We know the identities

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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)

 

  1. Chapter 14 Class 8 Factorisation
  2. Concept wise

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.