I just don't know how far can I simplify this question. x^8-1. Thank You very much.
2007-05-02 07:57:18 · 5 answers · asked by crobabe2182 1 in Science & Mathematics ➔ Mathematics
diff of squares:
x^8-1 = (x^4+1)(x^4-1)
x^4-1 = (x^2+1)(x^2-1)
x^2-1=(x+1)(x-1)
x^8-1 = (x^4+1)(x^2+1)(x+1)(x-1)
2007-05-02 08:02:07 · answer #1 · answered by holdm 7 · 0⤊ 0⤋
x^8-1
: perhaps the question asks factorization
x^8-1=(x^4+1))x^4 -1=(x^4+1))(x^2+1(x^2 -1)=(x^4+1)(x^2+1))(x+1(x-1) answer
2007-05-02 15:16:22 · answer #2 · answered by Anonymous · 0⤊ 0⤋
x^8 = (x^4)^2
so (x^4)^2 - 1 = (x^4 - 1)(x^4 + 1)
= (x^2 - 1)(x^2 + 1)(x^4 + 1)
= (x - 1)(x + 1)(x^2 + 1)(x^4 + 1)
2007-05-02 15:10:42 · answer #3 · answered by Anonymous · 0⤊ 0⤋
x^8-1
(x^4 + 1)(x^4 - 1)
(x^4 + 1)(x^2 + 1)(x^2 - 1)
(x^4 + 1)(x^2 + 1)(x + 1)(x - 1)
KrazyKyngeKorny
(Krazy, not stupid)
2007-05-02 15:02:43 · answer #4 · answered by krazykyngekorny 4 · 0⤊ 0⤋
(x^4-1)(x^4+1)
(x^2-1)(x^2+1)(x^4+1)
(x-1)(x+1)(x^2+1)(x^4+1)
(x^4+1)(x^2+1)(x-1)(x+1)
2007-05-02 15:03:13 · answer #5 · answered by Dave aka Spider Monkey 7 · 0⤊ 0⤋