Divide using synthetic division $$ ( 5x^{4}+x^{3}-x^{2}-2x+6 ) \div ( x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}3&5&1&-1&-2&6\\& & 15& 48& 141& \color{black}{417} \\ \hline &\color{blue}{5}&\color{blue}{16}&\color{blue}{47}&\color{blue}{139}&\color{orangered}{423} \end{array} $$The solution is:
$$ \frac{ 5x^{4}+x^{3}-x^{2}-2x+6 }{ x-3 } = \color{blue}{5x^{3}+16x^{2}+47x+139} ~+~ \frac{ \color{red}{ 423 } }{ x-3 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&5&1&-1&-2&6\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}3&\color{orangered}{ 5 }&1&-1&-2&6\\& & & & & \\ \hline &\color{orangered}{5}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 5 } = \color{blue}{ 15 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&5&1&-1&-2&6\\& & \color{blue}{15} & & & \\ \hline &\color{blue}{5}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ 15 } = \color{orangered}{ 16 } $
$$ \begin{array}{c|rrrrr}3&5&\color{orangered}{ 1 }&-1&-2&6\\& & \color{orangered}{15} & & & \\ \hline &5&\color{orangered}{16}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 16 } = \color{blue}{ 48 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&5&1&-1&-2&6\\& & 15& \color{blue}{48} & & \\ \hline &5&\color{blue}{16}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ 48 } = \color{orangered}{ 47 } $
$$ \begin{array}{c|rrrrr}3&5&1&\color{orangered}{ -1 }&-2&6\\& & 15& \color{orangered}{48} & & \\ \hline &5&16&\color{orangered}{47}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 47 } = \color{blue}{ 141 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&5&1&-1&-2&6\\& & 15& 48& \color{blue}{141} & \\ \hline &5&16&\color{blue}{47}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ 141 } = \color{orangered}{ 139 } $
$$ \begin{array}{c|rrrrr}3&5&1&-1&\color{orangered}{ -2 }&6\\& & 15& 48& \color{orangered}{141} & \\ \hline &5&16&47&\color{orangered}{139}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 139 } = \color{blue}{ 417 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&5&1&-1&-2&6\\& & 15& 48& 141& \color{blue}{417} \\ \hline &5&16&47&\color{blue}{139}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 6 } + \color{orangered}{ 417 } = \color{orangered}{ 423 } $
$$ \begin{array}{c|rrrrr}3&5&1&-1&-2&\color{orangered}{ 6 }\\& & 15& 48& 141& \color{orangered}{417} \\ \hline &\color{blue}{5}&\color{blue}{16}&\color{blue}{47}&\color{blue}{139}&\color{orangered}{423} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 5x^{3}+16x^{2}+47x+139 } $ with a remainder of $ \color{red}{ 423 } $.