Divide using synthetic division $$ ( x^{5}-13x^{4}-120x+80 ) \div ( x+3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrr}-3&1&-13&0&0&-120&80\\& & -3& 48& -144& 432& \color{black}{-936} \\ \hline &\color{blue}{1}&\color{blue}{-16}&\color{blue}{48}&\color{blue}{-144}&\color{blue}{312}&\color{orangered}{-856} \end{array} $$The solution is:
$$ \frac{ x^{5}-13x^{4}-120x+80 }{ x+3 } = \color{blue}{x^{4}-16x^{3}+48x^{2}-144x+312} \color{red}{~-~} \frac{ \color{red}{ 856 } }{ x+3 } $$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.
$$ \begin{array}{c|rrrrrr}\color{blue}{-3}&1&-13&0&0&-120&80\\& & & & & & \\ \hline &&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrr}-3&\color{orangered}{ 1 }&-13&0&0&-120&80\\& & & & & & \\ \hline &\color{orangered}{1}&&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-3}&1&-13&0&0&-120&80\\& & \color{blue}{-3} & & & & \\ \hline &\color{blue}{1}&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -13 } + \color{orangered}{ \left( -3 \right) } = \color{orangered}{ -16 } $
$$ \begin{array}{c|rrrrrr}-3&1&\color{orangered}{ -13 }&0&0&-120&80\\& & \color{orangered}{-3} & & & & \\ \hline &1&\color{orangered}{-16}&&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -16 \right) } = \color{blue}{ 48 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-3}&1&-13&0&0&-120&80\\& & -3& \color{blue}{48} & & & \\ \hline &1&\color{blue}{-16}&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 48 } = \color{orangered}{ 48 } $
$$ \begin{array}{c|rrrrrr}-3&1&-13&\color{orangered}{ 0 }&0&-120&80\\& & -3& \color{orangered}{48} & & & \\ \hline &1&-16&\color{orangered}{48}&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 48 } = \color{blue}{ -144 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-3}&1&-13&0&0&-120&80\\& & -3& 48& \color{blue}{-144} & & \\ \hline &1&-16&\color{blue}{48}&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -144 \right) } = \color{orangered}{ -144 } $
$$ \begin{array}{c|rrrrrr}-3&1&-13&0&\color{orangered}{ 0 }&-120&80\\& & -3& 48& \color{orangered}{-144} & & \\ \hline &1&-16&48&\color{orangered}{-144}&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -144 \right) } = \color{blue}{ 432 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-3}&1&-13&0&0&-120&80\\& & -3& 48& -144& \color{blue}{432} & \\ \hline &1&-16&48&\color{blue}{-144}&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -120 } + \color{orangered}{ 432 } = \color{orangered}{ 312 } $
$$ \begin{array}{c|rrrrrr}-3&1&-13&0&0&\color{orangered}{ -120 }&80\\& & -3& 48& -144& \color{orangered}{432} & \\ \hline &1&-16&48&-144&\color{orangered}{312}& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 312 } = \color{blue}{ -936 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-3}&1&-13&0&0&-120&80\\& & -3& 48& -144& 432& \color{blue}{-936} \\ \hline &1&-16&48&-144&\color{blue}{312}& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ 80 } + \color{orangered}{ \left( -936 \right) } = \color{orangered}{ -856 } $
$$ \begin{array}{c|rrrrrr}-3&1&-13&0&0&-120&\color{orangered}{ 80 }\\& & -3& 48& -144& 432& \color{orangered}{-936} \\ \hline &\color{blue}{1}&\color{blue}{-16}&\color{blue}{48}&\color{blue}{-144}&\color{blue}{312}&\color{orangered}{-856} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{4}-16x^{3}+48x^{2}-144x+312 } $ with a remainder of $ \color{red}{ -856 } $.