how do you factor 48w^5-3w completely

Question

how do you complete the following polynomials completely 

Answer 1

48w^5-3w=3w(16w^4-1)

3w(16w^4-1)=3w(4w^2-1(4w^2+1)

3w(4w^2-1)(4w^2+1)=3w(2w-1)(2w+1)(4w^2+1)

If you want complex numbers (4w^2+1)=(2w+i)(2w-i)because i^2=-1

Answer 2

48 w^5 - 3 w =

 

3W ( 16 W⁴ - 1 ) =

 

3W ( 4W^2 - 1) ( 4 W^2 +1 ) =

 

3W ( 2W + 1 ) ( 2W - 1) ( 4W^2 +1 )