Q(2)=3 nên \(a\cdot2^2+b\cdot2+c=3\)
=>4a+2b+c=3(1)
Q(-1)=6
=>\(a\cdot\left(-1\right)^2+b\cdot\left(-1\right)+c=6\)
=>a-b+c=6(2)
Q(1)=0
=>\(a\cdot1^2+b\cdot1+c=0\)
=>a+b+c=0(3)
Từ (1),(2),(3) ta có hệ phương trình:
\(\left\{{}\begin{matrix}4a+2b+c=3\\a-b+c=6\\a+b+c=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}4a+2b+c=3\\4a-4b+4c=24\\4a+4b+4c=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6b-3c=-21\\-2b-3c=3\\a+b+c=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}8b=-24\\-2b-3c=3\\a+b+c=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}b=-3\\3c=-2b-3=-2\cdot\left(-3\right)-3=6-3=3\\a+b+c=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}b=-3\\c=1\\a=-b-c=-\left(-3\right)-1=3-1=2\end{matrix}\right.\)
