Divide using synthetic division $$ ( 2x^{4}+5x^{3}+7x^{2}-4x+6 ) \div ( x+3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-3&2&5&7&-4&6\\& & -6& 3& -30& \color{black}{102} \\ \hline &\color{blue}{2}&\color{blue}{-1}&\color{blue}{10}&\color{blue}{-34}&\color{orangered}{108} \end{array} $$The solution is:
$$ \frac{ 2x^{4}+5x^{3}+7x^{2}-4x+6 }{ x+3 } = \color{blue}{2x^{3}-x^{2}+10x-34} ~+~ \frac{ \color{red}{ 108 } }{ x+3 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&2&5&7&-4&6\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-3&\color{orangered}{ 2 }&5&7&-4&6\\& & & & & \\ \hline &\color{orangered}{2}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 2 } = \color{blue}{ -6 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&2&5&7&-4&6\\& & \color{blue}{-6} & & & \\ \hline &\color{blue}{2}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ \left( -6 \right) } = \color{orangered}{ -1 } $
$$ \begin{array}{c|rrrrr}-3&2&\color{orangered}{ 5 }&7&-4&6\\& & \color{orangered}{-6} & & & \\ \hline &2&\color{orangered}{-1}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -1 \right) } = \color{blue}{ 3 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&2&5&7&-4&6\\& & -6& \color{blue}{3} & & \\ \hline &2&\color{blue}{-1}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 7 } + \color{orangered}{ 3 } = \color{orangered}{ 10 } $
$$ \begin{array}{c|rrrrr}-3&2&5&\color{orangered}{ 7 }&-4&6\\& & -6& \color{orangered}{3} & & \\ \hline &2&-1&\color{orangered}{10}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 10 } = \color{blue}{ -30 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&2&5&7&-4&6\\& & -6& 3& \color{blue}{-30} & \\ \hline &2&-1&\color{blue}{10}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -4 } + \color{orangered}{ \left( -30 \right) } = \color{orangered}{ -34 } $
$$ \begin{array}{c|rrrrr}-3&2&5&7&\color{orangered}{ -4 }&6\\& & -6& 3& \color{orangered}{-30} & \\ \hline &2&-1&10&\color{orangered}{-34}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -34 \right) } = \color{blue}{ 102 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&2&5&7&-4&6\\& & -6& 3& -30& \color{blue}{102} \\ \hline &2&-1&10&\color{blue}{-34}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 6 } + \color{orangered}{ 102 } = \color{orangered}{ 108 } $
$$ \begin{array}{c|rrrrr}-3&2&5&7&-4&\color{orangered}{ 6 }\\& & -6& 3& -30& \color{orangered}{102} \\ \hline &\color{blue}{2}&\color{blue}{-1}&\color{blue}{10}&\color{blue}{-34}&\color{orangered}{108} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{3}-x^{2}+10x-34 } $ with a remainder of $ \color{red}{ 108 } $.