Factorization when the given expression is a perfect square

(i) m⁴ – 10m²n² + 25n⁴

Solution:
Given expression is m⁴ – 10m²n² + 25n⁴
The given expression m⁴ – 10m²n² + 25n⁴ 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 = m², b = 5n²
Apply the formula and substitute the a and b values.
m⁴ – 10m²n² + 25n⁴
(m²)² – 2 (m²) (5n²) + (5n²)²
(m² – 5n²)²
(m² – 5n²) (m² – 5n²)

Factors of the m⁴ – 10m²n² + 25n⁴ are (m² – 5n²) (m² – 5n²)

(ii) b²+ 6b + 9

Solution:
Given expression is b²+ 6b + 9
The given expression b²+ 6b + 9 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 = b, b = 3
Apply the formula and substitute the a and b values.
b²+ 6b + 9
(b)² + 2 (b) (3) + (3)²
(b + 3)²
(b + 3) (b + 3)

Factors of the b²+ 6b + 9 are (b + 3) (b + 3)

(iii) p⁴ – 2p² q² + q⁴

Solution:
Given expression is p⁴ – 2p² q² + q⁴
The given expression p⁴ – 2p² q² + q⁴ 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 = p², b = q²
Apply the formula and substitute the a and b values.
p⁴ – 2p² q² + q⁴
(p²)² – 2 (p²) (q²) + (q²)²
(p² – q²)²
(p² – q²) (p² – q²)
From the formula (a² – b²) = (a + b) (a – b), rewrite the above equation.
(p + q) (p – q) (p + q) (p – q)

Factors of the p⁴ – 2p² q² + q⁴ are (p + q) (p – q) (p + q) (p – q)

2. Factor using the identity

(i) 25 – a² – 2ab – b²

Solution:
Given expression is 25 – a² – 2ab – b²
Rearrange the given expression as 25 – (a² + 2ab + b²)
a² + 2ab + b² = (a + b)² = (a + b) (a + b)
25 – (a + b)²
(5)²– (a + b)²
From the formula (a² – b²) = (a + b) (a – b), rewrite the above equation.
[(5 + a + b)(5 – a – b)]

(ii) 1- 2mn – (m² + n²)

Solution:
Given expression is 1- 2mn – (m² + n²)
1- 2mn – m² – n²
1 – (2mn + m² + n²)
1 – (m + n)²
(1)² – (m + n)²
From the formula (a² – b²) = (a + b) (a – b), rewrite the above equation.
(1 + m + n) (1 – m + n)

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 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 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).

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).