Divide using synthetic division $$ ( x^{3}+x^{2}-50x+48 ) \div ( x+6 ) $$
The synthetic division table is:
$$ \begin{array}{c|rrrr}-6&1&1&-50&48\\& & -6& 30& \color{black}{120} \\ \hline &\color{blue}{1}&\color{blue}{-5}&\color{blue}{-20}&\color{orangered}{168} \end{array} $$The solution is:
$$ \dfrac{ x^{3}+x^{2}-50x+48 }{ x+6 } = \color{blue}{x^{2}-5x-20} ~+~ \dfrac{ \color{red}{ 168 } }{ x+6 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{-6}&1&1&-50&48\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-6&\color{orangered}{ 1 }&1&-50&48\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -6 } \cdot \color{blue}{ 1 } = \color{blue}{ -6 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-6}&1&1&-50&48\\& & \color{blue}{-6} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ \left( -6 \right) } = \color{orangered}{ -5 } $
$$ \begin{array}{c|rrrr}-6&1&\color{orangered}{ 1 }&-50&48\\& & \color{orangered}{-6} & & \\ \hline &1&\color{orangered}{-5}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -6 } \cdot \color{blue}{ \left( -5 \right) } = \color{blue}{ 30 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-6}&1&1&-50&48\\& & -6& \color{blue}{30} & \\ \hline &1&\color{blue}{-5}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -50 } + \color{orangered}{ 30 } = \color{orangered}{ -20 } $
$$ \begin{array}{c|rrrr}-6&1&1&\color{orangered}{ -50 }&48\\& & -6& \color{orangered}{30} & \\ \hline &1&-5&\color{orangered}{-20}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -6 } \cdot \color{blue}{ \left( -20 \right) } = \color{blue}{ 120 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-6}&1&1&-50&48\\& & -6& 30& \color{blue}{120} \\ \hline &1&-5&\color{blue}{-20}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 48 } + \color{orangered}{ 120 } = \color{orangered}{ 168 } $
$$ \begin{array}{c|rrrr}-6&1&1&-50&\color{orangered}{ 48 }\\& & -6& 30& \color{orangered}{120} \\ \hline &\color{blue}{1}&\color{blue}{-5}&\color{blue}{-20}&\color{orangered}{168} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{2}-5x-20 } $ with a remainder of $ \color{red}{ 168 } $.