Divide using synthetic division $$ ( x^{5}-12x^{4}+38x^{3}+12x^{2}-135x ) \div ( -11 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrr}0&1&-12&38&12&-135&0\\& & 0& 0& 0& 0& \color{black}{0} \\ \hline &\color{blue}{1}&\color{blue}{-12}&\color{blue}{38}&\color{blue}{12}&\color{blue}{-135}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ x^{5}-12x^{4}+38x^{3}+12x^{2}-135x }{ x } = \color{blue}{x^{4}-12x^{3}+38x^{2}+12x-135} $$Explanation
Step 1 : Write down the coefficients of the dividend into division table.Put the zero at the left.
$$ \begin{array}{c|rrrrrr}\color{blue}{0}&1&-12&38&12&-135&0\\& & & & & & \\ \hline &&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrr}0&\color{orangered}{ 1 }&-12&38&12&-135&0\\& & & & & & \\ \hline &\color{orangered}{1}&&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{0}&1&-12&38&12&-135&0\\& & \color{blue}{0} & & & & \\ \hline &\color{blue}{1}&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -12 } + \color{orangered}{ 0 } = \color{orangered}{ -12 } $
$$ \begin{array}{c|rrrrrr}0&1&\color{orangered}{ -12 }&38&12&-135&0\\& & \color{orangered}{0} & & & & \\ \hline &1&\color{orangered}{-12}&&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ \left( -12 \right) } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{0}&1&-12&38&12&-135&0\\& & 0& \color{blue}{0} & & & \\ \hline &1&\color{blue}{-12}&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 38 } + \color{orangered}{ 0 } = \color{orangered}{ 38 } $
$$ \begin{array}{c|rrrrrr}0&1&-12&\color{orangered}{ 38 }&12&-135&0\\& & 0& \color{orangered}{0} & & & \\ \hline &1&-12&\color{orangered}{38}&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ 38 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{0}&1&-12&38&12&-135&0\\& & 0& 0& \color{blue}{0} & & \\ \hline &1&-12&\color{blue}{38}&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 12 } + \color{orangered}{ 0 } = \color{orangered}{ 12 } $
$$ \begin{array}{c|rrrrrr}0&1&-12&38&\color{orangered}{ 12 }&-135&0\\& & 0& 0& \color{orangered}{0} & & \\ \hline &1&-12&38&\color{orangered}{12}&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ 12 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{0}&1&-12&38&12&-135&0\\& & 0& 0& 0& \color{blue}{0} & \\ \hline &1&-12&38&\color{blue}{12}&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -135 } + \color{orangered}{ 0 } = \color{orangered}{ -135 } $
$$ \begin{array}{c|rrrrrr}0&1&-12&38&12&\color{orangered}{ -135 }&0\\& & 0& 0& 0& \color{orangered}{0} & \\ \hline &1&-12&38&12&\color{orangered}{-135}& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ \left( -135 \right) } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{0}&1&-12&38&12&-135&0\\& & 0& 0& 0& 0& \color{blue}{0} \\ \hline &1&-12&38&12&\color{blue}{-135}& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 0 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrrr}0&1&-12&38&12&-135&\color{orangered}{ 0 }\\& & 0& 0& 0& 0& \color{orangered}{0} \\ \hline &\color{blue}{1}&\color{blue}{-12}&\color{blue}{38}&\color{blue}{12}&\color{blue}{-135}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{4}-12x^{3}+38x^{2}+12x-135 } $ with a remainder of $ \color{red}{ 0 } $.