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

Question

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.

Answer 1

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.

Answer 2

Hint.

As $x^2-3x+2 = (x-2)^2+(x-2)$ calling $F(x) = f(x^2+x)$ we have

$$ F(x)+2F(x-2)=3x(3x-5) $$

Answer 3

We have the equation $f(x^2+x)+2f(x^2-3x+2)=9x^2-15x$.

Replacing $x$ by $1-x$, we get:

$$f(x^2-3x+2) + 2f(x^2+x)= 9x^2-3x -6$$

Eliminating $f(x^2-3x+2)$ from the equation

$$3f(x^2+x)=9x^2+9x-12$$

So:
$$f(x^2+x)=3(x^2+x)-4$$

Putting $x^2+x=2016$:

$$3 \cdot 2016 - 4 = 6044$$