Divide using synthetic division $$ ( x^{4}-6x^{3}-10x^{2}+20x+15 ) \div ( x+2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-2&1&-6&-10&20&15\\& & -2& 16& -12& \color{black}{-16} \\ \hline &\color{blue}{1}&\color{blue}{-8}&\color{blue}{6}&\color{blue}{8}&\color{orangered}{-1} \end{array} $$The solution is:
$$ \frac{ x^{4}-6x^{3}-10x^{2}+20x+15 }{ x+2 } = \color{blue}{x^{3}-8x^{2}+6x+8} \color{red}{~-~} \frac{ \color{red}{ 1 } }{ x+2 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&1&-6&-10&20&15\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-2&\color{orangered}{ 1 }&-6&-10&20&15\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&1&-6&-10&20&15\\& & \color{blue}{-2} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -6 } + \color{orangered}{ \left( -2 \right) } = \color{orangered}{ -8 } $
$$ \begin{array}{c|rrrrr}-2&1&\color{orangered}{ -6 }&-10&20&15\\& & \color{orangered}{-2} & & & \\ \hline &1&\color{orangered}{-8}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -8 \right) } = \color{blue}{ 16 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&1&-6&-10&20&15\\& & -2& \color{blue}{16} & & \\ \hline &1&\color{blue}{-8}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -10 } + \color{orangered}{ 16 } = \color{orangered}{ 6 } $
$$ \begin{array}{c|rrrrr}-2&1&-6&\color{orangered}{ -10 }&20&15\\& & -2& \color{orangered}{16} & & \\ \hline &1&-8&\color{orangered}{6}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 6 } = \color{blue}{ -12 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&1&-6&-10&20&15\\& & -2& 16& \color{blue}{-12} & \\ \hline &1&-8&\color{blue}{6}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 20 } + \color{orangered}{ \left( -12 \right) } = \color{orangered}{ 8 } $
$$ \begin{array}{c|rrrrr}-2&1&-6&-10&\color{orangered}{ 20 }&15\\& & -2& 16& \color{orangered}{-12} & \\ \hline &1&-8&6&\color{orangered}{8}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 8 } = \color{blue}{ -16 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&1&-6&-10&20&15\\& & -2& 16& -12& \color{blue}{-16} \\ \hline &1&-8&6&\color{blue}{8}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 15 } + \color{orangered}{ \left( -16 \right) } = \color{orangered}{ -1 } $
$$ \begin{array}{c|rrrrr}-2&1&-6&-10&20&\color{orangered}{ 15 }\\& & -2& 16& -12& \color{orangered}{-16} \\ \hline &\color{blue}{1}&\color{blue}{-8}&\color{blue}{6}&\color{blue}{8}&\color{orangered}{-1} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}-8x^{2}+6x+8 } $ with a remainder of $ \color{red}{ -1 } $.