Question

a⁴-b⁴+2b²-1 factorise

Answer 1

 Best Answer

Q.1: 
i) 4x² - 4a²x + a⁴ - b⁴ = 0 

ii) Grouping, {(2x)² - 2(2x)(a²) + (a²)²} - (b²)² = 0 
==> (2x - a²)² - (b²)² = 0 

iii) The above is of the form a² - b² = (a - b)(a + b), 
here a = (2x - a²) and b = b² 

So, (2x - a²)² - (b²)² = {(2x - a²) + (b²)}*{(2x - a²) - (b²)} 
= {2x - (a² - b²)}*{2x - (a² + b²)} = 0 

==> Either {2x - (a² - b²)} = 0 or {2x - (a² + b²)} = 0 

So, when {2x - (a² - b²)} = 0, x = (a² - b²)/2 

and when {2x - (a² + b²)} = 0, x = (a² + b²)/2 

Answer 2

Hi ,

a⁴ - b⁴ + 2b² - 1

= a⁴ - ( b⁴ - 2b² + 1 )

= a⁴ - [ ( b² )² - 2 × b² × 1 + 1² ]

= ( a² )² - ( b² - 1 )²

= ( a² + b² - 1 ) [ a² - ( b² - 1 ) ]

= ( a² + b² - 1 ) ( a² - b² + 1 )

I hope this helps you.

:)