17. A real-valued function $\mathrm{f}$ satisfies the relation
$$
f\left(x^{2}+x\right)+2 f\left(x^{2}-3 x+2\right)=9 x^{2}-15 x
$$
for all real values of $x$. Find $f(2011)$.
17. Answer. 6029
Solution. Given the original equation
$$
f\left(x^{2}+x\right)+2 f\left(x^{2}-3 x+2\right)=9 x^{2}-15 x,
$$
we replace $x$ by $1-x$ and obtain
$$
\mathrm{f}\left(x^{2}-3 x+2\right)+2 \mathrm{f}\left(x^{2}+x\right)=9(1-x)^{2}-15(1-x)=9 x^{2}-3 x-6 .
$$

Eliminating $\mathrm{f}\left(x^{2}-3 x+2\right)$ from the two equations, we obtain
$$
3 \mathrm{f}\left(x^{2}+x\right)=9 x^{2}+9 x-12 \text {, }
$$
thereby
$$
\mathrm{f}\left(x^{2}+x\right)=3 x^{2}+3 x-4=3\left(x^{2}+x\right)-4,
$$
hence $f(2011)=3(2011)-4=6029$.