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