Divide using synthetic division $$ ( x^{4}-6x^{3}-30x^{2}+48x+155 ) \div ( x+3 ) $$
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-3&1&-6&-30&48&155\\& & -3& 27& 9& \color{black}{-171} \\ \hline &\color{blue}{1}&\color{blue}{-9}&\color{blue}{-3}&\color{blue}{57}&\color{orangered}{-16} \end{array} $$The solution is:
$$ \dfrac{ x^{4}-6x^{3}-30x^{2}+48x+155 }{ x+3 } = \color{blue}{x^{3}-9x^{2}-3x+57} \color{red}{~-~} \dfrac{ \color{red}{ 16 } }{ x+3 } $$Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 3 = 0 $ ( $ x = \color{blue}{ -3 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&1&-6&-30&48&155\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-3&\color{orangered}{ 1 }&-6&-30&48&155\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 1 } = \color{blue}{ -3 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&1&-6&-30&48&155\\& & \color{blue}{-3} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -6 } + \color{orangered}{ \left( -3 \right) } = \color{orangered}{ -9 } $
$$ \begin{array}{c|rrrrr}-3&1&\color{orangered}{ -6 }&-30&48&155\\& & \color{orangered}{-3} & & & \\ \hline &1&\color{orangered}{-9}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -9 \right) } = \color{blue}{ 27 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&1&-6&-30&48&155\\& & -3& \color{blue}{27} & & \\ \hline &1&\color{blue}{-9}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -30 } + \color{orangered}{ 27 } = \color{orangered}{ -3 } $
$$ \begin{array}{c|rrrrr}-3&1&-6&\color{orangered}{ -30 }&48&155\\& & -3& \color{orangered}{27} & & \\ \hline &1&-9&\color{orangered}{-3}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -3 \right) } = \color{blue}{ 9 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&1&-6&-30&48&155\\& & -3& 27& \color{blue}{9} & \\ \hline &1&-9&\color{blue}{-3}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 48 } + \color{orangered}{ 9 } = \color{orangered}{ 57 } $
$$ \begin{array}{c|rrrrr}-3&1&-6&-30&\color{orangered}{ 48 }&155\\& & -3& 27& \color{orangered}{9} & \\ \hline &1&-9&-3&\color{orangered}{57}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 57 } = \color{blue}{ -171 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&1&-6&-30&48&155\\& & -3& 27& 9& \color{blue}{-171} \\ \hline &1&-9&-3&\color{blue}{57}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 155 } + \color{orangered}{ \left( -171 \right) } = \color{orangered}{ -16 } $
$$ \begin{array}{c|rrrrr}-3&1&-6&-30&48&\color{orangered}{ 155 }\\& & -3& 27& 9& \color{orangered}{-171} \\ \hline &\color{blue}{1}&\color{blue}{-9}&\color{blue}{-3}&\color{blue}{57}&\color{orangered}{-16} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}-9x^{2}-3x+57 } $ with a remainder of $ \color{red}{ -16 } $.