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
\((2x - y)(2x + y)(4x^2 - y^2)\)
Verified Correct Answer: A.
Step-by-Step Mathematical Solution
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
= \((2x - y)(2x + y)((2x)^2 + y^2)\)
∴ \((2x - y)(2x + y)(4x^2 + y^2)\)
Source: WAEC / JAMB / NECO Classroom Archive (Record #67282)