Factorize: \(16x^4 - y^4\)

A.

\((2x - y)(2x + y)(4x^2 + y^2)\)

B.

\((2x + y)(2x + y)(4x^2 + y^2)\)

C.

\((2x - y)(2x - y)(4x^2 + y^2)\)

D.

\((2x - y)(2x + y)(4x^2 - y^2)\)

Correct answer is A

16\(x^4 - y^4\)

= 2\(^4x^4 - y^4\)

= \((2x)^4 - y^4\)

= \(((2x)^2)^2 - (y^2)^2\)

Using a\(^2 - b^2\) = (a - b)(a + b) identity

= ((2x)\(^2 - y^2)((2x)^2 + y^2)\)

Using the identity one more time

= \((2x - y)(2x + y)((2x)^2 + y^2)\)

∴ \((2x - y)(2x + y)(4x^2 + y^2)\)