Đáp án:
\[a = - 4;\,\,\,b = 12\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
P\left( x \right) = {x^4} - 2{x^3} + 7{x^2} + ax + b - 2\\
= \left( {{x^4} + 2{x^2}} \right) - \left( {2{x^3} + 4x} \right) + \left( {5{x^2} + 10} \right) + ax + 4x + b - 12\\
= {x^2}\left( {{x^2} + 2} \right) - 2x\left( {{x^2} + 2} \right) + 5\left( {{x^2} + 2} \right) + \left( {a + 4} \right)x + \left( {b - 12} \right)\\
= \left( {{x^2} + 2} \right)\left( {{x^2} - 2x + 5} \right) + \left( {a + 4} \right)x + \left( {b - 12} \right)\\
= Q\left( x \right).\left( {{x^2} - 2x + 5} \right) + \left( {a + 4} \right)x + \left( {b - 12} \right)\\
P\left( x \right)\,\, \vdots \,\,Q\left( x \right) \Rightarrow \left[ {\left( {a + 4} \right)x + \left( {b - 12} \right)} \right]\,\, \vdots \,\,\left( {{x^2} + 2} \right)\\
\Rightarrow \left\{ \begin{array}{l}
\left( {a + 4} \right) = 0\\
b - 12 = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
a = - 4\\
b = 12
\end{array} \right.
\end{array}\)
Vậy \(a = - 4;\,\,\,b = 12\)
