Function exercise asked in mathematics teacher group

 Exercise:
   
     Find a function $f\left( x \right)$ such that ${{x}^{2}}=2f\left( x \right)+f\left( 2-x \right)$\
   
     Solution: Let $u=2-x\Leftrightarrow u-2=-x\Leftrightarrow x=2-u$\
   
     So ${{\left( 2-u \right)}^{2}}=2f\left( 2-u \right)+f\left( u \right)\Leftrightarrow f\left( x \right)+2f\left( 2-x \right)={{\left( 2-x \right)}^{2}}$\
   
     $\left\{ \begin{aligned}
     & 2f\left( x \right)+f\left( 2-x \right)={{x}^{2}}\,\,\,\,\,\,\,\,\,\,\,\,\,\times \left( 2 \right) \\
     & f\left( x \right)+2f\left( 2-x \right)={{\left( 2-x \right)}^{2}}\,\,\,\times \left( -1 \right) \\
     \end{aligned} \right.$
   
     $\left\{ \begin{aligned}
     4f\left( x \right)+2f\left( 2-x \right) &=2{{x}^{2}} \\
     -f\left( x \right)-2f\left( 2-x \right)&=-{{\left( 2-x \right)}^{2}} \\
     \end{aligned} \right.$
   
     $\Rightarrow 3f\left( x \right)=2{{x}^{2}}-{{\left( 2-x \right)}^{2}}=2{{x}^{2}}-\left( 4+{{x}^{2}}-4x \right)={{x}^{2}}+4x-4={{x}^{2}}+4x+4-4-4$
   
     $\Rightarrow 3f\left( x \right)={{\left( x+2 \right)}^{2}}-16=\left( x+2-4 \right)\left( x+2+4 \right)=\left( x-2 \right)\left( x+6 \right)$
   
     Thus $f\left( x \right)=\frac{1}{3}\left( x-2 \right)\left( x+6 \right)$