Divide using synthetic division $$ ( 2x^{6}+6x^{5}-3x^{3}-9x^{2}+5x+15 ) \div ( x+3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrrr}-3&2&6&0&-3&-9&5&15\\& & -6& 0& 0& 9& 0& \color{black}{-15} \\ \hline &\color{blue}{2}&\color{blue}{0}&\color{blue}{0}&\color{blue}{-3}&\color{blue}{0}&\color{blue}{5}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ 2x^{6}+6x^{5}-3x^{3}-9x^{2}+5x+15 }{ x+3 } = \color{blue}{2x^{5}-3x^{2}+5} $$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.
$$ \begin{array}{c|rrrrrrr}\color{blue}{-3}&2&6&0&-3&-9&5&15\\& & & & & & & \\ \hline &&&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrrr}-3&\color{orangered}{ 2 }&6&0&-3&-9&5&15\\& & & & & & & \\ \hline &\color{orangered}{2}&&&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrrrr}\color{blue}{-3}&2&6&0&-3&-9&5&15\\& & \color{blue}{-6} & & & & & \\ \hline &\color{blue}{2}&&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 6 } + \color{orangered}{ \left( -6 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrrrr}-3&2&\color{orangered}{ 6 }&0&-3&-9&5&15\\& & \color{orangered}{-6} & & & & & \\ \hline &2&\color{orangered}{0}&&&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 0 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrrrr}\color{blue}{-3}&2&6&0&-3&-9&5&15\\& & -6& \color{blue}{0} & & & & \\ \hline &2&\color{blue}{0}&&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 0 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrrrr}-3&2&6&\color{orangered}{ 0 }&-3&-9&5&15\\& & -6& \color{orangered}{0} & & & & \\ \hline &2&0&\color{orangered}{0}&&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 0 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrrrr}\color{blue}{-3}&2&6&0&-3&-9&5&15\\& & -6& 0& \color{blue}{0} & & & \\ \hline &2&0&\color{blue}{0}&&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -3 } + \color{orangered}{ 0 } = \color{orangered}{ -3 } $
$$ \begin{array}{c|rrrrrrr}-3&2&6&0&\color{orangered}{ -3 }&-9&5&15\\& & -6& 0& \color{orangered}{0} & & & \\ \hline &2&0&0&\color{orangered}{-3}&&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -3 \right) } = \color{blue}{ 9 } $.
$$ \begin{array}{c|rrrrrrr}\color{blue}{-3}&2&6&0&-3&-9&5&15\\& & -6& 0& 0& \color{blue}{9} & & \\ \hline &2&0&0&\color{blue}{-3}&&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -9 } + \color{orangered}{ 9 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrrrr}-3&2&6&0&-3&\color{orangered}{ -9 }&5&15\\& & -6& 0& 0& \color{orangered}{9} & & \\ \hline &2&0&0&-3&\color{orangered}{0}&& \end{array} $$Step 10 : 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 } $.
$$ \begin{array}{c|rrrrrrr}\color{blue}{-3}&2&6&0&-3&-9&5&15\\& & -6& 0& 0& 9& \color{blue}{0} & \\ \hline &2&0&0&-3&\color{blue}{0}&& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ 0 } = \color{orangered}{ 5 } $
$$ \begin{array}{c|rrrrrrr}-3&2&6&0&-3&-9&\color{orangered}{ 5 }&15\\& & -6& 0& 0& 9& \color{orangered}{0} & \\ \hline &2&0&0&-3&0&\color{orangered}{5}& \end{array} $$Step 12 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 5 } = \color{blue}{ -15 } $.
$$ \begin{array}{c|rrrrrrr}\color{blue}{-3}&2&6&0&-3&-9&5&15\\& & -6& 0& 0& 9& 0& \color{blue}{-15} \\ \hline &2&0&0&-3&0&\color{blue}{5}& \end{array} $$Step 13 : Add down last column: $ \color{orangered}{ 15 } + \color{orangered}{ \left( -15 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrrrr}-3&2&6&0&-3&-9&5&\color{orangered}{ 15 }\\& & -6& 0& 0& 9& 0& \color{orangered}{-15} \\ \hline &\color{blue}{2}&\color{blue}{0}&\color{blue}{0}&\color{blue}{-3}&\color{blue}{0}&\color{blue}{5}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{5}-3x^{2}+5 } $ with a remainder of $ \color{red}{ 0 } $.