Đáp án:
$\begin{array}{l}
a)6{x^2} + 5ax - 4\\
= 6{x^2} - 12x + \left( {5a + 12} \right).x - 2\left( {5a + 12} \right)\\
+ 2\left( {5a + 12} \right) - 4\\
= 6x\left( {x - 2} \right) + \left( {5a + 12} \right)\left( {x - 2} \right) + 10a + 20\\
= \left( {6x + 5a + 12} \right)\left( {x - 2} \right) + 10a + 20\\
Do:\left( {6x + 5a + 12} \right)\left( {x - 2} \right) \vdots \left( {x - 2} \right)\\
\Rightarrow 10a + 20 = 10\\
\Rightarrow a = - 1\\
b)f\left( x \right)\\
= {x^4} + 5{x^3} - 2{x^2} + a.x + 40\\
= {x^4} - 3{x^3} + 2{x^2} + 8{x^3} - 24{x^2} + 16x\\
+ 20{x^2} - 60x + 40 + \left( {a + 44} \right).x\\
= \left( {{x^2} - 3x + 2} \right)\left( {{x^2} + 8x + 20} \right) + \left( {a + 44} \right).x\\
Do:f\left( x \right) \vdots \left( {{x^2} - 3x + 2} \right)\\
\Rightarrow \left( {a + 44} \right)x = 0\\
\Rightarrow a = - 44\\
\Rightarrow \text{thương} = {x^2} + 8x + 20\\
= {\left( {x + 4} \right)^2} + 4 \ge 4\\
\Rightarrow GTNN = 4\,khi:x = - 4
\end{array}$
