Ta có: a = c + 3
\(f\left(x\right)=ax^2+bx+c\)
\(\Rightarrow f\left(-2\right)=4a+2b+c=0\)
\(\Rightarrow f\left(2\right)=4a-2b+c=0\)
\(\Rightarrow\left\{{}\begin{matrix}8a+2c=0\\4b=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}4a+c=0\\b=0\end{matrix}\right.\)
Thay a = c + 3 vào 4a + c = 0 ta có:
\(4c+12+c=0\)
\(\Rightarrow c=-2,4\)
\(\Rightarrow a=0,6\)
Vậy bộ số \(\left(a;b;c\right)\) là \(\left(0,6;0;-2,4\right)\)