Divide using synthetic division $$ ( x^{3}-12x^{2}+52x-80 ) \div ( 2x+4 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-4&1&-12&52&-80\\& & -4& 64& \color{black}{-464} \\ \hline &\color{blue}{1}&\color{blue}{-16}&\color{blue}{116}&\color{orangered}{-544} \end{array} $$The solution is:
$$ \frac{ x^{3}-12x^{2}+52x-80 }{ x+4 } = \color{blue}{x^{2}-16x+116} \color{red}{~-~} \frac{ \color{red}{ 544 } }{ x+4 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{-4}&1&-12&52&-80\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-4&\color{orangered}{ 1 }&-12&52&-80\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ 1 } = \color{blue}{ -4 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-4}&1&-12&52&-80\\& & \color{blue}{-4} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -12 } + \color{orangered}{ \left( -4 \right) } = \color{orangered}{ -16 } $
$$ \begin{array}{c|rrrr}-4&1&\color{orangered}{ -12 }&52&-80\\& & \color{orangered}{-4} & & \\ \hline &1&\color{orangered}{-16}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ \left( -16 \right) } = \color{blue}{ 64 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-4}&1&-12&52&-80\\& & -4& \color{blue}{64} & \\ \hline &1&\color{blue}{-16}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 52 } + \color{orangered}{ 64 } = \color{orangered}{ 116 } $
$$ \begin{array}{c|rrrr}-4&1&-12&\color{orangered}{ 52 }&-80\\& & -4& \color{orangered}{64} & \\ \hline &1&-16&\color{orangered}{116}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ 116 } = \color{blue}{ -464 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-4}&1&-12&52&-80\\& & -4& 64& \color{blue}{-464} \\ \hline &1&-16&\color{blue}{116}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -80 } + \color{orangered}{ \left( -464 \right) } = \color{orangered}{ -544 } $
$$ \begin{array}{c|rrrr}-4&1&-12&52&\color{orangered}{ -80 }\\& & -4& 64& \color{orangered}{-464} \\ \hline &\color{blue}{1}&\color{blue}{-16}&\color{blue}{116}&\color{orangered}{-544} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{2}-16x+116 } $ with a remainder of $ \color{red}{ -544 } $.