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25 February, 13:03

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)

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Answers (1)
  1. 25 February, 15:19
    0
    16x^8 - 1 = (4x^4 + 1) (2x^2 + 1) (2x^2 - 1)

    Step-by-step 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)
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