Saudi Arabia Mathematical Competitions

Assume that $P(t)=a_n{t}^n+\dots +a_{1}t+a_0$, where $a_n\ne 0$. Taking $x=y=z$ in the given relation, it follows $$3P(x)+P(3x)=3P(2x).$$ Looking for the coefficient of $x^n$ in this equation, we obtain the relation $3+3^n=3\cdot 2^n$, that is $3^{n-1}=2^n-1$. This relation holds for $n=1$ and $n=2$. For $n\ge 3$ we prove by induction that $3^{n-1}>2^n-1$.

If $n=0$, that is $P$ is a constant polynomial $P(t)=c$, then it follows, for example from the above, that $c=0$.

If $n=1$, that is $P(t)=bt$, $b\ne 0$, then an easy checking shows that it satisfies the relation in the problem.

If $n=2$, we can choose $P(t)=a{t}^2$, $a\ne 0$, which also satisfies the relation.

Finally, we find that all desired polynomials are $$P(t)=a{t}^2+bt,$$ where $a,b\in \mathbb{R}$.