Brent is right if the situation is a^2 + b^2 and 2ab is not a perfect square. But here the case is all different. Our 2ab is a perfect square; hence we can go for it.Aman verma wrote:Q: Factorize (x^4 + 4 ):
(A) (x^2 + 2)^2
(B) (x^2 + 2x + 2)(x^2 - 2x + 2 )
(C) (x^2 + 2)(x^2 - 2)
(D) (x^2 + 4)(x^2 - 4)
(E) (x^2 - 1)(x^2 + 1)
NB: Now Brent has earlier mentioned in my previous post that sum of squares can not be factored. I would like to know how much that proposition is applicable to this problem.
Plugging values is the best approach when stuck for a missing concept. That has already been shown by few.
For your curiosity...
x^4 + 4
= x^4 + 4x^2 + 4 - 4x^2
= (x^2 + 2)^2 - (2x)^2 {this changed to a^2 - b^2 form}
= (x^2 + 2 - 2x)( x^2 + 2 + 2x)

















