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