2006-06-22 17:29:29 · 8 answers · asked by riel_carlo 2 in Science & Mathematics ➔ Mathematics
I mean factor the expression to make it a simpler one... Or you could also say it's equal to zero
2006-06-22 17:39:25 · update #1
I don't think he asked for a solution of an eqn. but for factors
x^4 + y^4 = (x^2 + y^2)^2 - 2.x^2.y^2
RHS can be expressed as difference of 2 squares
And so on. I don't feel like doing the rest. you get the idea.
2006-06-22 18:14:55 · answer #1 · answered by The_Dark_Knight 4 · 1⤊ 5⤋
This can be factored over the complex numbers.
Setting x^4+y^4 = 0,
x^4 = -y^4
or x^2 = ± i*y^2
or x = ±(â2/2)(1 ± i)y ;
i.e.,
x^4 + y^4 =
[x + (â2/2)(1+i)y] * [x - (â2/2)(1+i)y] *
[x + (â2/2)(1-i)y] * [x - (â2/2)(1-i)y]
which is the factorization you requested.
2006-06-23 01:51:17 · answer #2 · answered by Scott R 6 · 0⤊ 0⤋
It cannot be written any simpler.
However, you can factor this over the non-reals.
x^4 + y^4
= (x^2 + i*y^2)(x^2 - i*y^2)
where i = sqrt(-1)
The only real solution to x^4 + y^4 = 0 is (0,0).
2006-06-23 00:46:49 · answer #3 · answered by MsMath 7 · 0⤊ 0⤋
There are no factors of the the expression given.
2006-06-23 00:34:50 · answer #4 · answered by bashah1939 4 · 0⤊ 0⤋
x^4+y^4=0 then
x=0
y=0
put the values and get the answer
2006-06-23 00:45:11 · answer #5 · answered by alooo... 4 · 0⤊ 0⤋
Try Differentiating it once with respect to x n again with y
2006-06-23 15:12:55 · answer #6 · answered by Drixx 1 · 0⤊ 0⤋
Start by getting the equation right. What you have exists only at (0,0)
2006-06-23 00:33:20 · answer #7 · answered by Bill S 6 · 0⤊ 0⤋
No more factors... It is already in its simplest form...
2006-06-23 03:25:07 · answer #8 · answered by chandyman21 3 · 0⤊ 0⤋