Divide using synthetic division $$ ( x^{5}-3x^{4}-19x^{3}+7x^{2}+39x-26 ) \div ( x+3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrr}-3&1&-3&-19&7&39&-26\\& & -3& 18& 3& -30& \color{black}{-27} \\ \hline &\color{blue}{1}&\color{blue}{-6}&\color{blue}{-1}&\color{blue}{10}&\color{blue}{9}&\color{orangered}{-53} \end{array} $$The solution is:
$$ \frac{ x^{5}-3x^{4}-19x^{3}+7x^{2}+39x-26 }{ x+3 } = \color{blue}{x^{4}-6x^{3}-x^{2}+10x+9} \color{red}{~-~} \frac{ \color{red}{ 53 } }{ x+3 } $$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.
$$ \begin{array}{c|rrrrrr}\color{blue}{-3}&1&-3&-19&7&39&-26\\& & & & & & \\ \hline &&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrr}-3&\color{orangered}{ 1 }&-3&-19&7&39&-26\\& & & & & & \\ \hline &\color{orangered}{1}&&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 1 } = \color{blue}{ -3 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-3}&1&-3&-19&7&39&-26\\& & \color{blue}{-3} & & & & \\ \hline &\color{blue}{1}&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -3 } + \color{orangered}{ \left( -3 \right) } = \color{orangered}{ -6 } $
$$ \begin{array}{c|rrrrrr}-3&1&\color{orangered}{ -3 }&-19&7&39&-26\\& & \color{orangered}{-3} & & & & \\ \hline &1&\color{orangered}{-6}&&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -6 \right) } = \color{blue}{ 18 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-3}&1&-3&-19&7&39&-26\\& & -3& \color{blue}{18} & & & \\ \hline &1&\color{blue}{-6}&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -19 } + \color{orangered}{ 18 } = \color{orangered}{ -1 } $
$$ \begin{array}{c|rrrrrr}-3&1&-3&\color{orangered}{ -19 }&7&39&-26\\& & -3& \color{orangered}{18} & & & \\ \hline &1&-6&\color{orangered}{-1}&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -1 \right) } = \color{blue}{ 3 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-3}&1&-3&-19&7&39&-26\\& & -3& 18& \color{blue}{3} & & \\ \hline &1&-6&\color{blue}{-1}&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 7 } + \color{orangered}{ 3 } = \color{orangered}{ 10 } $
$$ \begin{array}{c|rrrrrr}-3&1&-3&-19&\color{orangered}{ 7 }&39&-26\\& & -3& 18& \color{orangered}{3} & & \\ \hline &1&-6&-1&\color{orangered}{10}&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 10 } = \color{blue}{ -30 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-3}&1&-3&-19&7&39&-26\\& & -3& 18& 3& \color{blue}{-30} & \\ \hline &1&-6&-1&\color{blue}{10}&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 39 } + \color{orangered}{ \left( -30 \right) } = \color{orangered}{ 9 } $
$$ \begin{array}{c|rrrrrr}-3&1&-3&-19&7&\color{orangered}{ 39 }&-26\\& & -3& 18& 3& \color{orangered}{-30} & \\ \hline &1&-6&-1&10&\color{orangered}{9}& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 9 } = \color{blue}{ -27 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-3}&1&-3&-19&7&39&-26\\& & -3& 18& 3& -30& \color{blue}{-27} \\ \hline &1&-6&-1&10&\color{blue}{9}& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ -26 } + \color{orangered}{ \left( -27 \right) } = \color{orangered}{ -53 } $
$$ \begin{array}{c|rrrrrr}-3&1&-3&-19&7&39&\color{orangered}{ -26 }\\& & -3& 18& 3& -30& \color{orangered}{-27} \\ \hline &\color{blue}{1}&\color{blue}{-6}&\color{blue}{-1}&\color{blue}{10}&\color{blue}{9}&\color{orangered}{-53} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{4}-6x^{3}-x^{2}+10x+9 } $ with a remainder of $ \color{red}{ -53 } $.