Divide using synthetic division $$ ( 3x^{4}-2x^{3}-7x^{2}+4x+2 ) \div ( x-2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}2&3&-2&-7&4&2\\& & 6& 8& 2& \color{black}{12} \\ \hline &\color{blue}{3}&\color{blue}{4}&\color{blue}{1}&\color{blue}{6}&\color{orangered}{14} \end{array} $$The solution is:
$$ \frac{ 3x^{4}-2x^{3}-7x^{2}+4x+2 }{ x-2 } = \color{blue}{3x^{3}+4x^{2}+x+6} ~+~ \frac{ \color{red}{ 14 } }{ x-2 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -2 = 0 $ ( $ x = \color{blue}{ 2 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&3&-2&-7&4&2\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}2&\color{orangered}{ 3 }&-2&-7&4&2\\& & & & & \\ \hline &\color{orangered}{3}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 3 } = \color{blue}{ 6 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&3&-2&-7&4&2\\& & \color{blue}{6} & & & \\ \hline &\color{blue}{3}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ 6 } = \color{orangered}{ 4 } $
$$ \begin{array}{c|rrrrr}2&3&\color{orangered}{ -2 }&-7&4&2\\& & \color{orangered}{6} & & & \\ \hline &3&\color{orangered}{4}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 4 } = \color{blue}{ 8 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&3&-2&-7&4&2\\& & 6& \color{blue}{8} & & \\ \hline &3&\color{blue}{4}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -7 } + \color{orangered}{ 8 } = \color{orangered}{ 1 } $
$$ \begin{array}{c|rrrrr}2&3&-2&\color{orangered}{ -7 }&4&2\\& & 6& \color{orangered}{8} & & \\ \hline &3&4&\color{orangered}{1}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 1 } = \color{blue}{ 2 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&3&-2&-7&4&2\\& & 6& 8& \color{blue}{2} & \\ \hline &3&4&\color{blue}{1}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 4 } + \color{orangered}{ 2 } = \color{orangered}{ 6 } $
$$ \begin{array}{c|rrrrr}2&3&-2&-7&\color{orangered}{ 4 }&2\\& & 6& 8& \color{orangered}{2} & \\ \hline &3&4&1&\color{orangered}{6}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 6 } = \color{blue}{ 12 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&3&-2&-7&4&2\\& & 6& 8& 2& \color{blue}{12} \\ \hline &3&4&1&\color{blue}{6}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ 12 } = \color{orangered}{ 14 } $
$$ \begin{array}{c|rrrrr}2&3&-2&-7&4&\color{orangered}{ 2 }\\& & 6& 8& 2& \color{orangered}{12} \\ \hline &\color{blue}{3}&\color{blue}{4}&\color{blue}{1}&\color{blue}{6}&\color{orangered}{14} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 3x^{3}+4x^{2}+x+6 } $ with a remainder of $ \color{red}{ 14 } $.