Divide using synthetic division $$ ( 3x^{4}+2x^{3}-9x^{2}-8 ) \div ( x-2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}2&3&2&-9&0&-8\\& & 6& 16& 14& \color{black}{28} \\ \hline &\color{blue}{3}&\color{blue}{8}&\color{blue}{7}&\color{blue}{14}&\color{orangered}{20} \end{array} $$The solution is:
$$ \frac{ 3x^{4}+2x^{3}-9x^{2}-8 }{ x-2 } = \color{blue}{3x^{3}+8x^{2}+7x+14} ~+~ \frac{ \color{red}{ 20 } }{ x-2 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&3&2&-9&0&-8\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}2&\color{orangered}{ 3 }&2&-9&0&-8\\& & & & & \\ \hline &\color{orangered}{3}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&3&2&-9&0&-8\\& & \color{blue}{6} & & & \\ \hline &\color{blue}{3}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ 6 } = \color{orangered}{ 8 } $
$$ \begin{array}{c|rrrrr}2&3&\color{orangered}{ 2 }&-9&0&-8\\& & \color{orangered}{6} & & & \\ \hline &3&\color{orangered}{8}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 8 } = \color{blue}{ 16 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&3&2&-9&0&-8\\& & 6& \color{blue}{16} & & \\ \hline &3&\color{blue}{8}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -9 } + \color{orangered}{ 16 } = \color{orangered}{ 7 } $
$$ \begin{array}{c|rrrrr}2&3&2&\color{orangered}{ -9 }&0&-8\\& & 6& \color{orangered}{16} & & \\ \hline &3&8&\color{orangered}{7}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 7 } = \color{blue}{ 14 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&3&2&-9&0&-8\\& & 6& 16& \color{blue}{14} & \\ \hline &3&8&\color{blue}{7}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 14 } = \color{orangered}{ 14 } $
$$ \begin{array}{c|rrrrr}2&3&2&-9&\color{orangered}{ 0 }&-8\\& & 6& 16& \color{orangered}{14} & \\ \hline &3&8&7&\color{orangered}{14}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 14 } = \color{blue}{ 28 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&3&2&-9&0&-8\\& & 6& 16& 14& \color{blue}{28} \\ \hline &3&8&7&\color{blue}{14}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -8 } + \color{orangered}{ 28 } = \color{orangered}{ 20 } $
$$ \begin{array}{c|rrrrr}2&3&2&-9&0&\color{orangered}{ -8 }\\& & 6& 16& 14& \color{orangered}{28} \\ \hline &\color{blue}{3}&\color{blue}{8}&\color{blue}{7}&\color{blue}{14}&\color{orangered}{20} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 3x^{3}+8x^{2}+7x+14 } $ with a remainder of $ \color{red}{ 20 } $.