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Factor completely 16x^8 − 1.

(4x^4 − 1)(4x^4 + 1)
(2x^2 − 1)(2x^2 + 1)(4x^4 + 1)
(2x^2 − 1)(2x^2 + 1)(2x^2 + 1)(2x^2 + 1)
(2x^2 − 1)(2x^2 + 1)(4x^4 − 1)

1 Answer

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Answer:

16x^8 − 1 = (4x^4 + 1)(2x^2 + 1)(2x^2 - 1)

Explanation:

The difference of squares:

((ax)^2n - b) = (ax^n + b)(ax^n - b)

16x^8 − 1 = (4x^4 + 1)(4x^4 - 1)

(4x^4 - 1) = (2x^2 + 1)(2x^2 - 1)

16x^8 − 1 = (4x^4 + 1)(2x^2 + 1)(2x^2 - 1)

User MrWaqasAhmed
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