Factorisation using identities

We know the identities

Let’s factorise using these identities

x ² + 4x + 4

x ² + 4x + 4

= x ² + 4 + 4x

= x ² + (2) ² + 2 × 2 × x

Using (a + b) ² = a ² + b ² + 2ab

Where a = x , b = 2

= (x + 2) ²

4x ² + 12x + 9

4x ² + 12x + 9

  = 4x ² + 9 + 12x

  = (2x) ² + 3 ² + 2 × 2x × 3

Using (a+b) ² = a ² + b ² + 2ab

Where a = 2x, b = 3

  = (2 x + 3 ) ²

x ² − 4x + 4

x ² – 4x + 4

= x ² + 4 – 4x

= x ² + (2) ² – 2 × 2 × x

Using (a - b) ² = a ² + b ² – 2ab

Where a = x, b = 2

= (x - 2) ²

4x ² – 12x + 9

4x ² – 12x + 9

           = 4x ² + 9 – 12x

           = (2x) ² + 3 ² – 2 × 2x × 3

Using (a - b) ² = a ² + b ² – 2ab

Where a = 2x, b = 3

            = (2 x - 3 ) ²

4x ² − 9

4x ² − 9

= (2x) ² − 3 ²

Using a ² - b ² = (a - b)  (a + b)

Where a = 2x, b = 3

  = (2𝑥 − 3) (2𝑥 + 3)

y ⁴ − 16

y ⁴ − 16

= (y ² ) ² − (4) ²

Using a ² - b ² = (a - b) (a + b)

Where a = y ² , b = 4

= (y ² - 4) (y ² + 4)

= (y ² - 2 ² ) (y ² + 4)

Using a ² - b ² = (a - b)  (a + b)

Where a = y ² , b = 4

  = (y − 2) (y + 2) (y ² + 4)