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Factorise:

x^4 - 256

2007-06-17 21:17:39 · 9 answers · asked by ilikebigengines 2 in Science & Mathematics Mathematics

9 answers

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 · answer #1 · answered by Wooly 4 · 0 0

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 · answer #2 · answered by Swamy 7 · 0 0

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 · answer #3 · answered by jurassicko 4 · 0 1

x^4=16^2
=(4^2)^2
=4^4 ,
x=4

2007-06-17 21:40:26 · answer #4 · answered by student 1 · 0 1

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 · answer #5 · answered by veeraa1729 2 · 1 0

here we go!

x^4=16^2
=(4^2)^2
=4^4 ,
x=4

2007-06-17 22:12:35 · answer #6 · answered by matahari 4 · 0 0

Take the 4th root of both sides...

4rt(x^4) = 4rt(256)
x = +/-4

2007-06-17 21:54:22 · answer #7 · answered by Anonymous · 0 1

x^4 = 256
x^3 = 64
x^2 = 16
x = 4

2007-06-21 20:09:31 · answer #8 · answered by Jun Agruda 7 · 2 0

take two times under root of this
x^4=256
so it will be 4

2007-06-17 21:23:39 · answer #9 · answered by ghulamalimurtaza 3 · 0 1

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