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