Divide using synthetic division $$ ( -x^{5}+7x^{3}+x+1 ) \div ( x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrr}3&-1&0&7&0&1&1\\& & -3& -9& -6& -18& \color{black}{-51} \\ \hline &\color{blue}{-1}&\color{blue}{-3}&\color{blue}{-2}&\color{blue}{-6}&\color{blue}{-17}&\color{orangered}{-50} \end{array} $$The solution is:
$$ \frac{ -x^{5}+7x^{3}+x+1 }{ x-3 } = \color{blue}{-x^{4}-3x^{3}-2x^{2}-6x-17} \color{red}{~-~} \frac{ \color{red}{ 50 } }{ x-3 } $$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.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&-1&0&7&0&1&1\\& & & & & & \\ \hline &&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrr}3&\color{orangered}{ -1 }&0&7&0&1&1\\& & & & & & \\ \hline &\color{orangered}{-1}&&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -1 \right) } = \color{blue}{ -3 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&-1&0&7&0&1&1\\& & \color{blue}{-3} & & & & \\ \hline &\color{blue}{-1}&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -3 \right) } = \color{orangered}{ -3 } $
$$ \begin{array}{c|rrrrrr}3&-1&\color{orangered}{ 0 }&7&0&1&1\\& & \color{orangered}{-3} & & & & \\ \hline &-1&\color{orangered}{-3}&&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -3 \right) } = \color{blue}{ -9 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&-1&0&7&0&1&1\\& & -3& \color{blue}{-9} & & & \\ \hline &-1&\color{blue}{-3}&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 7 } + \color{orangered}{ \left( -9 \right) } = \color{orangered}{ -2 } $
$$ \begin{array}{c|rrrrrr}3&-1&0&\color{orangered}{ 7 }&0&1&1\\& & -3& \color{orangered}{-9} & & & \\ \hline &-1&-3&\color{orangered}{-2}&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -2 \right) } = \color{blue}{ -6 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&-1&0&7&0&1&1\\& & -3& -9& \color{blue}{-6} & & \\ \hline &-1&-3&\color{blue}{-2}&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -6 \right) } = \color{orangered}{ -6 } $
$$ \begin{array}{c|rrrrrr}3&-1&0&7&\color{orangered}{ 0 }&1&1\\& & -3& -9& \color{orangered}{-6} & & \\ \hline &-1&-3&-2&\color{orangered}{-6}&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -6 \right) } = \color{blue}{ -18 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&-1&0&7&0&1&1\\& & -3& -9& -6& \color{blue}{-18} & \\ \hline &-1&-3&-2&\color{blue}{-6}&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ \left( -18 \right) } = \color{orangered}{ -17 } $
$$ \begin{array}{c|rrrrrr}3&-1&0&7&0&\color{orangered}{ 1 }&1\\& & -3& -9& -6& \color{orangered}{-18} & \\ \hline &-1&-3&-2&-6&\color{orangered}{-17}& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -17 \right) } = \color{blue}{ -51 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&-1&0&7&0&1&1\\& & -3& -9& -6& -18& \color{blue}{-51} \\ \hline &-1&-3&-2&-6&\color{blue}{-17}& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ \left( -51 \right) } = \color{orangered}{ -50 } $
$$ \begin{array}{c|rrrrrr}3&-1&0&7&0&1&\color{orangered}{ 1 }\\& & -3& -9& -6& -18& \color{orangered}{-51} \\ \hline &\color{blue}{-1}&\color{blue}{-3}&\color{blue}{-2}&\color{blue}{-6}&\color{blue}{-17}&\color{orangered}{-50} \end{array} $$Bottom line represents the quotient $ \color{blue}{ -x^{4}-3x^{3}-2x^{2}-6x-17 } $ with a remainder of $ \color{red}{ -50 } $.