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