If $f : \mathbb R \rightarrow \mathbb R $ such that $f(x^2+x)+2f(x^2-3x+2) = 9x^2-15x$. Find $f(2016)$. 
Determine all $f : \mathbb R \rightarrow \mathbb R $ such that $$f(x^2+x)+2f(x^2-3x+2) = 9x^2-15x$$ for all $x$. Find $f(2016)$.

Similar problem appeared on this site before: $f(x^2 + x)+2f(x^2 - 3x + 2)=9x^2 - 15x$ then find $f(2016)$. (The question is now deleted.) The same problem with finding $2011$ (instead of $2016$) appeared in 2011 Singapore Mathematical Olympiad as problem 17 (Wayback Machine).
I’ve tried put $x=0,1$ and got \begin{align*}
f(0)+2f(2)&=0\\
f(2)+2f(0)&=-6
\end{align*}
which gives me $f(0)=-4$, $f(2)=2$. 
Similarly, if we notice that $x^2+x=x^2-3x+2$ holds for $x=\frac12$, we can find the value at the point $\frac34=\left(\frac12\right)^2+\frac12$.
But the above doesn’t seem to help for other values. 
Thank you very much for helping.
 A: First (observation):
Note that we can determine $f(0), f(2)$ easily:
$$
x=0 \qquad \rightarrow \qquad f(0)+2f(2)=0;\\ 
x=1 \qquad \rightarrow \qquad f(2)+2f(0)=-6;
$$
so
$$
f(0)=-4,\quad f(2)=2.
$$
Same way we can determine $f(6), f(20)$ (substituting $x=-3, x=4$).
Same way we can determine $f(56), f(30)$ (substituting $x=-6, x=7$).
 ...
Second (solution):
Let's focus on $x=-a, x=a+1$, where $a\in\mathbb{R}$:
$$
x=-a \qquad \rightarrow \qquad f(a^2-a)+2f(a^2+3a+2) = 9a^2+15a; \\
x=a+1 \qquad \rightarrow \qquad f(a^2+3a+2)+2f(a^2-a) = 9a^2+3a-6;
$$
so (when denote $A=f(a^2-a)$, $B=f(a^2+3a+2)$):
$$
\left\{ \begin{array}{l}A+2B = 9a^2+15a; \\ B+2A = 9a^2+3a-6;\end{array} \right.$$
$$
\left\{ \begin{array}{l}B+A = 6a^2+6a-2;\\ B-A = 12a+6;\end{array} \right.
$$
and
$$
\left\{ \begin{array}{l}f(a^2-a) = A = 3a^2-3a-4; \\
f(a^2+3a+2) = B = 3a^2+9a+2. \end{array}\right.\tag{1}
$$
From $(1)$ we conclude that for each $z$ which can be written in the form
$$
z = a^2-a, \qquad a \in\mathbb{R} \tag{2}
$$
(in fact, for $z\ge -\frac{1}{4}$)
we have
$$
f(z) = 3z-4.
$$
Therefore $f(z)$ is linear function for $z\ge -\frac{1}{4}$.
Since $z=2016$ admits representation $(2)$, then $f(2016)=3\cdot 2016-4 = 6044.$
A: Replace $x$ by $1-x$ and then you can see how the equation transforms (I'll let you see it yourself).
Then you solve the equations. 
Tell me if you need more help.
A: First, we solve $x^2 + x = 2016$ and (separately) $x^2 - 3x + 2 = 2016$ and write down the solutions. Then observe that, luckily,
When $x = \dfrac{-1 - \sqrt{8065}}{2}$:


*

*$x^2 + x = 2016$

*$x^2 - 3x + 2 = 2020 + 2\sqrt{8065} = a$ (say)

*$9x^2 - 15x = 18156 + 12\sqrt{8065}$


$$f(2016) + 2f(a) = 18156 + 12\sqrt{8065}$$
When $x = \dfrac{3 + \sqrt{8065}}{2}$:


*

*$x^2 + x = a$

*$x^2 - 3x + 2 = 2016$

*$9x^2 - 15x = 18144 + 6\sqrt{8065}$.


$$f(a) + 2f(2016) = 18144 + 6\sqrt{8065}$$
From the two equations,
$$4f(2016) - f(2016) = 2(18144) - 18156$$
$$\boxed{f(2016) = 6044}$$
A: Consider a linear function $ f(x)=ax+b$
$$ f(x^2+x) = ax^2+ax+b$$
$$ f(x^2-3x+2)= ax^2-3ax +2a+b$$
$$ f(x^2+x)+2f(x^2-3x+2)=3ax^2-5ax +4a+3b = 9x^2 -15x$$
$$a=3, b=-4$$
$$ f(x) = 3x-4$$
$$f(2016)=6044$$
A: Assuming $f $ is a polynomial, consider the degree picture:
If $ \deg [f (x)]=n $, then $\deg [f (ax^2+bx+c)]=2n$, and on the RHS we have $\deg [9x^2+15x]=2$
So in solving $2n=2$, we have that the degree of $f=1$......This shows that you can assume $f $ takes the form
$$f (x)=ax+b $$
A: Hint.
As  $x^2-3x+2 = (x-2)^2+(x-2)$ calling $F(x) = f(x^2+x)$ we have
$$
F(x)+2F(x-2)=3x(3x-5)
$$
A: Denote: $x^2+x=a$. Then:
$$f(a)+2f(a-4x+2)=9a-24x.$$
Plug $x=\frac12$ to get:
$$f(a)+2f(a)=9a-12 \Rightarrow f(a)=3a-4.$$
Hence:
$$f(2016)=3\cdot 2016-4=6044.$$
