2007-11-20 16:37:01 · 10 answers · asked by bebixiu 1 in Science & Mathematics ➔ Mathematics
(x^4+8)(x^4-8)
2007-11-20 18:41:20 · answer #1 · answered by Internet Explorer 2 · 0⤊ 0⤋
x^8 - 64 =
(x^4 + 8)(x^4 - 8)
2007-11-21 00:42:28 · answer #2 · answered by Philo 7 · 2⤊ 0⤋
Part of your instructions should say "factor over the integers" or "factor over the reals"
If it's 'integers', all the x^4+8 and x^4-8 are right.
If it's 'reals', then you need to keep going until all eight factors have x's to the first power, and your 8's will be fourth roots.
2007-11-21 00:58:23 · answer #3 · answered by Lana B 1 · 0⤊ 0⤋
=(x^4-8)(x^4+8)
2007-11-23 05:21:53 · answer #4 · answered by Sumita T 3 · 0⤊ 0⤋
(x^4+8)(x^4-8)
2007-11-21 00:42:18 · answer #5 · answered by Stinkypuppy 3 · 1⤊ 0⤋
(x^4-8)(x^4+8)
2007-11-21 00:44:57 · answer #6 · answered by dinesh320 2 · 2⤊ 0⤋
(x^4-8)(x^4+8)
2007-11-21 00:42:31 · answer #7 · answered by Small Victories 4 · 2⤊ 0⤋
x^8-64
=(x^4 + 8)(x^4 - 8)
=(x^4 + 8)(x^2 + 8^(1/2))(x^2 - 8^(1/2))
=(x^4 + 8)(x^2 + 8^(1/2))(x + 8^(1/4))(x - 8^(1/4))
also
=(x^4 + 2^3)(x^2 + 2^(3/2))(x + 2^(3/4))(x - 2^(3/4))
roots are x = +/- 2^(3/4)
2007-11-21 00:51:42 · answer #8 · answered by teleksterling 1 · 0⤊ 2⤋
great question...i wish that i could answer it, but i barely passed my college math class, however, i wanted to comment that the problem is a good one. good luck
2007-11-21 00:50:04 · answer #9 · answered by msbalisong 4 · 0⤊ 1⤋
x+8
x -8
2007-11-21 00:39:53 · answer #10 · answered by towanda 7 · 0⤊ 6⤋