Divide using synthetic division $$ ( x^{4}-x^{3}-15x^{2}+48x-20 ) \div ( x+2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-2&1&-1&-15&48&-20\\& & -2& 6& 18& \color{black}{-132} \\ \hline &\color{blue}{1}&\color{blue}{-3}&\color{blue}{-9}&\color{blue}{66}&\color{orangered}{-152} \end{array} $$The solution is:
$$ \frac{ x^{4}-x^{3}-15x^{2}+48x-20 }{ x+2 } = \color{blue}{x^{3}-3x^{2}-9x+66} \color{red}{~-~} \frac{ \color{red}{ 152 } }{ x+2 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&1&-1&-15&48&-20\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-2&\color{orangered}{ 1 }&-1&-15&48&-20\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 1 } = \color{blue}{ -2 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&1&-1&-15&48&-20\\& & \color{blue}{-2} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ \left( -2 \right) } = \color{orangered}{ -3 } $
$$ \begin{array}{c|rrrrr}-2&1&\color{orangered}{ -1 }&-15&48&-20\\& & \color{orangered}{-2} & & & \\ \hline &1&\color{orangered}{-3}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -3 \right) } = \color{blue}{ 6 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&1&-1&-15&48&-20\\& & -2& \color{blue}{6} & & \\ \hline &1&\color{blue}{-3}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -15 } + \color{orangered}{ 6 } = \color{orangered}{ -9 } $
$$ \begin{array}{c|rrrrr}-2&1&-1&\color{orangered}{ -15 }&48&-20\\& & -2& \color{orangered}{6} & & \\ \hline &1&-3&\color{orangered}{-9}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -9 \right) } = \color{blue}{ 18 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&1&-1&-15&48&-20\\& & -2& 6& \color{blue}{18} & \\ \hline &1&-3&\color{blue}{-9}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 48 } + \color{orangered}{ 18 } = \color{orangered}{ 66 } $
$$ \begin{array}{c|rrrrr}-2&1&-1&-15&\color{orangered}{ 48 }&-20\\& & -2& 6& \color{orangered}{18} & \\ \hline &1&-3&-9&\color{orangered}{66}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 66 } = \color{blue}{ -132 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&1&-1&-15&48&-20\\& & -2& 6& 18& \color{blue}{-132} \\ \hline &1&-3&-9&\color{blue}{66}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -20 } + \color{orangered}{ \left( -132 \right) } = \color{orangered}{ -152 } $
$$ \begin{array}{c|rrrrr}-2&1&-1&-15&48&\color{orangered}{ -20 }\\& & -2& 6& 18& \color{orangered}{-132} \\ \hline &\color{blue}{1}&\color{blue}{-3}&\color{blue}{-9}&\color{blue}{66}&\color{orangered}{-152} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}-3x^{2}-9x+66 } $ with a remainder of $ \color{red}{ -152 } $.