Divide using synthetic division $$ ( x^{5}-10x^{3}-x^{2}-6 ) \div ( x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrr}3&1&0&-10&-1&0&-6\\& & 3& 9& -3& -12& \color{black}{-36} \\ \hline &\color{blue}{1}&\color{blue}{3}&\color{blue}{-1}&\color{blue}{-4}&\color{blue}{-12}&\color{orangered}{-42} \end{array} $$The solution is:
$$ \frac{ x^{5}-10x^{3}-x^{2}-6 }{ x-3 } = \color{blue}{x^{4}+3x^{3}-x^{2}-4x-12} \color{red}{~-~} \frac{ \color{red}{ 42 } }{ x-3 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -3 = 0 $ ( $ x = \color{blue}{ 3 } $ ) at the left.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&1&0&-10&-1&0&-6\\& & & & & & \\ \hline &&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrr}3&\color{orangered}{ 1 }&0&-10&-1&0&-6\\& & & & & & \\ \hline &\color{orangered}{1}&&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 1 } = \color{blue}{ 3 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&1&0&-10&-1&0&-6\\& & \color{blue}{3} & & & & \\ \hline &\color{blue}{1}&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 3 } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrrrr}3&1&\color{orangered}{ 0 }&-10&-1&0&-6\\& & \color{orangered}{3} & & & & \\ \hline &1&\color{orangered}{3}&&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 3 } = \color{blue}{ 9 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&1&0&-10&-1&0&-6\\& & 3& \color{blue}{9} & & & \\ \hline &1&\color{blue}{3}&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -10 } + \color{orangered}{ 9 } = \color{orangered}{ -1 } $
$$ \begin{array}{c|rrrrrr}3&1&0&\color{orangered}{ -10 }&-1&0&-6\\& & 3& \color{orangered}{9} & & & \\ \hline &1&3&\color{orangered}{-1}&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -1 \right) } = \color{blue}{ -3 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&1&0&-10&-1&0&-6\\& & 3& 9& \color{blue}{-3} & & \\ \hline &1&3&\color{blue}{-1}&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ \left( -3 \right) } = \color{orangered}{ -4 } $
$$ \begin{array}{c|rrrrrr}3&1&0&-10&\color{orangered}{ -1 }&0&-6\\& & 3& 9& \color{orangered}{-3} & & \\ \hline &1&3&-1&\color{orangered}{-4}&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -4 \right) } = \color{blue}{ -12 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&1&0&-10&-1&0&-6\\& & 3& 9& -3& \color{blue}{-12} & \\ \hline &1&3&-1&\color{blue}{-4}&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -12 \right) } = \color{orangered}{ -12 } $
$$ \begin{array}{c|rrrrrr}3&1&0&-10&-1&\color{orangered}{ 0 }&-6\\& & 3& 9& -3& \color{orangered}{-12} & \\ \hline &1&3&-1&-4&\color{orangered}{-12}& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -12 \right) } = \color{blue}{ -36 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&1&0&-10&-1&0&-6\\& & 3& 9& -3& -12& \color{blue}{-36} \\ \hline &1&3&-1&-4&\color{blue}{-12}& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ -6 } + \color{orangered}{ \left( -36 \right) } = \color{orangered}{ -42 } $
$$ \begin{array}{c|rrrrrr}3&1&0&-10&-1&0&\color{orangered}{ -6 }\\& & 3& 9& -3& -12& \color{orangered}{-36} \\ \hline &\color{blue}{1}&\color{blue}{3}&\color{blue}{-1}&\color{blue}{-4}&\color{blue}{-12}&\color{orangered}{-42} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{4}+3x^{3}-x^{2}-4x-12 } $ with a remainder of $ \color{red}{ -42 } $.