The function $f : \mathbb{R} \to \mathbb{R}$ satisfies $f(x) f(y) = f(x + y) + xy$for all real numbers $x$ and $y.$ Find all possible functions $f.$ The function $f : \mathbb{R} \to \mathbb{R}$ satisfies
$f(x) f(y) = f(x + y) + xy$ for all real numbers $x$ and $y.$ Find all possible functions $f.$
I started by letting $x = y = 0$, which made $f(0) = 1$.  I'm not sure what to do now. Any help is appreciated!! :)
Now, let $(x,y)=(1,-1).$
Thus, $f(1)f(-1)=0.$
If $f(1)=0$ so let $y=1.$
We obtain: $$f(x+1)+x=0$$ or $$f(x)=1-x$$ and easy to check that it's a solution.
If $f(-1)=0$ so let $y=-1$.
We obtain:
$$f(x-1)=x,$$ which gives $$f(x)=x+1$$ and we can check again that it's a solution.
Now set $y=-x$ which, since we have established that $f(0)=1$ gives $$f(x)f(-x) = 1-x^2 = (1+x)(1-x)$$ so $f(x) = 1+x$ and by symmetry $f(x) = 1-x$ are both solutions.