Theo đề, ta có hệ phương trình:
\(\left\{{}\begin{matrix}a\cdot4-2b+c=0\\a\cdot4+2b+c=0\\a-c=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4a-2b+c=0\\4a+2b+c=0\\a+0b-c=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{3}{5}\\b=0\\c=-\dfrac{12}{5}\end{matrix}\right.\)