Divide using synthetic division $$ ( 4x^{5}+6x^{4}+5x^{2}-x-10 ) \div ( x ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrr}0&4&6&0&5&-1&-10\\& & 0& 0& 0& 0& \color{black}{0} \\ \hline &\color{blue}{4}&\color{blue}{6}&\color{blue}{0}&\color{blue}{5}&\color{blue}{-1}&\color{orangered}{-10} \end{array} $$The solution is:
$$ \frac{ 4x^{5}+6x^{4}+5x^{2}-x-10 }{ x } = \color{blue}{4x^{4}+6x^{3}+5x-1} \color{red}{~-~} \frac{ \color{red}{ 10 } }{ x } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table.Put the zero at the left.
$$ \begin{array}{c|rrrrrr}\color{blue}{0}&4&6&0&5&-1&-10\\& & & & & & \\ \hline &&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrr}0&\color{orangered}{ 4 }&6&0&5&-1&-10\\& & & & & & \\ \hline &\color{orangered}{4}&&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ 4 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{0}&4&6&0&5&-1&-10\\& & \color{blue}{0} & & & & \\ \hline &\color{blue}{4}&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 6 } + \color{orangered}{ 0 } = \color{orangered}{ 6 } $
$$ \begin{array}{c|rrrrrr}0&4&\color{orangered}{ 6 }&0&5&-1&-10\\& & \color{orangered}{0} & & & & \\ \hline &4&\color{orangered}{6}&&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ 6 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{0}&4&6&0&5&-1&-10\\& & 0& \color{blue}{0} & & & \\ \hline &4&\color{blue}{6}&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 0 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrrr}0&4&6&\color{orangered}{ 0 }&5&-1&-10\\& & 0& \color{orangered}{0} & & & \\ \hline &4&6&\color{orangered}{0}&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ 0 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{0}&4&6&0&5&-1&-10\\& & 0& 0& \color{blue}{0} & & \\ \hline &4&6&\color{blue}{0}&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ 0 } = \color{orangered}{ 5 } $
$$ \begin{array}{c|rrrrrr}0&4&6&0&\color{orangered}{ 5 }&-1&-10\\& & 0& 0& \color{orangered}{0} & & \\ \hline &4&6&0&\color{orangered}{5}&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ 5 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{0}&4&6&0&5&-1&-10\\& & 0& 0& 0& \color{blue}{0} & \\ \hline &4&6&0&\color{blue}{5}&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ 0 } = \color{orangered}{ -1 } $
$$ \begin{array}{c|rrrrrr}0&4&6&0&5&\color{orangered}{ -1 }&-10\\& & 0& 0& 0& \color{orangered}{0} & \\ \hline &4&6&0&5&\color{orangered}{-1}& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ \left( -1 \right) } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{0}&4&6&0&5&-1&-10\\& & 0& 0& 0& 0& \color{blue}{0} \\ \hline &4&6&0&5&\color{blue}{-1}& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ -10 } + \color{orangered}{ 0 } = \color{orangered}{ -10 } $
$$ \begin{array}{c|rrrrrr}0&4&6&0&5&-1&\color{orangered}{ -10 }\\& & 0& 0& 0& 0& \color{orangered}{0} \\ \hline &\color{blue}{4}&\color{blue}{6}&\color{blue}{0}&\color{blue}{5}&\color{blue}{-1}&\color{orangered}{-10} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 4x^{4}+6x^{3}+5x-1 } $ with a remainder of $ \color{red}{ -10 } $.