Divide using synthetic division $$ ( 5x^{4}-17x^{3}+19x+2 ) \div ( x-4 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}4&5&-17&0&19&2\\& & 20& 12& 48& \color{black}{268} \\ \hline &\color{blue}{5}&\color{blue}{3}&\color{blue}{12}&\color{blue}{67}&\color{orangered}{270} \end{array} $$The solution is:
$$ \frac{ 5x^{4}-17x^{3}+19x+2 }{ x-4 } = \color{blue}{5x^{3}+3x^{2}+12x+67} ~+~ \frac{ \color{red}{ 270 } }{ x-4 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&5&-17&0&19&2\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}4&\color{orangered}{ 5 }&-17&0&19&2\\& & & & & \\ \hline &\color{orangered}{5}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 5 } = \color{blue}{ 20 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&5&-17&0&19&2\\& & \color{blue}{20} & & & \\ \hline &\color{blue}{5}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -17 } + \color{orangered}{ 20 } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrrr}4&5&\color{orangered}{ -17 }&0&19&2\\& & \color{orangered}{20} & & & \\ \hline &5&\color{orangered}{3}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 3 } = \color{blue}{ 12 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&5&-17&0&19&2\\& & 20& \color{blue}{12} & & \\ \hline &5&\color{blue}{3}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 12 } = \color{orangered}{ 12 } $
$$ \begin{array}{c|rrrrr}4&5&-17&\color{orangered}{ 0 }&19&2\\& & 20& \color{orangered}{12} & & \\ \hline &5&3&\color{orangered}{12}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 12 } = \color{blue}{ 48 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&5&-17&0&19&2\\& & 20& 12& \color{blue}{48} & \\ \hline &5&3&\color{blue}{12}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 19 } + \color{orangered}{ 48 } = \color{orangered}{ 67 } $
$$ \begin{array}{c|rrrrr}4&5&-17&0&\color{orangered}{ 19 }&2\\& & 20& 12& \color{orangered}{48} & \\ \hline &5&3&12&\color{orangered}{67}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 67 } = \color{blue}{ 268 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&5&-17&0&19&2\\& & 20& 12& 48& \color{blue}{268} \\ \hline &5&3&12&\color{blue}{67}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ 268 } = \color{orangered}{ 270 } $
$$ \begin{array}{c|rrrrr}4&5&-17&0&19&\color{orangered}{ 2 }\\& & 20& 12& 48& \color{orangered}{268} \\ \hline &\color{blue}{5}&\color{blue}{3}&\color{blue}{12}&\color{blue}{67}&\color{orangered}{270} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 5x^{3}+3x^{2}+12x+67 } $ with a remainder of $ \color{red}{ 270 } $.