Divide using synthetic division $$ ( x^{5}-13x^{4}+49x^{3}-40x^{2}-16x+31 ) \div ( x-6 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrr}6&1&-13&49&-40&-16&31\\& & 6& -42& 42& 12& \color{black}{-24} \\ \hline &\color{blue}{1}&\color{blue}{-7}&\color{blue}{7}&\color{blue}{2}&\color{blue}{-4}&\color{orangered}{7} \end{array} $$The solution is:
$$ \frac{ x^{5}-13x^{4}+49x^{3}-40x^{2}-16x+31 }{ x-6 } = \color{blue}{x^{4}-7x^{3}+7x^{2}+2x-4} ~+~ \frac{ \color{red}{ 7 } }{ x-6 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -6 = 0 $ ( $ x = \color{blue}{ 6 } $ ) at the left.
$$ \begin{array}{c|rrrrrr}\color{blue}{6}&1&-13&49&-40&-16&31\\& & & & & & \\ \hline &&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrr}6&\color{orangered}{ 1 }&-13&49&-40&-16&31\\& & & & & & \\ \hline &\color{orangered}{1}&&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 6 } \cdot \color{blue}{ 1 } = \color{blue}{ 6 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{6}&1&-13&49&-40&-16&31\\& & \color{blue}{6} & & & & \\ \hline &\color{blue}{1}&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -13 } + \color{orangered}{ 6 } = \color{orangered}{ -7 } $
$$ \begin{array}{c|rrrrrr}6&1&\color{orangered}{ -13 }&49&-40&-16&31\\& & \color{orangered}{6} & & & & \\ \hline &1&\color{orangered}{-7}&&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 6 } \cdot \color{blue}{ \left( -7 \right) } = \color{blue}{ -42 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{6}&1&-13&49&-40&-16&31\\& & 6& \color{blue}{-42} & & & \\ \hline &1&\color{blue}{-7}&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 49 } + \color{orangered}{ \left( -42 \right) } = \color{orangered}{ 7 } $
$$ \begin{array}{c|rrrrrr}6&1&-13&\color{orangered}{ 49 }&-40&-16&31\\& & 6& \color{orangered}{-42} & & & \\ \hline &1&-7&\color{orangered}{7}&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 6 } \cdot \color{blue}{ 7 } = \color{blue}{ 42 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{6}&1&-13&49&-40&-16&31\\& & 6& -42& \color{blue}{42} & & \\ \hline &1&-7&\color{blue}{7}&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -40 } + \color{orangered}{ 42 } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrrrr}6&1&-13&49&\color{orangered}{ -40 }&-16&31\\& & 6& -42& \color{orangered}{42} & & \\ \hline &1&-7&7&\color{orangered}{2}&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 6 } \cdot \color{blue}{ 2 } = \color{blue}{ 12 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{6}&1&-13&49&-40&-16&31\\& & 6& -42& 42& \color{blue}{12} & \\ \hline &1&-7&7&\color{blue}{2}&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -16 } + \color{orangered}{ 12 } = \color{orangered}{ -4 } $
$$ \begin{array}{c|rrrrrr}6&1&-13&49&-40&\color{orangered}{ -16 }&31\\& & 6& -42& 42& \color{orangered}{12} & \\ \hline &1&-7&7&2&\color{orangered}{-4}& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 6 } \cdot \color{blue}{ \left( -4 \right) } = \color{blue}{ -24 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{6}&1&-13&49&-40&-16&31\\& & 6& -42& 42& 12& \color{blue}{-24} \\ \hline &1&-7&7&2&\color{blue}{-4}& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ 31 } + \color{orangered}{ \left( -24 \right) } = \color{orangered}{ 7 } $
$$ \begin{array}{c|rrrrrr}6&1&-13&49&-40&-16&\color{orangered}{ 31 }\\& & 6& -42& 42& 12& \color{orangered}{-24} \\ \hline &\color{blue}{1}&\color{blue}{-7}&\color{blue}{7}&\color{blue}{2}&\color{blue}{-4}&\color{orangered}{7} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{4}-7x^{3}+7x^{2}+2x-4 } $ with a remainder of $ \color{red}{ 7 } $.