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)

Question

Andres Whitaker

Mathematics

26 May, 06:45

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)

+4

Answers (1)

Know the Answer?

Antoine Mckenzie · Answer 1

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)

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)

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)