If g(x) satisfies 4g(x) + g(1-x) = 3x^{2}, what is what is g(x)?
I'm really stuck, what is a possible solution?
I expect since this particular combination of functions gives you a quadratic, you will probably find that g(x) is a quadratic. So let $\displaystyle \displaystyle \begin{align*} g(x) = a\,x^2 + b\,x + c \end{align*}$, then $\displaystyle \displaystyle \begin{align*} g(1 - x) = a(1 - x)^2 + b(1 - x) + c = a - 2a\,x + a\,x^2 + b - b\,x + c \end{align*}$. So
$\displaystyle \displaystyle \begin{align*} 4g(x) + g(1 - x) &= 3x^2 \\ 4a\,x^2 + 4b\,x + 4c + a - 2a\,x + a\,x^2 + b - b\,x + c &= 3x^2 \\ 5a\,x^2 + (3b - 2a)x + a + b + 5c &= 3x^2 + 0x + 0 \end{align*}$
So this means $\displaystyle \displaystyle \begin{align*} 5a = 3 \implies a = \frac{3}{5}, 3b - 2a = 0 \implies b = \frac{2}{5}, a + b + 5c &= 0 \implies c = -\frac{1}{5} \end{align*}$.
Therefore a function which satisfies your original equation is $\displaystyle \displaystyle \begin{align*} g(x) = \frac{3}{5}x^2 + \frac{2}{5}x - \frac{1}{5} \end{align*}$.
Simple solution:
Substitute $\displaystyle x\to 1-x$...you get a second equation, eliminate g(1-x) between them to obtain g(x):
$\displaystyle g(x)\text{:=}\frac{1}{15} \left(9 x^2+2 x-3\right)$
Replace $\displaystyle x$ with $\displaystyle 1-x$ to obtain
$\displaystyle 4g(1-x) + g(x) = 3(1-x)^2$. We already know
$\displaystyle 4g(x) + g(1-x) = 3x^2$
Multiply the second equation by -4 and add them.
$\displaystyle -15g(x) = 3(1-x)^2 - 12x^2$
Divide both sides by -15 and simplify.