Divide using synthetic division $$ ( x^{4}-2x^{3}-52x^{2}+36x-28 ) \div ( x-8 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}8&1&-2&-52&36&-28\\& & 8& 48& -32& \color{black}{32} \\ \hline &\color{blue}{1}&\color{blue}{6}&\color{blue}{-4}&\color{blue}{4}&\color{orangered}{4} \end{array} $$The solution is:
$$ \frac{ x^{4}-2x^{3}-52x^{2}+36x-28 }{ x-8 } = \color{blue}{x^{3}+6x^{2}-4x+4} ~+~ \frac{ \color{red}{ 4 } }{ x-8 } $$Explanation
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.
$$ \begin{array}{c|rrrrr}\color{blue}{8}&1&-2&-52&36&-28\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}8&\color{orangered}{ 1 }&-2&-52&36&-28\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{8}&1&-2&-52&36&-28\\& & \color{blue}{8} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ 8 } = \color{orangered}{ 6 } $
$$ \begin{array}{c|rrrrr}8&1&\color{orangered}{ -2 }&-52&36&-28\\& & \color{orangered}{8} & & & \\ \hline &1&\color{orangered}{6}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 8 } \cdot \color{blue}{ 6 } = \color{blue}{ 48 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{8}&1&-2&-52&36&-28\\& & 8& \color{blue}{48} & & \\ \hline &1&\color{blue}{6}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -52 } + \color{orangered}{ 48 } = \color{orangered}{ -4 } $
$$ \begin{array}{c|rrrrr}8&1&-2&\color{orangered}{ -52 }&36&-28\\& & 8& \color{orangered}{48} & & \\ \hline &1&6&\color{orangered}{-4}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 8 } \cdot \color{blue}{ \left( -4 \right) } = \color{blue}{ -32 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{8}&1&-2&-52&36&-28\\& & 8& 48& \color{blue}{-32} & \\ \hline &1&6&\color{blue}{-4}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 36 } + \color{orangered}{ \left( -32 \right) } = \color{orangered}{ 4 } $
$$ \begin{array}{c|rrrrr}8&1&-2&-52&\color{orangered}{ 36 }&-28\\& & 8& 48& \color{orangered}{-32} & \\ \hline &1&6&-4&\color{orangered}{4}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 8 } \cdot \color{blue}{ 4 } = \color{blue}{ 32 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{8}&1&-2&-52&36&-28\\& & 8& 48& -32& \color{blue}{32} \\ \hline &1&6&-4&\color{blue}{4}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -28 } + \color{orangered}{ 32 } = \color{orangered}{ 4 } $
$$ \begin{array}{c|rrrrr}8&1&-2&-52&36&\color{orangered}{ -28 }\\& & 8& 48& -32& \color{orangered}{32} \\ \hline &\color{blue}{1}&\color{blue}{6}&\color{blue}{-4}&\color{blue}{4}&\color{orangered}{4} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}+6x^{2}-4x+4 } $ with a remainder of $ \color{red}{ 4 } $.