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

Alena

14 June, 09:34

+4

Answers (1)

Know the Answer?

Vaughn Lozano · Answer 1

Vaughn Lozano 14 June, 12:42

0

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