How would you apply a factoring formula to completely factor the expression "x4 - 1"?
2007-10-23 05:00:09 · 7 answers · asked by Snoopy Baby 1 in Science & Mathematics ➔ Mathematics
How would you apply a factoring formula to completely factor the expression "x^4 - 1"?
2007-10-23 05:00:27 · update #1
Difference of squares formula:
(a² - b²) = (a + b)(a - b)
x^4 - 1 = (x² + 1)(x² - 1)
Next recognize that the factor x² - 1 is also a difference of squares, so that
x^4 - 1 = (x² + 1)(x² - 1) = (x² + 1)(x + 1)(x - 1)
2007-10-23 05:02:51 · answer #1 · answered by MamaMia © 7 · 3⤊ 3⤋
We know the algebraic identity:
a^2 - b^2 = (a + b)(a - b)
Here, a = x^2, b = 1
Applying the identity:
x^4 - 1 = (x^2 + 1)(x^2 - 1)
We can further factor (x^2 - 1) using the identity to get the answer as:
x^4 - 1 = (x^2 + 1)(x + 1)(x - 1)
If you want to, we can pitch this a bit further into the world of complex numbers.
i stands for iota.
i = sqrt (-1)
We can write (x^2 + 1) as [x^2 - (-1)]
This is also a difference of squares! We just saw that i = sqrt (-1) so i^2 = -1. Using i^2 = -1, we can rewrite (x^2 + 1) as:
(x^2 - i^2) = (x + i)(x - i)
So, applying our identity yet again, we arrive at the final answer:
x^4 - 1 = (x + i)(x - i)(x + 1)(x - 1)
2007-10-23 05:17:58 · answer #2 · answered by Akilesh - Internet Undertaker 7 · 0⤊ 0⤋
x^4-1 = (x^2-1)(x^2+1) = (x-1)(x+1)(x^2+1)
2007-10-23 05:03:12 · answer #3 · answered by qq5121971 2 · 2⤊ 2⤋
x^4 - 1= (x^2-1)*(x^2+1)=(x-1)*(x+1)*(x-i)*(x+i)
2007-10-23 05:14:40 · answer #4 · answered by ? 5 · 1⤊ 1⤋
That is the difference of squares, and one of its factors is the difference of squares.
2007-10-23 05:03:20 · answer #5 · answered by Anonymous · 2⤊ 3⤋
The answer is 69 trust me
2007-10-23 05:03:23 · answer #6 · answered by ben k 2 · 0⤊ 6⤋
well it would be 4x-1 i think
2007-10-23 05:04:24 · answer #7 · answered by baggieboy13 1 · 0⤊ 5⤋