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) 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