Divide using synthetic division $$ ( 6x^{3}+2x+131 ) \div ( 7 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}0&6&0&2&131\\& & 0& 0& \color{black}{0} \\ \hline &\color{blue}{6}&\color{blue}{0}&\color{blue}{2}&\color{orangered}{131} \end{array} $$The solution is:
$$ \frac{ 6x^{3}+2x+131 }{ x } = \color{blue}{6x^{2}+2} ~+~ \frac{ \color{red}{ 131 } }{ x } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table.Put the zero at the left.
$$ \begin{array}{c|rrrr}\color{blue}{0}&6&0&2&131\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}0&\color{orangered}{ 6 }&0&2&131\\& & & & \\ \hline &\color{orangered}{6}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ 6 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrr}\color{blue}{0}&6&0&2&131\\& & \color{blue}{0} & & \\ \hline &\color{blue}{6}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 0 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrr}0&6&\color{orangered}{ 0 }&2&131\\& & \color{orangered}{0} & & \\ \hline &6&\color{orangered}{0}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ 0 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrr}\color{blue}{0}&6&0&2&131\\& & 0& \color{blue}{0} & \\ \hline &6&\color{blue}{0}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ 0 } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrr}0&6&0&\color{orangered}{ 2 }&131\\& & 0& \color{orangered}{0} & \\ \hline &6&0&\color{orangered}{2}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ 2 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrr}\color{blue}{0}&6&0&2&131\\& & 0& 0& \color{blue}{0} \\ \hline &6&0&\color{blue}{2}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 131 } + \color{orangered}{ 0 } = \color{orangered}{ 131 } $
$$ \begin{array}{c|rrrr}0&6&0&2&\color{orangered}{ 131 }\\& & 0& 0& \color{orangered}{0} \\ \hline &\color{blue}{6}&\color{blue}{0}&\color{blue}{2}&\color{orangered}{131} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 6x^{2}+2 } $ with a remainder of $ \color{red}{ 131 } $.