Divide using synthetic division $$ ( 4x^{4}+2x^{3}-2x-1 ) \div ( x+3 ) $$
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-3&4&2&0&-2&-1\\& & -12& 30& -90& \color{black}{276} \\ \hline &\color{blue}{4}&\color{blue}{-10}&\color{blue}{30}&\color{blue}{-92}&\color{orangered}{275} \end{array} $$The solution is:
$$ \dfrac{ 4x^{4}+2x^{3}-2x-1 }{ x+3 } = \color{blue}{4x^{3}-10x^{2}+30x-92} ~+~ \dfrac{ \color{red}{ 275 } }{ x+3 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&4&2&0&-2&-1\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-3&\color{orangered}{ 4 }&2&0&-2&-1\\& & & & & \\ \hline &\color{orangered}{4}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&4&2&0&-2&-1\\& & \color{blue}{-12} & & & \\ \hline &\color{blue}{4}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ \left( -12 \right) } = \color{orangered}{ -10 } $
$$ \begin{array}{c|rrrrr}-3&4&\color{orangered}{ 2 }&0&-2&-1\\& & \color{orangered}{-12} & & & \\ \hline &4&\color{orangered}{-10}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -10 \right) } = \color{blue}{ 30 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&4&2&0&-2&-1\\& & -12& \color{blue}{30} & & \\ \hline &4&\color{blue}{-10}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 30 } = \color{orangered}{ 30 } $
$$ \begin{array}{c|rrrrr}-3&4&2&\color{orangered}{ 0 }&-2&-1\\& & -12& \color{orangered}{30} & & \\ \hline &4&-10&\color{orangered}{30}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 30 } = \color{blue}{ -90 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&4&2&0&-2&-1\\& & -12& 30& \color{blue}{-90} & \\ \hline &4&-10&\color{blue}{30}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ \left( -90 \right) } = \color{orangered}{ -92 } $
$$ \begin{array}{c|rrrrr}-3&4&2&0&\color{orangered}{ -2 }&-1\\& & -12& 30& \color{orangered}{-90} & \\ \hline &4&-10&30&\color{orangered}{-92}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -92 \right) } = \color{blue}{ 276 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&4&2&0&-2&-1\\& & -12& 30& -90& \color{blue}{276} \\ \hline &4&-10&30&\color{blue}{-92}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ 276 } = \color{orangered}{ 275 } $
$$ \begin{array}{c|rrrrr}-3&4&2&0&-2&\color{orangered}{ -1 }\\& & -12& 30& -90& \color{orangered}{276} \\ \hline &\color{blue}{4}&\color{blue}{-10}&\color{blue}{30}&\color{blue}{-92}&\color{orangered}{275} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 4x^{3}-10x^{2}+30x-92 } $ with a remainder of $ \color{red}{ 275 } $.