Giải thích các bước giải:
Ta có:
\[\begin{array}{l}
f\left( { - 2} \right) + f\left( 3 \right) = \left( {{{\left( { - 2} \right)}^2}a - 2b + c} \right) + \left( {{3^2}a + 3b + c} \right)\\
= \left( {4a - 2b + c} \right) + \left( {9a + 3b + c} \right)\\
= 13a + b + 2c = 0
\end{array}\]
Suy ra \[\left[ \begin{array}{l}
\left\{ \begin{array}{l}
f\left( { - 2} \right) > 0\\
f\left( 3 \right) < 0
\end{array} \right.\\
\left\{ \begin{array}{l}
f\left( { - 2} \right) < 0\\
f\left( 3 \right) > 0
\end{array} \right.
\end{array} \right. \Rightarrow f\left( { - 2} \right).f\left( 3 \right) < 0\]
