How to verify that x + y is a factor of x^2(y^2 – 1) – y^2(1 + x^2) + x^2 + y^2?

Answer 1

Start by simplifying the expression:
x^2(y^2 – 1) – y^2(1 + x^2) + x^2 + y^2

x^2*y^2 - x^2 -y^2 -x^2*y^2 + x^2 + y^2

the x^2 terms cancel. The y^2 terms cancel. the x^2*y^2 terms cancel. You are left with

0

Anything and everything is a factor of 0.

(x+y) * 0 = 0


This is a pretty degenerate problem.

Answer 2

The expression can be written as x^2(y^2 – 1) – y^2(1 + x^2) + x^2 + y^2 = x^2 y^2 - x^2 - y^2 - y^2 x^2 + x^2 + y^2 = 0. We have an identity. So, x + y is trivially a factor of the given expression, 0 = (x +y) 0

Answer 3

For simplicity mulitply x^2 and y^2 through there respectives pices of the equation:

(x^2 y^2) - x^2 - y^2 - (x^2 y^2) + x^2 + y^2

Which really, when the whole thing breaks down equals zero... But you can write it out as:
-x^2 -y^2 +x^2 +y^2
(-1 + 1)(x^2 +y^2)

Answer 4

let f(x) = x^2(y^2-1) - y^2(1+x^2) +x ^2 + y^2
= x^2(y^2-1) + x^2 - y^2(1+x^2) + y^2
= x^2(y^2-1+1)- y^2(1+x^2-1)
= x^2y^2 - y^2x^2 = 0

so x+y is a factor

Answer 5

uhhhhhhhhhhhh, hmmmmmmmmm, well, let's see...........
Don't know!
Good luck