Divide using synthetic division $$ ( 4x^{3}+5x^{2}+x-10 ) \div ( x+2 ) $$
The synthetic division table is:
$$ \begin{array}{c|rrrr}-2&4&5&1&-10\\& & -8& 6& \color{black}{-14} \\ \hline &\color{blue}{4}&\color{blue}{-3}&\color{blue}{7}&\color{orangered}{-24} \end{array} $$The solution is:
$$ \dfrac{ 4x^{3}+5x^{2}+x-10 }{ x+2 } = \color{blue}{4x^{2}-3x+7} \color{red}{~-~} \dfrac{ \color{red}{ 24 } }{ x+2 } $$Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 2 = 0 $ ( $ x = \color{blue}{ -2 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&4&5&1&-10\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-2&\color{orangered}{ 4 }&5&1&-10\\& & & & \\ \hline &\color{orangered}{4}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 4 } = \color{blue}{ -8 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&4&5&1&-10\\& & \color{blue}{-8} & & \\ \hline &\color{blue}{4}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ \left( -8 \right) } = \color{orangered}{ -3 } $
$$ \begin{array}{c|rrrr}-2&4&\color{orangered}{ 5 }&1&-10\\& & \color{orangered}{-8} & & \\ \hline &4&\color{orangered}{-3}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -3 \right) } = \color{blue}{ 6 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&4&5&1&-10\\& & -8& \color{blue}{6} & \\ \hline &4&\color{blue}{-3}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ 6 } = \color{orangered}{ 7 } $
$$ \begin{array}{c|rrrr}-2&4&5&\color{orangered}{ 1 }&-10\\& & -8& \color{orangered}{6} & \\ \hline &4&-3&\color{orangered}{7}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 7 } = \color{blue}{ -14 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&4&5&1&-10\\& & -8& 6& \color{blue}{-14} \\ \hline &4&-3&\color{blue}{7}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -10 } + \color{orangered}{ \left( -14 \right) } = \color{orangered}{ -24 } $
$$ \begin{array}{c|rrrr}-2&4&5&1&\color{orangered}{ -10 }\\& & -8& 6& \color{orangered}{-14} \\ \hline &\color{blue}{4}&\color{blue}{-3}&\color{blue}{7}&\color{orangered}{-24} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 4x^{2}-3x+7 } $ with a remainder of $ \color{red}{ -24 } $.