Divide using synthetic division $$ ( x^{3}+6x^{2}-30x+102 ) \div ( x+10 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-10&1&6&-30&102\\& & -10& 40& \color{black}{-100} \\ \hline &\color{blue}{1}&\color{blue}{-4}&\color{blue}{10}&\color{orangered}{2} \end{array} $$The solution is:
$$ \frac{ x^{3}+6x^{2}-30x+102 }{ x+10 } = \color{blue}{x^{2}-4x+10} ~+~ \frac{ \color{red}{ 2 } }{ x+10 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 10 = 0 $ ( $ x = \color{blue}{ -10 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{-10}&1&6&-30&102\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-10&\color{orangered}{ 1 }&6&-30&102\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -10 } \cdot \color{blue}{ 1 } = \color{blue}{ -10 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-10}&1&6&-30&102\\& & \color{blue}{-10} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 6 } + \color{orangered}{ \left( -10 \right) } = \color{orangered}{ -4 } $
$$ \begin{array}{c|rrrr}-10&1&\color{orangered}{ 6 }&-30&102\\& & \color{orangered}{-10} & & \\ \hline &1&\color{orangered}{-4}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -10 } \cdot \color{blue}{ \left( -4 \right) } = \color{blue}{ 40 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-10}&1&6&-30&102\\& & -10& \color{blue}{40} & \\ \hline &1&\color{blue}{-4}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -30 } + \color{orangered}{ 40 } = \color{orangered}{ 10 } $
$$ \begin{array}{c|rrrr}-10&1&6&\color{orangered}{ -30 }&102\\& & -10& \color{orangered}{40} & \\ \hline &1&-4&\color{orangered}{10}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -10 } \cdot \color{blue}{ 10 } = \color{blue}{ -100 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-10}&1&6&-30&102\\& & -10& 40& \color{blue}{-100} \\ \hline &1&-4&\color{blue}{10}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 102 } + \color{orangered}{ \left( -100 \right) } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrr}-10&1&6&-30&\color{orangered}{ 102 }\\& & -10& 40& \color{orangered}{-100} \\ \hline &\color{blue}{1}&\color{blue}{-4}&\color{blue}{10}&\color{orangered}{2} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{2}-4x+10 } $ with a remainder of $ \color{red}{ 2 } $.