Ta có: \(f\left(x\right)=ax^2+bx+c\)
\(\Rightarrow f\left(0\right)=c⋮3\Rightarrow c⋮3\)
\(\left\{{}\begin{matrix}f\left(1\right)=a+b+c⋮3\\f\left(-1\right)=a-b+c⋮3\end{matrix}\right.\)
Mà \(c⋮5\)
\(\Rightarrow\left\{{}\begin{matrix}a+b⋮3\\a-b⋮3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2a⋮3\\2b⋮3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a⋮3\\b⋮3\end{matrix}\right.\) ( do \(\left(2;3\right)=1\) )
Vậy \(a,b,c⋮3\)