Divide using synthetic division $$ ( 3x^{4}+14x^{3}+9x^{2}-12x+4 ) \div ( x+2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-2&3&14&9&-12&4\\& & -6& -16& 14& \color{black}{-4} \\ \hline &\color{blue}{3}&\color{blue}{8}&\color{blue}{-7}&\color{blue}{2}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ 3x^{4}+14x^{3}+9x^{2}-12x+4 }{ x+2 } = \color{blue}{3x^{3}+8x^{2}-7x+2} $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&3&14&9&-12&4\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-2&\color{orangered}{ 3 }&14&9&-12&4\\& & & & & \\ \hline &\color{orangered}{3}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&3&14&9&-12&4\\& & \color{blue}{-6} & & & \\ \hline &\color{blue}{3}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 14 } + \color{orangered}{ \left( -6 \right) } = \color{orangered}{ 8 } $
$$ \begin{array}{c|rrrrr}-2&3&\color{orangered}{ 14 }&9&-12&4\\& & \color{orangered}{-6} & & & \\ \hline &3&\color{orangered}{8}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&3&14&9&-12&4\\& & -6& \color{blue}{-16} & & \\ \hline &3&\color{blue}{8}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 9 } + \color{orangered}{ \left( -16 \right) } = \color{orangered}{ -7 } $
$$ \begin{array}{c|rrrrr}-2&3&14&\color{orangered}{ 9 }&-12&4\\& & -6& \color{orangered}{-16} & & \\ \hline &3&8&\color{orangered}{-7}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -7 \right) } = \color{blue}{ 14 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&3&14&9&-12&4\\& & -6& -16& \color{blue}{14} & \\ \hline &3&8&\color{blue}{-7}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -12 } + \color{orangered}{ 14 } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrrr}-2&3&14&9&\color{orangered}{ -12 }&4\\& & -6& -16& \color{orangered}{14} & \\ \hline &3&8&-7&\color{orangered}{2}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 2 } = \color{blue}{ -4 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&3&14&9&-12&4\\& & -6& -16& 14& \color{blue}{-4} \\ \hline &3&8&-7&\color{blue}{2}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 4 } + \color{orangered}{ \left( -4 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}-2&3&14&9&-12&\color{orangered}{ 4 }\\& & -6& -16& 14& \color{orangered}{-4} \\ \hline &\color{blue}{3}&\color{blue}{8}&\color{blue}{-7}&\color{blue}{2}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 3x^{3}+8x^{2}-7x+2 } $ with a remainder of $ \color{red}{ 0 } $.