Divide using synthetic division $$ ( 2x^{3}+4x^{2}+6x+3 ) \div ( x-2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}2&2&4&6&3\\& & 4& 16& \color{black}{44} \\ \hline &\color{blue}{2}&\color{blue}{8}&\color{blue}{22}&\color{orangered}{47} \end{array} $$The solution is:
$$ \frac{ 2x^{3}+4x^{2}+6x+3 }{ x-2 } = \color{blue}{2x^{2}+8x+22} ~+~ \frac{ \color{red}{ 47 } }{ x-2 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{2}&2&4&6&3\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}2&\color{orangered}{ 2 }&4&6&3\\& & & & \\ \hline &\color{orangered}{2}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 2 } = \color{blue}{ 4 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&2&4&6&3\\& & \color{blue}{4} & & \\ \hline &\color{blue}{2}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 4 } + \color{orangered}{ 4 } = \color{orangered}{ 8 } $
$$ \begin{array}{c|rrrr}2&2&\color{orangered}{ 4 }&6&3\\& & \color{orangered}{4} & & \\ \hline &2&\color{orangered}{8}&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&2&4&6&3\\& & 4& \color{blue}{16} & \\ \hline &2&\color{blue}{8}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 6 } + \color{orangered}{ 16 } = \color{orangered}{ 22 } $
$$ \begin{array}{c|rrrr}2&2&4&\color{orangered}{ 6 }&3\\& & 4& \color{orangered}{16} & \\ \hline &2&8&\color{orangered}{22}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 22 } = \color{blue}{ 44 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&2&4&6&3\\& & 4& 16& \color{blue}{44} \\ \hline &2&8&\color{blue}{22}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ 44 } = \color{orangered}{ 47 } $
$$ \begin{array}{c|rrrr}2&2&4&6&\color{orangered}{ 3 }\\& & 4& 16& \color{orangered}{44} \\ \hline &\color{blue}{2}&\color{blue}{8}&\color{blue}{22}&\color{orangered}{47} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{2}+8x+22 } $ with a remainder of $ \color{red}{ 47 } $.