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

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) a2Â + 2ab + b2Â = (a + b)2Â = (a + b) (a + b)
(ii) a2Â – 2ab + b2Â = (a – b)2Â = (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 + 40abÂ is in the form a2Â + 2ab + b2.
So find the factors of given expression using a2Â + 2ab + b2Â = (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 â€“ 42xy + 49y2

Solution:
Given expression is 9x2 â€“ 42xy + 49y2
The given expression 9x2 â€“ 42xy + 49y2Â 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 â€“ 42xy + 49y2Â are (9x – 7y) (9x – 7y)

(iii) 25m2 + 80m + 64

Solution:
Given expression is 25m2 + 80m + 64
The given expression 25m2 + 80m + 64Â is in the form a2Â + 2ab + b2.
So find the factors of given expression using a2Â + 2ab + b2Â = (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 + 64Â 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 + 9Â is in the form a2Â + 2ab + b2.
So find the factors of the expression using a2Â + 2ab + b2Â = (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) 4x4Â + 1

Solution:
Given expression is 4x4Â + 1.
The Given expression is 4x4Â + 1Â is not a perfect square.
Add and subtract 4xÂ² to make the given expression 4x4Â + 1Â is not a perfect square.
4x4 + 1 + 4xÂ² – 4xÂ²
4x4 + 4xÂ² + 1 – 4xÂ²
The above expression 4x4 + 4xÂ² + 1Â is in the form a2Â + 2ab + b2.
So find the factors of the expression using a2Â + 2ab + b2Â = (a + b)2 = (a + b) (a + b) where a = 2xÂ², b = 1
Apply the formula and substitute the a and b values.
4x4 + 4xÂ² + 1
(2xÂ²)2 + 2 (2xÂ²) (1) + (1)2
(2xÂ² + 1)2
(2xÂ² + 1)2 – 4xÂ²
(2xÂ² + 1)2 – (2x)2
(2xÂ² + 1 + 2x) (2xÂ² + 1 – 2x)
(2xÂ² + 2x + 1) (2xÂ² – 2x + 1)

Factors of the 4×4 + 1 are (2xÂ² + 2x + 1) (2xÂ² – 2x + 1)

(ii) (a + 2b)2 + 2(a + 2b) (3b â€“ a) + (3b – a)2

Solution:
Given expression is (a + 2b)2 + 2(a + 2b) (3b â€“ a) + (3b – a)2
The given expression (a + 2b)2 + 2(a + 2b) (3b â€“ a) + (3b – a)2Â is in the form a2Â + 2ab + b2.
So find the factors of given expression using a2Â + 2ab + b2Â = (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 â€“ a) + (3b – a)2
(a + 2b)2 + 2 (a + 2b) (3b â€“ a) + (3b – a)2
(a + 2b + 3b – a)2
(5b)2
25b2

Factors of the (a + 2b)2 + 2(a + 2b) (3b â€“ a) + (3b – a)2Â are 25b2

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