Giải thích các bước giải:
$\begin{array}{l}
a)A = 4{x^3} + 15x + 24x + 3 + a\\
= 4x\left( {{x^2} + 4x + 7} \right) - \left( {{x^2} + 4x + 7} \right) + a + 10\\
= \left( {{x^2} + 4x + 7} \right)\left( {4x - 1} \right) + a + 10
\end{array}$
Để $A \vdots \left( {{x^2} + 4x + 7} \right)$
$\begin{array}{l}
\Leftrightarrow a + 10 = 0\\
\Leftrightarrow a = - 10
\end{array}$
Vậy $a = - 10$ thỏa mãn đề.
$\begin{array}{l}
b)A = {x^4} + 3{x^3} - {x^2} + \left( {2a - 3} \right)x + 3b + a\\
= {x^2}\left( {{x^2} + 3x - 1} \right) + \left( {2a - 3} \right)x + 3b + a
\end{array}$
Để $A \vdots \left( {{x^2} + 3x - 1} \right)$
$\begin{array}{l}
\Leftrightarrow \left( {2a - 3} \right)x + 3b + a = 0\\
\Leftrightarrow \left\{ \begin{array}{l}
2a - 3 = 0\\
3b + a = 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
a = \dfrac{3}{2}\\
b = \dfrac{{ - a}}{3}
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
a = \dfrac{3}{2}\\
b = \dfrac{{ - 1}}{2}
\end{array} \right.
\end{array}$
$ \Rightarrow \left( {a;b} \right) = \left( {\dfrac{3}{2};\dfrac{{ - 1}}{2}} \right)$
Vậy $\left( {a;b} \right) = \left( {\dfrac{3}{2};\dfrac{{ - 1}}{2}} \right)$ thỏa mãn đề.
