F(x) holds for all real numbers, and so does y=0, that is, f (x)-f (0) = (x+ 1) x and f (x) = x 2+x-2.

* * * Attachment: However, it is also valid to take x=0, that is, f (y)-f (0) = 0 and f (y) =-2.

We know that the function representation has nothing to do with the sign of variables, so it can also be written as f(x)=-2, which is obviously inconsistent with the conclusion deduced above, so this problem has loopholes.

I think the condition should be changed to "define all non-zero real numbers: f(x+y)-f(y)=(x+2y+ 1)x, and f (1) = 0, and f (0) =-2".

Therefore, let y= 1, x≦- 1, f (x+1)-f (1) = f (x+1) = x2+3x = (x+66)

This is the correct question and answer!