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