Divide using synthetic division $$ ( 7x^{4}-13x^{3}-3x^{2}+4x+15 ) \div ( x-1 ) $$
The synthetic division table is:
$$ \begin{array}{c|rrrrr}1&7&-13&-3&4&15\\& & 7& -6& -9& \color{black}{-5} \\ \hline &\color{blue}{7}&\color{blue}{-6}&\color{blue}{-9}&\color{blue}{-5}&\color{orangered}{10} \end{array} $$The solution is:
$$ \dfrac{ 7x^{4}-13x^{3}-3x^{2}+4x+15 }{ x-1 } = \color{blue}{7x^{3}-6x^{2}-9x-5} ~+~ \dfrac{ \color{red}{ 10 } }{ x-1 } $$Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -1 = 0 $ ( $ x = \color{blue}{ 1 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&7&-13&-3&4&15\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}1&\color{orangered}{ 7 }&-13&-3&4&15\\& & & & & \\ \hline &\color{orangered}{7}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 7 } = \color{blue}{ 7 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&7&-13&-3&4&15\\& & \color{blue}{7} & & & \\ \hline &\color{blue}{7}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -13 } + \color{orangered}{ 7 } = \color{orangered}{ -6 } $
$$ \begin{array}{c|rrrrr}1&7&\color{orangered}{ -13 }&-3&4&15\\& & \color{orangered}{7} & & & \\ \hline &7&\color{orangered}{-6}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ \left( -6 \right) } = \color{blue}{ -6 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&7&-13&-3&4&15\\& & 7& \color{blue}{-6} & & \\ \hline &7&\color{blue}{-6}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -3 } + \color{orangered}{ \left( -6 \right) } = \color{orangered}{ -9 } $
$$ \begin{array}{c|rrrrr}1&7&-13&\color{orangered}{ -3 }&4&15\\& & 7& \color{orangered}{-6} & & \\ \hline &7&-6&\color{orangered}{-9}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ \left( -9 \right) } = \color{blue}{ -9 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&7&-13&-3&4&15\\& & 7& -6& \color{blue}{-9} & \\ \hline &7&-6&\color{blue}{-9}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 4 } + \color{orangered}{ \left( -9 \right) } = \color{orangered}{ -5 } $
$$ \begin{array}{c|rrrrr}1&7&-13&-3&\color{orangered}{ 4 }&15\\& & 7& -6& \color{orangered}{-9} & \\ \hline &7&-6&-9&\color{orangered}{-5}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ \left( -5 \right) } = \color{blue}{ -5 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&7&-13&-3&4&15\\& & 7& -6& -9& \color{blue}{-5} \\ \hline &7&-6&-9&\color{blue}{-5}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 15 } + \color{orangered}{ \left( -5 \right) } = \color{orangered}{ 10 } $
$$ \begin{array}{c|rrrrr}1&7&-13&-3&4&\color{orangered}{ 15 }\\& & 7& -6& -9& \color{orangered}{-5} \\ \hline &\color{blue}{7}&\color{blue}{-6}&\color{blue}{-9}&\color{blue}{-5}&\color{orangered}{10} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 7x^{3}-6x^{2}-9x-5 } $ with a remainder of $ \color{red}{ 10 } $.