Which equation shows p (x) = x^6 - 1 factored completely over the integers? (Hint : You will need to use more than one method to complete this problem.)

A. p (x) = (x - 1) (x^2 + x + 1) (x + 1) (x^2 - x + 1)

B. p (x) = (x^3 - 1) (x^3 - 1)

C. p (x) = (x^2 - 1) (x^4 + x^2 + 1)

D. p (x) = (x - 1) (x + 1) (x^4 + x^2 + 1)

Question

Beckham Copeland

12 February, 09:38

+4

Answers (1)

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Kirsten Conley · Answer 1

Kirsten Conley 12 February, 11:58

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