Divide using synthetic division $$ ( 2x^{5}+5x^{4}+12x^{2}+15x-2 ) \div ( x+3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrr}-3&2&5&0&12&15&-2\\& & -6& 3& -9& -9& \color{black}{-18} \\ \hline &\color{blue}{2}&\color{blue}{-1}&\color{blue}{3}&\color{blue}{3}&\color{blue}{6}&\color{orangered}{-20} \end{array} $$The solution is:
$$ \frac{ 2x^{5}+5x^{4}+12x^{2}+15x-2 }{ x+3 } = \color{blue}{2x^{4}-x^{3}+3x^{2}+3x+6} \color{red}{~-~} \frac{ \color{red}{ 20 } }{ x+3 } $$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.
$$ \begin{array}{c|rrrrrr}\color{blue}{-3}&2&5&0&12&15&-2\\& & & & & & \\ \hline &&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrr}-3&\color{orangered}{ 2 }&5&0&12&15&-2\\& & & & & & \\ \hline &\color{orangered}{2}&&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-3}&2&5&0&12&15&-2\\& & \color{blue}{-6} & & & & \\ \hline &\color{blue}{2}&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ \left( -6 \right) } = \color{orangered}{ -1 } $
$$ \begin{array}{c|rrrrrr}-3&2&\color{orangered}{ 5 }&0&12&15&-2\\& & \color{orangered}{-6} & & & & \\ \hline &2&\color{orangered}{-1}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-3}&2&5&0&12&15&-2\\& & -6& \color{blue}{3} & & & \\ \hline &2&\color{blue}{-1}&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 3 } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrrrr}-3&2&5&\color{orangered}{ 0 }&12&15&-2\\& & -6& \color{orangered}{3} & & & \\ \hline &2&-1&\color{orangered}{3}&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 3 } = \color{blue}{ -9 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-3}&2&5&0&12&15&-2\\& & -6& 3& \color{blue}{-9} & & \\ \hline &2&-1&\color{blue}{3}&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 12 } + \color{orangered}{ \left( -9 \right) } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrrrr}-3&2&5&0&\color{orangered}{ 12 }&15&-2\\& & -6& 3& \color{orangered}{-9} & & \\ \hline &2&-1&3&\color{orangered}{3}&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 3 } = \color{blue}{ -9 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-3}&2&5&0&12&15&-2\\& & -6& 3& -9& \color{blue}{-9} & \\ \hline &2&-1&3&\color{blue}{3}&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 15 } + \color{orangered}{ \left( -9 \right) } = \color{orangered}{ 6 } $
$$ \begin{array}{c|rrrrrr}-3&2&5&0&12&\color{orangered}{ 15 }&-2\\& & -6& 3& -9& \color{orangered}{-9} & \\ \hline &2&-1&3&3&\color{orangered}{6}& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 6 } = \color{blue}{ -18 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-3}&2&5&0&12&15&-2\\& & -6& 3& -9& -9& \color{blue}{-18} \\ \hline &2&-1&3&3&\color{blue}{6}& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ \left( -18 \right) } = \color{orangered}{ -20 } $
$$ \begin{array}{c|rrrrrr}-3&2&5&0&12&15&\color{orangered}{ -2 }\\& & -6& 3& -9& -9& \color{orangered}{-18} \\ \hline &\color{blue}{2}&\color{blue}{-1}&\color{blue}{3}&\color{blue}{3}&\color{blue}{6}&\color{orangered}{-20} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{4}-x^{3}+3x^{2}+3x+6 } $ with a remainder of $ \color{red}{ -20 } $.