Divide using synthetic division $$ ( x^{4}-x^{3}-x^{2}+x+3 ) \div ( x-3 ) $$
The synthetic division table is:
$$ \begin{array}{c|rrrrr}3&1&-1&-1&1&3\\& & 3& 6& 15& \color{black}{48} \\ \hline &\color{blue}{1}&\color{blue}{2}&\color{blue}{5}&\color{blue}{16}&\color{orangered}{51} \end{array} $$The solution is:
$$ \dfrac{ x^{4}-x^{3}-x^{2}+x+3 }{ x-3 } = \color{blue}{x^{3}+2x^{2}+5x+16} ~+~ \dfrac{ \color{red}{ 51 } }{ x-3 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&1&-1&-1&1&3\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}3&\color{orangered}{ 1 }&-1&-1&1&3\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&1&-1&-1&1&3\\& & \color{blue}{3} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ 3 } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrrr}3&1&\color{orangered}{ -1 }&-1&1&3\\& & \color{orangered}{3} & & & \\ \hline &1&\color{orangered}{2}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 2 } = \color{blue}{ 6 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&1&-1&-1&1&3\\& & 3& \color{blue}{6} & & \\ \hline &1&\color{blue}{2}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ 6 } = \color{orangered}{ 5 } $
$$ \begin{array}{c|rrrrr}3&1&-1&\color{orangered}{ -1 }&1&3\\& & 3& \color{orangered}{6} & & \\ \hline &1&2&\color{orangered}{5}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 5 } = \color{blue}{ 15 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&1&-1&-1&1&3\\& & 3& 6& \color{blue}{15} & \\ \hline &1&2&\color{blue}{5}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ 15 } = \color{orangered}{ 16 } $
$$ \begin{array}{c|rrrrr}3&1&-1&-1&\color{orangered}{ 1 }&3\\& & 3& 6& \color{orangered}{15} & \\ \hline &1&2&5&\color{orangered}{16}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 16 } = \color{blue}{ 48 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&1&-1&-1&1&3\\& & 3& 6& 15& \color{blue}{48} \\ \hline &1&2&5&\color{blue}{16}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ 48 } = \color{orangered}{ 51 } $
$$ \begin{array}{c|rrrrr}3&1&-1&-1&1&\color{orangered}{ 3 }\\& & 3& 6& 15& \color{orangered}{48} \\ \hline &\color{blue}{1}&\color{blue}{2}&\color{blue}{5}&\color{blue}{16}&\color{orangered}{51} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}+2x^{2}+5x+16 } $ with a remainder of $ \color{red}{ 51 } $.