Divide using synthetic division $$ ( 2x^{3}+9x^{2}-7x+1 ) \div ( x-3 ) $$
The synthetic division table is:
$$ \begin{array}{c|rrrr}3&2&9&-7&1\\& & 6& 45& \color{black}{114} \\ \hline &\color{blue}{2}&\color{blue}{15}&\color{blue}{38}&\color{orangered}{115} \end{array} $$The solution is:
$$ \dfrac{ 2x^{3}+9x^{2}-7x+1 }{ x-3 } = \color{blue}{2x^{2}+15x+38} ~+~ \dfrac{ \color{red}{ 115 } }{ x-3 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{3}&2&9&-7&1\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}3&\color{orangered}{ 2 }&9&-7&1\\& & & & \\ \hline &\color{orangered}{2}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 2 } = \color{blue}{ 6 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&2&9&-7&1\\& & \color{blue}{6} & & \\ \hline &\color{blue}{2}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 9 } + \color{orangered}{ 6 } = \color{orangered}{ 15 } $
$$ \begin{array}{c|rrrr}3&2&\color{orangered}{ 9 }&-7&1\\& & \color{orangered}{6} & & \\ \hline &2&\color{orangered}{15}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 15 } = \color{blue}{ 45 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&2&9&-7&1\\& & 6& \color{blue}{45} & \\ \hline &2&\color{blue}{15}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -7 } + \color{orangered}{ 45 } = \color{orangered}{ 38 } $
$$ \begin{array}{c|rrrr}3&2&9&\color{orangered}{ -7 }&1\\& & 6& \color{orangered}{45} & \\ \hline &2&15&\color{orangered}{38}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 38 } = \color{blue}{ 114 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&2&9&-7&1\\& & 6& 45& \color{blue}{114} \\ \hline &2&15&\color{blue}{38}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ 114 } = \color{orangered}{ 115 } $
$$ \begin{array}{c|rrrr}3&2&9&-7&\color{orangered}{ 1 }\\& & 6& 45& \color{orangered}{114} \\ \hline &\color{blue}{2}&\color{blue}{15}&\color{blue}{38}&\color{orangered}{115} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{2}+15x+38 } $ with a remainder of $ \color{red}{ 115 } $.