Giải thích các bước giải:
Ta có:
\[\begin{array}{l}
f,\\
3{x^3} + a{x^2} + bx + 9 = 3\left( {{x^3} - 9x} \right) + a\left( {{x^2} - 9} \right) + \left( {\left( {b + 27} \right)x + 9a + 9} \right)\\
= \left( {{x^2} - 9} \right)\left( {3x + a} \right) + \left( {\left( {b + 27} \right)x + 9a + 9} \right)\\
\Rightarrow \left( {3{x^3} + a{x^2} + bx + 9} \right) \vdots \left( {{x^2} - 9} \right) \Leftrightarrow \left( {\left( {b + 27} \right)x + 9a + 9} \right) \vdots \left( {{x^2} - 9} \right)\\
\Rightarrow \left\{ \begin{array}{l}
b + 27 = 0\\
9a + 9 = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
b = - 27\\
a = - 1
\end{array} \right.\\
g,\\
{x^4} + a{x^3} + bx - 1 = \left( {{x^4} - 1} \right) + x\left( {a{x^2} + bx} \right) = \left( {{x^2} - 1} \right)\left( {{x^2} + 1} \right) + ax\left( {{x^2} - 1} \right) + \left( {a + b} \right)x\\
= \left( {{x^2} - 1} \right)\left( {{x^2} + 1 + ax} \right) + \left( {a + b} \right)x\\
\Rightarrow \left( {{x^4} + a{x^3} + bx - 1} \right) \vdots \left( {{x^2} - 1} \right) \Leftrightarrow \left( {a + b} \right)x \vdots \left( {{x^2} - 1} \right)\\
\Rightarrow a + b = 0 \Leftrightarrow a = - b
\end{array}\]
h, thiếu đề
