Answer
The value of $\left( 2{{x}^{2}}+x-10 \right)\text{ }\!\!\div\!\!\text{ }\left( x-2 \right)$ is $2x+5$.
Work Step by Step
Consider the provided expression,
$\left( 2{{x}^{2}}+x-10 \right)\text{ }\!\!\div\!\!\text{ }\left( x-2 \right)$
Arrange the polynomial in descending powers, with a 0 coefficient for any missing term as shown below,
$x-2\overline{\left){2{{x}^{2}}+x-10}\right.}$
Now write 2 for the divisor, $x-2$ to the right and write the coefficients of the dividend as shown below,
$\left. {\underline {\,
2 \,}}\! \right| \begin{matrix}
2 & 1 & -10 \\
\end{matrix}$
Now bring down the leading coefficient of the dividend on the bottom row as shown below,
$\begin{align}
& \left. {\underline {\,
2 \,}}\! \right| \begin{matrix}
\text{ }2 & 1 & -10 \\
\end{matrix} \\
& \text{ }\underline{\downarrow \text{Bring down 2}} \\
& \text{ 2} \\
\end{align}$
Multiply 2 with the bottom row value and write the product in the next column in the second row as shown below,
$\begin{align}
& \left. {\underline {\,
2 \,}}\! \right| \begin{matrix}
\text{ }2 & 1 & -10 \\
\end{matrix} \\
& \text{ }\underline{\downarrow \text{ 4}}\text{ } \\
& \text{ 2 } \\
\end{align}$
Add the column values and write down the sum in the bottom row as shown below,
$\begin{align}
& \left. {\underline {\,
2 \,}}\! \right| \begin{matrix}
\text{ }2 & 1 & -10 \\
\end{matrix} \\
& \text{ }\underline{\downarrow \text{ 4}}\text{ } \\
& \text{ 2 5 } \\
\end{align}$
Similarly, repeat this series of multiplications and additions until the columns are filled in,
$\begin{align}
& \left. {\underline {\,
2 \,}}\! \right| \begin{matrix}
\text{ }2 & 1 & -10 \\
\end{matrix} \\
& \text{ }\underline{\downarrow \text{ 4 10}}\text{ } \\
& \text{ 2 5 0} \\
\end{align}$
Thus, the quotient is $2x+5$ and the remainder is $0$.
