x^4 - 1 (x to the 4th power minus 1)
not sure if I have it- thanks!!
2007-11-28 11:24:54 · 8 answers · asked by Smilingcheek 4 in Science & Mathematics ➔ Mathematics
x^4-1
(x^2-1)(x^2+1)
(x-1)(x+1)(x^2+1)
2007-11-28 11:28:44 · answer #1 · answered by Dave aka Spider Monkey 7 · 1⤊ 0⤋
difference of squares
(x^2 - 1)(x^2 + 1) the first factor is still the difference of squares
(x + 1)(x - 1)(x^2 +1)
2007-11-28 19:28:32 · answer #2 · answered by Linda K 5 · 1⤊ 0⤋
x^4-1
= (x^2)^2-(1)^2
= (x^2+1)(x^2-1)
= (x^2+1)(x+1)(x-1)
If you are factoring over C (the complex numbers), you can get
(x+1)(x-1)(x+i)(x-i)
2007-11-28 19:27:38 · answer #3 · answered by 7 · 1⤊ 0⤋
x^4-1=
(x^2-1) (x^2+1)
square root each, and you get
x-1 or x+1
I think, I'm not sure, but that's what I think it is.
2007-11-28 19:30:04 · answer #4 · answered by Anonymous · 1⤊ 0⤋
(x-1)(x+1)(x^2+1)=(x-1)(x+1)(x+i)(x-i)
hope this helps
2007-11-28 19:27:46 · answer #5 · answered by wictably 2 · 1⤊ 0⤋
[07]
x^4-1
=(x^2)^2-(1)^2
=(x^2-1)(x^2+1)
={(x)^2-(1)^2}(x^2+1)
=(x-1)(x+1)(x^2+1)
2007-11-28 19:29:08 · answer #6 · answered by alpha 7 · 2⤊ 0⤋
you can't factor it any lower
2007-11-28 19:30:50 · answer #7 · answered by Anonymous · 1⤊ 0⤋
I think that's the furthest you can go with it....
2007-11-28 19:32:19 · answer #8 · answered by Sophie 1 · 1⤊ 0⤋