Divide using synthetic division $$ ( x^{4}-6x^{3}-12x ) \div ( 0 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}0&1&-6&0&-12&0\\& & 0& 0& 0& \color{black}{0} \\ \hline &\color{blue}{1}&\color{blue}{-6}&\color{blue}{0}&\color{blue}{-12}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ x^{4}-6x^{3}-12x }{ x } = \color{blue}{x^{3}-6x^{2}-12} $$Explanation
Step 1 : Write down the coefficients of the dividend into division table.Put the zero at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{0}&1&-6&0&-12&0\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}0&\color{orangered}{ 1 }&-6&0&-12&0\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{0}&1&-6&0&-12&0\\& & \color{blue}{0} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -6 } + \color{orangered}{ 0 } = \color{orangered}{ -6 } $
$$ \begin{array}{c|rrrrr}0&1&\color{orangered}{ -6 }&0&-12&0\\& & \color{orangered}{0} & & & \\ \hline &1&\color{orangered}{-6}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{0}&1&-6&0&-12&0\\& & 0& \color{blue}{0} & & \\ \hline &1&\color{blue}{-6}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 0 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}0&1&-6&\color{orangered}{ 0 }&-12&0\\& & 0& \color{orangered}{0} & & \\ \hline &1&-6&\color{orangered}{0}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ 0 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{0}&1&-6&0&-12&0\\& & 0& 0& \color{blue}{0} & \\ \hline &1&-6&\color{blue}{0}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -12 } + \color{orangered}{ 0 } = \color{orangered}{ -12 } $
$$ \begin{array}{c|rrrrr}0&1&-6&0&\color{orangered}{ -12 }&0\\& & 0& 0& \color{orangered}{0} & \\ \hline &1&-6&0&\color{orangered}{-12}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ \left( -12 \right) } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{0}&1&-6&0&-12&0\\& & 0& 0& 0& \color{blue}{0} \\ \hline &1&-6&0&\color{blue}{-12}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 0 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}0&1&-6&0&-12&\color{orangered}{ 0 }\\& & 0& 0& 0& \color{orangered}{0} \\ \hline &\color{blue}{1}&\color{blue}{-6}&\color{blue}{0}&\color{blue}{-12}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}-6x^{2}-12 } $ with a remainder of $ \color{red}{ 0 } $.