15. A function $\mathrm{f}: \mathbf{R} \rightarrow \mathbf{R}$ satisfies the relation $\mathrm{f}(x) \mathrm{f}(y)-\mathrm{f}(2 x y+3)+3 \mathrm{f}(x+y)-3 \mathrm{f}(x)+6 x$, where $x, y \in \mathbf{R}$. Find the value of $\mathrm{f}(2009)$.
15. Answer: 4021
Given $\mathrm{f}(x) \mathrm{f}(y)=\mathrm{f}(2 x y+3)+3 \mathrm{f}(x+y)-3 \mathrm{f}(x)+6 x$, so if interchanging $\mathrm{x}$ and $y$ we have
$$
\mathrm{f}(y) \mathrm{f}(x)-\mathrm{f}(2 x y+3)+3 \mathrm{f}(x+y)-3 \mathrm{f}(y)+6 y \text {. }
$$

Subtracting, we have $-3 \mathrm{f}(x)+6 x=-3 \mathrm{f}(y)+6 y$ for all $x, y \in \mathbf{R}$, showing that $\mathrm{f}(x)-2 x$ is a constant, let it be $k$. So, $\mathrm{f}(x)=2 x+k$.
Substitute back to the given functional equation, we have
$$
\begin{array}{l}
(2 x+k)(2 y+k)=2(2 x y+3)+k+3[2(x+y)+k]-3(2 x+k)+6 x \\
4 x y+2 k x+2 k y+k^{2}=4 x y+6+k+6 x+6 y+3 k-6 x-3 k+6 x \\
2 k(x+y)-6(x+y)=k-k^{2}+6 \\
(k-3)(k+2)=2(x+y)(3-k) \\
(k-3)(k+2+2 x+2 y)=0 \text { for all } x, y \in \mathbf{R} .
\end{array}
$$

Thus $k=3$. Hence $\mathrm{f}(2009)=2(2009)+3=4021$.