Divide using synthetic division $$ ( x^{3}-6x^{2}-15x+100 ) \div ( 4 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}0&1&-6&-15&100\\& & 0& 0& \color{black}{0} \\ \hline &\color{blue}{1}&\color{blue}{-6}&\color{blue}{-15}&\color{orangered}{100} \end{array} $$The solution is:
$$ \frac{ x^{3}-6x^{2}-15x+100 }{ x } = \color{blue}{x^{2}-6x-15} ~+~ \frac{ \color{red}{ 100 } }{ x } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table.Put the zero at the left.
$$ \begin{array}{c|rrrr}\color{blue}{0}&1&-6&-15&100\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}0&\color{orangered}{ 1 }&-6&-15&100\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{0}&1&-6&-15&100\\& & \color{blue}{0} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -6 } + \color{orangered}{ 0 } = \color{orangered}{ -6 } $
$$ \begin{array}{c|rrrr}0&1&\color{orangered}{ -6 }&-15&100\\& & \color{orangered}{0} & & \\ \hline &1&\color{orangered}{-6}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ \left( -6 \right) } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrr}\color{blue}{0}&1&-6&-15&100\\& & 0& \color{blue}{0} & \\ \hline &1&\color{blue}{-6}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -15 } + \color{orangered}{ 0 } = \color{orangered}{ -15 } $
$$ \begin{array}{c|rrrr}0&1&-6&\color{orangered}{ -15 }&100\\& & 0& \color{orangered}{0} & \\ \hline &1&-6&\color{orangered}{-15}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ \left( -15 \right) } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrr}\color{blue}{0}&1&-6&-15&100\\& & 0& 0& \color{blue}{0} \\ \hline &1&-6&\color{blue}{-15}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 100 } + \color{orangered}{ 0 } = \color{orangered}{ 100 } $
$$ \begin{array}{c|rrrr}0&1&-6&-15&\color{orangered}{ 100 }\\& & 0& 0& \color{orangered}{0} \\ \hline &\color{blue}{1}&\color{blue}{-6}&\color{blue}{-15}&\color{orangered}{100} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{2}-6x-15 } $ with a remainder of $ \color{red}{ 100 } $.