Divide using synthetic division $$ ( x^{4}+6x^{3}+8x^{2}-30x-65 ) \div ( 2x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}3&1&6&8&-30&-65\\& & 3& 27& 105& \color{black}{225} \\ \hline &\color{blue}{1}&\color{blue}{9}&\color{blue}{35}&\color{blue}{75}&\color{orangered}{160} \end{array} $$The solution is:
$$ \frac{ x^{4}+6x^{3}+8x^{2}-30x-65 }{ x-3 } = \color{blue}{x^{3}+9x^{2}+35x+75} ~+~ \frac{ \color{red}{ 160 } }{ x-3 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&1&6&8&-30&-65\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}3&\color{orangered}{ 1 }&6&8&-30&-65\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&1&6&8&-30&-65\\& & \color{blue}{3} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 6 } + \color{orangered}{ 3 } = \color{orangered}{ 9 } $
$$ \begin{array}{c|rrrrr}3&1&\color{orangered}{ 6 }&8&-30&-65\\& & \color{orangered}{3} & & & \\ \hline &1&\color{orangered}{9}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 9 } = \color{blue}{ 27 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&1&6&8&-30&-65\\& & 3& \color{blue}{27} & & \\ \hline &1&\color{blue}{9}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 8 } + \color{orangered}{ 27 } = \color{orangered}{ 35 } $
$$ \begin{array}{c|rrrrr}3&1&6&\color{orangered}{ 8 }&-30&-65\\& & 3& \color{orangered}{27} & & \\ \hline &1&9&\color{orangered}{35}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 35 } = \color{blue}{ 105 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&1&6&8&-30&-65\\& & 3& 27& \color{blue}{105} & \\ \hline &1&9&\color{blue}{35}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -30 } + \color{orangered}{ 105 } = \color{orangered}{ 75 } $
$$ \begin{array}{c|rrrrr}3&1&6&8&\color{orangered}{ -30 }&-65\\& & 3& 27& \color{orangered}{105} & \\ \hline &1&9&35&\color{orangered}{75}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 75 } = \color{blue}{ 225 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&1&6&8&-30&-65\\& & 3& 27& 105& \color{blue}{225} \\ \hline &1&9&35&\color{blue}{75}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -65 } + \color{orangered}{ 225 } = \color{orangered}{ 160 } $
$$ \begin{array}{c|rrrrr}3&1&6&8&-30&\color{orangered}{ -65 }\\& & 3& 27& 105& \color{orangered}{225} \\ \hline &\color{blue}{1}&\color{blue}{9}&\color{blue}{35}&\color{blue}{75}&\color{orangered}{160} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}+9x^{2}+35x+75 } $ with a remainder of $ \color{red}{ 160 } $.