Divide using synthetic division $$ ( x^{4}-4x^{3}-3x+13 ) \div ( x-4 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}4&1&-4&0&-3&13\\& & 4& 0& 0& \color{black}{-12} \\ \hline &\color{blue}{1}&\color{blue}{0}&\color{blue}{0}&\color{blue}{-3}&\color{orangered}{1} \end{array} $$The solution is:
$$ \frac{ x^{4}-4x^{3}-3x+13 }{ x-4 } = \color{blue}{x^{3}-3} ~+~ \frac{ \color{red}{ 1 } }{ x-4 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -4 = 0 $ ( $ x = \color{blue}{ 4 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&1&-4&0&-3&13\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}4&\color{orangered}{ 1 }&-4&0&-3&13\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 1 } = \color{blue}{ 4 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&1&-4&0&-3&13\\& & \color{blue}{4} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -4 } + \color{orangered}{ 4 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}4&1&\color{orangered}{ -4 }&0&-3&13\\& & \color{orangered}{4} & & & \\ \hline &1&\color{orangered}{0}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 0 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&1&-4&0&-3&13\\& & 4& \color{blue}{0} & & \\ \hline &1&\color{blue}{0}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 0 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}4&1&-4&\color{orangered}{ 0 }&-3&13\\& & 4& \color{orangered}{0} & & \\ \hline &1&0&\color{orangered}{0}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 0 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&1&-4&0&-3&13\\& & 4& 0& \color{blue}{0} & \\ \hline &1&0&\color{blue}{0}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -3 } + \color{orangered}{ 0 } = \color{orangered}{ -3 } $
$$ \begin{array}{c|rrrrr}4&1&-4&0&\color{orangered}{ -3 }&13\\& & 4& 0& \color{orangered}{0} & \\ \hline &1&0&0&\color{orangered}{-3}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ \left( -3 \right) } = \color{blue}{ -12 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&1&-4&0&-3&13\\& & 4& 0& 0& \color{blue}{-12} \\ \hline &1&0&0&\color{blue}{-3}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 13 } + \color{orangered}{ \left( -12 \right) } = \color{orangered}{ 1 } $
$$ \begin{array}{c|rrrrr}4&1&-4&0&-3&\color{orangered}{ 13 }\\& & 4& 0& 0& \color{orangered}{-12} \\ \hline &\color{blue}{1}&\color{blue}{0}&\color{blue}{0}&\color{blue}{-3}&\color{orangered}{1} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}-3 } $ with a remainder of $ \color{red}{ 1 } $.