Divide using synthetic division $$ ( -x^{4}+4x^{3}-x+3 ) \div ( x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}3&-1&4&0&-1&3\\& & -3& 3& 9& \color{black}{24} \\ \hline &\color{blue}{-1}&\color{blue}{1}&\color{blue}{3}&\color{blue}{8}&\color{orangered}{27} \end{array} $$The solution is:
$$ \frac{ -x^{4}+4x^{3}-x+3 }{ x-3 } = \color{blue}{-x^{3}+x^{2}+3x+8} ~+~ \frac{ \color{red}{ 27 } }{ x-3 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&-1&4&0&-1&3\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}3&\color{orangered}{ -1 }&4&0&-1&3\\& & & & & \\ \hline &\color{orangered}{-1}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -1 \right) } = \color{blue}{ -3 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&-1&4&0&-1&3\\& & \color{blue}{-3} & & & \\ \hline &\color{blue}{-1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 4 } + \color{orangered}{ \left( -3 \right) } = \color{orangered}{ 1 } $
$$ \begin{array}{c|rrrrr}3&-1&\color{orangered}{ 4 }&0&-1&3\\& & \color{orangered}{-3} & & & \\ \hline &-1&\color{orangered}{1}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 1 } = \color{blue}{ 3 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&-1&4&0&-1&3\\& & -3& \color{blue}{3} & & \\ \hline &-1&\color{blue}{1}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 3 } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrrr}3&-1&4&\color{orangered}{ 0 }&-1&3\\& & -3& \color{orangered}{3} & & \\ \hline &-1&1&\color{orangered}{3}&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&-1&4&0&-1&3\\& & -3& 3& \color{blue}{9} & \\ \hline &-1&1&\color{blue}{3}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ 9 } = \color{orangered}{ 8 } $
$$ \begin{array}{c|rrrrr}3&-1&4&0&\color{orangered}{ -1 }&3\\& & -3& 3& \color{orangered}{9} & \\ \hline &-1&1&3&\color{orangered}{8}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 8 } = \color{blue}{ 24 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&-1&4&0&-1&3\\& & -3& 3& 9& \color{blue}{24} \\ \hline &-1&1&3&\color{blue}{8}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ 24 } = \color{orangered}{ 27 } $
$$ \begin{array}{c|rrrrr}3&-1&4&0&-1&\color{orangered}{ 3 }\\& & -3& 3& 9& \color{orangered}{24} \\ \hline &\color{blue}{-1}&\color{blue}{1}&\color{blue}{3}&\color{blue}{8}&\color{orangered}{27} \end{array} $$Bottom line represents the quotient $ \color{blue}{ -x^{3}+x^{2}+3x+8 } $ with a remainder of $ \color{red}{ 27 } $.