Divide using synthetic division $$ ( 5x^{3}-18x^{2}-x ) \div ( -2x ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}0&5&-18&-1&0\\& & 0& 0& \color{black}{0} \\ \hline &\color{blue}{5}&\color{blue}{-18}&\color{blue}{-1}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ 5x^{3}-18x^{2}-x }{ x } = \color{blue}{5x^{2}-18x-1} $$Explanation
Step 1 : Write down the coefficients of the dividend into division table.Put the zero at the left.
$$ \begin{array}{c|rrrr}\color{blue}{0}&5&-18&-1&0\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}0&\color{orangered}{ 5 }&-18&-1&0\\& & & & \\ \hline &\color{orangered}{5}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ 5 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrr}\color{blue}{0}&5&-18&-1&0\\& & \color{blue}{0} & & \\ \hline &\color{blue}{5}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -18 } + \color{orangered}{ 0 } = \color{orangered}{ -18 } $
$$ \begin{array}{c|rrrr}0&5&\color{orangered}{ -18 }&-1&0\\& & \color{orangered}{0} & & \\ \hline &5&\color{orangered}{-18}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ \left( -18 \right) } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrr}\color{blue}{0}&5&-18&-1&0\\& & 0& \color{blue}{0} & \\ \hline &5&\color{blue}{-18}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ 0 } = \color{orangered}{ -1 } $
$$ \begin{array}{c|rrrr}0&5&-18&\color{orangered}{ -1 }&0\\& & 0& \color{orangered}{0} & \\ \hline &5&-18&\color{orangered}{-1}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ \left( -1 \right) } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrr}\color{blue}{0}&5&-18&-1&0\\& & 0& 0& \color{blue}{0} \\ \hline &5&-18&\color{blue}{-1}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 0 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrr}0&5&-18&-1&\color{orangered}{ 0 }\\& & 0& 0& \color{orangered}{0} \\ \hline &\color{blue}{5}&\color{blue}{-18}&\color{blue}{-1}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 5x^{2}-18x-1 } $ with a remainder of $ \color{red}{ 0 } $.