Divide using synthetic division $$ ( 16x^{3}+12x^{2}+6x+10 ) \div ( x ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}0&16&12&6&10\\& & 0& 0& \color{black}{0} \\ \hline &\color{blue}{16}&\color{blue}{12}&\color{blue}{6}&\color{orangered}{10} \end{array} $$The solution is:
$$ \frac{ 16x^{3}+12x^{2}+6x+10 }{ x } = \color{blue}{16x^{2}+12x+6} ~+~ \frac{ \color{red}{ 10 } }{ x } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table.Put the zero at the left.
$$ \begin{array}{c|rrrr}\color{blue}{0}&16&12&6&10\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}0&\color{orangered}{ 16 }&12&6&10\\& & & & \\ \hline &\color{orangered}{16}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ 16 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrr}\color{blue}{0}&16&12&6&10\\& & \color{blue}{0} & & \\ \hline &\color{blue}{16}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 12 } + \color{orangered}{ 0 } = \color{orangered}{ 12 } $
$$ \begin{array}{c|rrrr}0&16&\color{orangered}{ 12 }&6&10\\& & \color{orangered}{0} & & \\ \hline &16&\color{orangered}{12}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ 12 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrr}\color{blue}{0}&16&12&6&10\\& & 0& \color{blue}{0} & \\ \hline &16&\color{blue}{12}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 6 } + \color{orangered}{ 0 } = \color{orangered}{ 6 } $
$$ \begin{array}{c|rrrr}0&16&12&\color{orangered}{ 6 }&10\\& & 0& \color{orangered}{0} & \\ \hline &16&12&\color{orangered}{6}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ 6 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrr}\color{blue}{0}&16&12&6&10\\& & 0& 0& \color{blue}{0} \\ \hline &16&12&\color{blue}{6}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 10 } + \color{orangered}{ 0 } = \color{orangered}{ 10 } $
$$ \begin{array}{c|rrrr}0&16&12&6&\color{orangered}{ 10 }\\& & 0& 0& \color{orangered}{0} \\ \hline &\color{blue}{16}&\color{blue}{12}&\color{blue}{6}&\color{orangered}{10} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 16x^{2}+12x+6 } $ with a remainder of $ \color{red}{ 10 } $.