If $f : \mathbb R \rightarrow \mathbb R $ such that $f(x^2+x)+2f(x^2-3x+2) = 9x^2-15x$. Find $f(2016)$.
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 $x=-6, x=7$).
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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 $f(2016)=3\cdot 2016-4 = 6044.$
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.
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, $$4f(2016) - f(2016) = 2(18144) - 18156$$
$$\boxed{f(2016) = 6044}$$