Divide using synthetic division $$ ( 4x^{4}-8x^{3}-63x^{2}+45x+162 ) \div ( x+3 ) $$
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-3&4&-8&-63&45&162\\& & -12& 60& 9& \color{black}{-162} \\ \hline &\color{blue}{4}&\color{blue}{-20}&\color{blue}{-3}&\color{blue}{54}&\color{orangered}{0} \end{array} $$The solution is:
$$ \dfrac{ 4x^{4}-8x^{3}-63x^{2}+45x+162 }{ x+3 } = \color{blue}{4x^{3}-20x^{2}-3x+54} $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&4&-8&-63&45&162\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-3&\color{orangered}{ 4 }&-8&-63&45&162\\& & & & & \\ \hline &\color{orangered}{4}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 4 } = \color{blue}{ -12 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&4&-8&-63&45&162\\& & \color{blue}{-12} & & & \\ \hline &\color{blue}{4}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -8 } + \color{orangered}{ \left( -12 \right) } = \color{orangered}{ -20 } $
$$ \begin{array}{c|rrrrr}-3&4&\color{orangered}{ -8 }&-63&45&162\\& & \color{orangered}{-12} & & & \\ \hline &4&\color{orangered}{-20}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -20 \right) } = \color{blue}{ 60 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&4&-8&-63&45&162\\& & -12& \color{blue}{60} & & \\ \hline &4&\color{blue}{-20}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -63 } + \color{orangered}{ 60 } = \color{orangered}{ -3 } $
$$ \begin{array}{c|rrrrr}-3&4&-8&\color{orangered}{ -63 }&45&162\\& & -12& \color{orangered}{60} & & \\ \hline &4&-20&\color{orangered}{-3}&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&4&-8&-63&45&162\\& & -12& 60& \color{blue}{9} & \\ \hline &4&-20&\color{blue}{-3}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 45 } + \color{orangered}{ 9 } = \color{orangered}{ 54 } $
$$ \begin{array}{c|rrrrr}-3&4&-8&-63&\color{orangered}{ 45 }&162\\& & -12& 60& \color{orangered}{9} & \\ \hline &4&-20&-3&\color{orangered}{54}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 54 } = \color{blue}{ -162 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&4&-8&-63&45&162\\& & -12& 60& 9& \color{blue}{-162} \\ \hline &4&-20&-3&\color{blue}{54}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 162 } + \color{orangered}{ \left( -162 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}-3&4&-8&-63&45&\color{orangered}{ 162 }\\& & -12& 60& 9& \color{orangered}{-162} \\ \hline &\color{blue}{4}&\color{blue}{-20}&\color{blue}{-3}&\color{blue}{54}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 4x^{3}-20x^{2}-3x+54 } $ with a remainder of $ \color{red}{ 0 } $.