The synthetic division table is:
$$ \begin{array}{c|rrrr}8&1&-13&46&-48\\& & 8& -40& \color{black}{48} \\ \hline &\color{blue}{1}&\color{blue}{-5}&\color{blue}{6}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ x^{3}-13x^{2}+46x-48 }{ x-8 } = \color{blue}{x^{2}-5x+6} $$Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -8 = 0 $ ( $ x = \color{blue}{ 8 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{8}&1&-13&46&-48\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}8&\color{orangered}{ 1 }&-13&46&-48\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 8 } \cdot \color{blue}{ 1 } = \color{blue}{ 8 } $.
$$ \begin{array}{c|rrrr}\color{blue}{8}&1&-13&46&-48\\& & \color{blue}{8} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -13 } + \color{orangered}{ 8 } = \color{orangered}{ -5 } $
$$ \begin{array}{c|rrrr}8&1&\color{orangered}{ -13 }&46&-48\\& & \color{orangered}{8} & & \\ \hline &1&\color{orangered}{-5}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 8 } \cdot \color{blue}{ \left( -5 \right) } = \color{blue}{ -40 } $.
$$ \begin{array}{c|rrrr}\color{blue}{8}&1&-13&46&-48\\& & 8& \color{blue}{-40} & \\ \hline &1&\color{blue}{-5}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 46 } + \color{orangered}{ \left( -40 \right) } = \color{orangered}{ 6 } $
$$ \begin{array}{c|rrrr}8&1&-13&\color{orangered}{ 46 }&-48\\& & 8& \color{orangered}{-40} & \\ \hline &1&-5&\color{orangered}{6}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 8 } \cdot \color{blue}{ 6 } = \color{blue}{ 48 } $.
$$ \begin{array}{c|rrrr}\color{blue}{8}&1&-13&46&-48\\& & 8& -40& \color{blue}{48} \\ \hline &1&-5&\color{blue}{6}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -48 } + \color{orangered}{ 48 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrr}8&1&-13&46&\color{orangered}{ -48 }\\& & 8& -40& \color{orangered}{48} \\ \hline &\color{blue}{1}&\color{blue}{-5}&\color{blue}{6}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{2}-5x+6 } $ with a remainder of $ \color{red}{ 0 } $.