using the principle of perfect squares
( where a^2 - b^2 = (a + b)(a - b) )
x^4 - 256
= (x^2 - 16)(x^2 + 16)
= (x - 4)(x + 4)(x^2 + 16)
2007-06-17 21:25:19
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answer #1
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answered by Wooly 4
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x^4 - 256 = x^2 + 16 and x^2 - 16 are two factors.
x^2 - 16 = x + 4 and x -4 are the factors.
So, we can write:
x^4 - 256 = (x + 4) (x - 4) (x^2 + 16)
2007-06-17 22:06:05
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answer #2
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answered by Swamy 7
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x^4 - 256
let x^4 - 256 = 0
x^4 = 256
(x²)² = 16 x 16
x² = square root (16 x 16) = 4 x 4 = 16
x = square root (16) = +/- 4
2007-06-17 21:52:20
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answer #3
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answered by jurassicko 4
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x^4=16^2
=(4^2)^2
=4^4 ,
x=4
2007-06-17 21:40:26
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answer #4
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answered by student 1
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using the formula a^2 - b^2 = (a+b)(a-b),
we have x^4 - 256 = (x^2)^2 - (4^2)^2 = (x^2 + 16)(x^2 - 16)
using again the formula,
= (x + 4i)(x-4i)(x+4)(x-4)
where two factors are complex factors.
2007-06-17 22:00:56
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answer #5
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answered by veeraa1729 2
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here we go!
x^4=16^2
=(4^2)^2
=4^4 ,
x=4
2007-06-17 22:12:35
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answer #6
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answered by matahari 4
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Take the 4th root of both sides...
4rt(x^4) = 4rt(256)
x = +/-4
2007-06-17 21:54:22
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answer #7
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answered by Anonymous
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x^4 = 256
x^3 = 64
x^2 = 16
x = 4
2007-06-21 20:09:31
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answer #8
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answered by Jun Agruda 7
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take two times under root of this
x^4=256
so it will be 4
2007-06-17 21:23:39
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answer #9
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answered by ghulamalimurtaza 3
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