Divide using synthetic division $$ ( 4x^{3}-8x^{2}-10 ) \div ( 4x+3 ) $$
The synthetic division table is:
$$ \begin{array}{c|rrrr}-3&4&-8&0&-10\\& & -12& 60& \color{black}{-180} \\ \hline &\color{blue}{4}&\color{blue}{-20}&\color{blue}{60}&\color{orangered}{-190} \end{array} $$The solution is:
$$ \dfrac{ 4x^{3}-8x^{2}-10 }{ x+3 } = \color{blue}{4x^{2}-20x+60} \color{red}{~-~} \dfrac{ \color{red}{ 190 } }{ x+3 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&4&-8&0&-10\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-3&\color{orangered}{ 4 }&-8&0&-10\\& & & & \\ \hline &\color{orangered}{4}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&4&-8&0&-10\\& & \color{blue}{-12} & & \\ \hline &\color{blue}{4}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -8 } + \color{orangered}{ \left( -12 \right) } = \color{orangered}{ -20 } $
$$ \begin{array}{c|rrrr}-3&4&\color{orangered}{ -8 }&0&-10\\& & \color{orangered}{-12} & & \\ \hline &4&\color{orangered}{-20}&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&4&-8&0&-10\\& & -12& \color{blue}{60} & \\ \hline &4&\color{blue}{-20}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 60 } = \color{orangered}{ 60 } $
$$ \begin{array}{c|rrrr}-3&4&-8&\color{orangered}{ 0 }&-10\\& & -12& \color{orangered}{60} & \\ \hline &4&-20&\color{orangered}{60}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 60 } = \color{blue}{ -180 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&4&-8&0&-10\\& & -12& 60& \color{blue}{-180} \\ \hline &4&-20&\color{blue}{60}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -10 } + \color{orangered}{ \left( -180 \right) } = \color{orangered}{ -190 } $
$$ \begin{array}{c|rrrr}-3&4&-8&0&\color{orangered}{ -10 }\\& & -12& 60& \color{orangered}{-180} \\ \hline &\color{blue}{4}&\color{blue}{-20}&\color{blue}{60}&\color{orangered}{-190} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 4x^{2}-20x+60 } $ with a remainder of $ \color{red}{ -190 } $.