Đáp án:
$a = \dfrac{{ - 34}}{3};b = 2;c = \dfrac{{28}}{3}$
Giải thích các bước giải:
Đặt $P\left( x \right) = 2{x^4} + a{x^2} + bx + c$
+) Ta có:
$\begin{array}{l}
P\left( x \right) \vdots \left( {x - 2} \right)\\
\Leftrightarrow P\left( 2 \right) = 0\\
\Leftrightarrow {2.2^4} + a{.2^2} + b.2 + c = 0\\
\Leftrightarrow 4a + 2b + c = - 32\left( 1 \right)
\end{array}$
+) Lại có:
$P\left( x \right)$ chia cho ${x^2} - 1$ dư $2x$
$ \Leftrightarrow \left( {P\left( x \right) - 2x} \right) \vdots \left( {{x^2} - 1} \right)$
$\begin{array}{l}
\Leftrightarrow \left( {P\left( x \right) - 2x} \right) \vdots \left( {{x^2} - 1} \right)\\
\Leftrightarrow \left\{ \begin{array}{l}
P\left( 1 \right) - 2 = 0\\
P\left( { - 1} \right) + 2 = 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
{2.1^4} + a{.1^2} + b.1 + c - 2 = 0\\
2.{\left( { - 1} \right)^4} + a.{\left( { - 1} \right)^2} + b.\left( { - 1} \right) + c + 2 = 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
a + b + c = 0\left( 2 \right)\\
a - b + c = - 4\left( 3 \right)
\end{array} \right.
\end{array}$
Từ $(1),(2),(3)$ $ \Rightarrow a = \dfrac{{ - 34}}{3};b = 2;c = \dfrac{{28}}{3}$
Vậy $a = \dfrac{{ - 34}}{3};b = 2;c = \dfrac{{28}}{3}$
