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