Divide using synthetic division $$ ( x^{4}-x^{3}+4 ) \div ( x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}3&1&-1&0&0&4\\& & 3& 6& 18& \color{black}{54} \\ \hline &\color{blue}{1}&\color{blue}{2}&\color{blue}{6}&\color{blue}{18}&\color{orangered}{58} \end{array} $$The solution is:
$$ \frac{ x^{4}-x^{3}+4 }{ x-3 } = \color{blue}{x^{3}+2x^{2}+6x+18} ~+~ \frac{ \color{red}{ 58 } }{ x-3 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&1&-1&0&0&4\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}3&\color{orangered}{ 1 }&-1&0&0&4\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&1&-1&0&0&4\\& & \color{blue}{3} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ 3 } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrrr}3&1&\color{orangered}{ -1 }&0&0&4\\& & \color{orangered}{3} & & & \\ \hline &1&\color{orangered}{2}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&1&-1&0&0&4\\& & 3& \color{blue}{6} & & \\ \hline &1&\color{blue}{2}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 6 } = \color{orangered}{ 6 } $
$$ \begin{array}{c|rrrrr}3&1&-1&\color{orangered}{ 0 }&0&4\\& & 3& \color{orangered}{6} & & \\ \hline &1&2&\color{orangered}{6}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 6 } = \color{blue}{ 18 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&1&-1&0&0&4\\& & 3& 6& \color{blue}{18} & \\ \hline &1&2&\color{blue}{6}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 18 } = \color{orangered}{ 18 } $
$$ \begin{array}{c|rrrrr}3&1&-1&0&\color{orangered}{ 0 }&4\\& & 3& 6& \color{orangered}{18} & \\ \hline &1&2&6&\color{orangered}{18}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 18 } = \color{blue}{ 54 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&1&-1&0&0&4\\& & 3& 6& 18& \color{blue}{54} \\ \hline &1&2&6&\color{blue}{18}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 4 } + \color{orangered}{ 54 } = \color{orangered}{ 58 } $
$$ \begin{array}{c|rrrrr}3&1&-1&0&0&\color{orangered}{ 4 }\\& & 3& 6& 18& \color{orangered}{54} \\ \hline &\color{blue}{1}&\color{blue}{2}&\color{blue}{6}&\color{blue}{18}&\color{orangered}{58} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}+2x^{2}+6x+18 } $ with a remainder of $ \color{red}{ 58 } $.