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