Divide using synthetic division $$ ( x^{3}+x^{2}-10 ) \div ( x+3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-3&1&1&0&-10\\& & -3& 6& \color{black}{-18} \\ \hline &\color{blue}{1}&\color{blue}{-2}&\color{blue}{6}&\color{orangered}{-28} \end{array} $$The solution is:
$$ \frac{ x^{3}+x^{2}-10 }{ x+3 } = \color{blue}{x^{2}-2x+6} \color{red}{~-~} \frac{ \color{red}{ 28 } }{ x+3 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 3 = 0 $ ( $ x = \color{blue}{ -3 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&1&1&0&-10\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-3&\color{orangered}{ 1 }&1&0&-10\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 1 } = \color{blue}{ -3 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&1&1&0&-10\\& & \color{blue}{-3} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ \left( -3 \right) } = \color{orangered}{ -2 } $
$$ \begin{array}{c|rrrr}-3&1&\color{orangered}{ 1 }&0&-10\\& & \color{orangered}{-3} & & \\ \hline &1&\color{orangered}{-2}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -2 \right) } = \color{blue}{ 6 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&1&1&0&-10\\& & -3& \color{blue}{6} & \\ \hline &1&\color{blue}{-2}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 6 } = \color{orangered}{ 6 } $
$$ \begin{array}{c|rrrr}-3&1&1&\color{orangered}{ 0 }&-10\\& & -3& \color{orangered}{6} & \\ \hline &1&-2&\color{orangered}{6}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 6 } = \color{blue}{ -18 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&1&1&0&-10\\& & -3& 6& \color{blue}{-18} \\ \hline &1&-2&\color{blue}{6}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -10 } + \color{orangered}{ \left( -18 \right) } = \color{orangered}{ -28 } $
$$ \begin{array}{c|rrrr}-3&1&1&0&\color{orangered}{ -10 }\\& & -3& 6& \color{orangered}{-18} \\ \hline &\color{blue}{1}&\color{blue}{-2}&\color{blue}{6}&\color{orangered}{-28} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{2}-2x+6 } $ with a remainder of $ \color{red}{ -28 } $.