Finding value of the function $f(x)f(y) = f(x) + f(y) + f(xy) - 2$? So here is the question 

If $f(x)$ is a polynomial satisfying $$f(x)f(y) = f(x) + f(y) + f(xy) - 2$$ for all real $x$ and $y$  and $f(3) = 10$, then $f(4)$ is equal to ?

here what i have tried

Putting $x=y=1$ in the given solution,$$(f(1)^2) = 3f(1) - 2$$ on solving it we get $$f(1) = 2$$
  or$$f(1)= 1$$ so putting $y=1$in the eqation $$f(x)f(y) = f(x) + f(y) + f(xy) - 2$$we get $$f(x) = 1$$ but it not true as $$f(3) = 10$$ so $f(1) = 2$

i am stuck here i don't know what to do next please help me
Akash
Thanks

Your equation translates to $(f(x)-1)(f(y)-1)=f(xy)-1$. Define $g(x)=f(x)-1$ and you obtain that $g(x)g(y)=g(xy)$. This can be treated as a Cauchy functional equation. 
Define $h(x)=\log(g(e^x))$. Then $h(x+y)=\log(g(e^x e^y))=\log(g(e^x))+\log(g(e^y))=h(x)+h(y)$. Since $h$ is continuous the only solution has the form $h(x)=ax$ so
$$ \log(g(e^x))=ax \Leftrightarrow g(e^x)=e^{ax} \Leftrightarrow g(x)=x^a (x>0). $$
