Divide using synthetic division $$ ( x^{3}-14x-8 ) \div ( 4 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}0&1&0&-14&-8\\& & 0& 0& \color{black}{0} \\ \hline &\color{blue}{1}&\color{blue}{0}&\color{blue}{-14}&\color{orangered}{-8} \end{array} $$The solution is:
$$ \frac{ x^{3}-14x-8 }{ x } = \color{blue}{x^{2}-14} \color{red}{~-~} \frac{ \color{red}{ 8 } }{ x } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table.Put the zero at the left.
$$ \begin{array}{c|rrrr}\color{blue}{0}&1&0&-14&-8\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}0&\color{orangered}{ 1 }&0&-14&-8\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ 1 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrr}\color{blue}{0}&1&0&-14&-8\\& & \color{blue}{0} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 0 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrr}0&1&\color{orangered}{ 0 }&-14&-8\\& & \color{orangered}{0} & & \\ \hline &1&\color{orangered}{0}&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{0}&1&0&-14&-8\\& & 0& \color{blue}{0} & \\ \hline &1&\color{blue}{0}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -14 } + \color{orangered}{ 0 } = \color{orangered}{ -14 } $
$$ \begin{array}{c|rrrr}0&1&0&\color{orangered}{ -14 }&-8\\& & 0& \color{orangered}{0} & \\ \hline &1&0&\color{orangered}{-14}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ \left( -14 \right) } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrr}\color{blue}{0}&1&0&-14&-8\\& & 0& 0& \color{blue}{0} \\ \hline &1&0&\color{blue}{-14}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -8 } + \color{orangered}{ 0 } = \color{orangered}{ -8 } $
$$ \begin{array}{c|rrrr}0&1&0&-14&\color{orangered}{ -8 }\\& & 0& 0& \color{orangered}{0} \\ \hline &\color{blue}{1}&\color{blue}{0}&\color{blue}{-14}&\color{orangered}{-8} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{2}-14 } $ with a remainder of $ \color{red}{ -8 } $.