Đáp án:
$a = - 23;b = 2$
Giải thích các bước giải:
Ta có:
$\begin{array}{l}
12{x^4} - 11{x^3} + a{x^2} + bx - 12\\
= 4{x^2}\left( {3{x^2} + x - 6} \right) - 5x\left( {3{x^2} + x - 6} \right) + \dfrac{{a + 29}}{3}\left( {3{x^2} + x - 6} \right) + \left( {b - \dfrac{{a + 29}}{3}} \right)x + 2\left( {a + 29} \right) - 12\\
= \left( {3{x^2} + x - 6} \right)\left( {4{x^2} - 5x + \dfrac{{a + 29}}{3}} \right) + \left( {b - \dfrac{{a + 29}}{3}} \right)x + 2\left( {a + 29} \right) - 12
\end{array}$
Để $\left( {12{x^4} - 11{x^3} + a{x^2} + bx - 12} \right) \vdots \left( {3{x^2} + x - 6} \right)$
$\begin{array}{l}
\Leftrightarrow \left( {b - \dfrac{{a + 29}}{3}} \right)x + 2\left( {a + 29} \right) - 12 = 0\\
\Leftrightarrow \left\{ \begin{array}{l}
b - \dfrac{{a + 29}}{3} = 0\\
2\left( {a + 29} \right) - 12 = 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
b = \dfrac{{a + 29}}{3}\\
a = - 23
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
b = 2\\
a = - 23
\end{array} \right.
\end{array}$
Vậy $a = - 23;b = 2$
